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PHYSICS
CHAPTER 8
CHAPTER 8: Rotational of rigid body (8 Hours)
1
PHYSICS CHAPTER 8 Learning Outcome: http//:kms.matrik.edu.my/
8.1 Rotational Kinematics (2 hour) At the end of this chapter, students should be able to: a) Define and describe: angular displacement (θ) average angular velocity (ωav)
instantaneous angular velocity (ω) average angular acceleration (αav)
instantaneous angular acceleration (α). b) Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use :
2
v s=rθ ; v=rω ; a t = rα ; a c =rω 2 = r
c) Use equations for rotational motion with constant angular acceleration.
ω=ω0 αt
1 2 θ=ω 0 t αt 2
2
ω =ω22 2 αθ 0
PHYSICS
CHAPTER 8
8.1 Parameters in rotational motion 8.1.1 Angular displacement,θ
is defined as an angle through which a point or line has been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad). (rad) Figure 8.1 shows a point P on a rotating compact disc (CD) moves through an arc length s on a circular path of radius r about a fixed axis through point O.
Figure 8.1 3
PHYSICS
CHAPTER 8
From Figure 8.1, thus
s θ= r where
OR
s=rθ
θ : angle angular displacement in radian s: arc length r : radius of the circle
Others unit for angular displacement is degree (°) and revolution (rev). (rev) Conversion factor :
1 rev=2π rad=360
°
Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise. anticlockwise Negative – if the rotational motion is clockwise. clockwise 4
PHYSICS
CHAPTER 8
8.1.2 Angular velocity Average angular velocity, ωav
is defined as the rate of change of angular displacement. displacement Equation :
θ 2−θ1 Δθ ω av = = t 2−t 1 Δt
where
θ 2 : final angular displacement in radian θ1 : initial angular displacement in radian
Δt : time interval Instantaneous angular velocity, ω
is defined as the instantaneous rate of change of angular displacement. displacement Equation : Δθ dθ
ω=limit Δt 0
Δt
=
dt 5
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The unit of angular velocity is radian per second (rad s-1) Others unit is revolution per minute (rev min−1 or rpm) Conversion factor:
2π −1 π −1 1 rpm = rad s = rad s 60 30 Note : Every part of a rotating rigid body has the same angular velocity. velocity Direction of the angular velocity Its direction can be determine by using right hand grip rule where
Thumb Curl fingers
: direction of angular velocity : direction of rotation 6
PHYSICS
CHAPTER 8
Figures 8.2 and 8.3 show the right hand grip rule for determining the direction of the angular velocity.
ω
Figure 8.2
ω Figure 8.3 7
PHYSICS
CHAPTER 8
Example 8.1 : The angular displacement,θ of the wheel is given by 2
θ=5t −t
where θ in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine a. the angle, θ in degree, at time 2.2 s and 4.8 s, b. the distance that a particle on the rim moves during that time interval, c. the average angular velocity, in rad s−1 and in rev min−1 (rpm), between 2.2 s and 4.8 s, d. the instantaneous angular velocity at time 3.0 s.
8
PHYSICS Solution :
CHAPTER 8 d 0 .56 r= = =0 . 28 m 2 2
a. At time, t1 =2.2 s :
2
θ1 =5 2. 2 − 2. 2
θ1 =22 rad
At time, t2 =4.8 s :
2
θ 2 =5 4 . 8 − 4 . 8
θ 2 =110 rad
9
PHYSICS Solution :
CHAPTER 8 d 0 .56 r= = =0 . 28 m 2 2
b. By applying the equation of arc length,
s=rθ
Therefore
Δs=rΔθ=r θ 2 −θ1
Δs=0.28 110−22 c. The average angular velocity in rad s−1 is given by
Δθ θ 2 −θ 1 ω av = = Δt t 2 −t 1
ω av =
110−22
4 . 8−2. 2 10
PHYSICS
CHAPTER 8
Solution : c. and the average angular velocity in rev min−1 is
33 . 9 rad 1 rev 60 s ω av = 1s 2π rad 1 min
d. The instantaneous angular velocity as a function of time is
dθ ω= dt d 2 ω= 5t −t dt ω=10 t −1
At time, t =3.0 s :
ω=10 3.0 −1 11
PHYSICS
CHAPTER 8
Example 8.2 : A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the diver’s average angular (rotational) velocity during a dive. (Given g = 9.81 m s−2) Solution :
u y =0 θ 0 =0
10 m
θ1 =2. 5 rev
water
12
PHYSICS
CHAPTER 8
Solution : θ1 =2. 5×2π=5π rad From the diagram, s =−10 m y Thus 1 2
s y=u y t− gt 2 1 2 −10=0− 9 . 81 t 2
Therefore the diver’s average angular velocity is
ω av =
θ1 −θ 0 t
5π−0 ω av = 1 . 43 13
PHYSICS
CHAPTER 8
8.1.3 Angular acceleration Average angular acceleration, αav
is defined as the rate of change of angular velocity. velocity Equation : ω −ω
Δω α av = = t 2 −t 1 Δt where ω 2 : final angular velocity ω 1 : initial angular velocity 2
1
Δt : time interval Instantaneous angular acceleration, α
is defined as the instantaneous rate of change of angular velocity. velocity Equation : Δω dω
α=limit Δt 0
Δt
=
dt 14
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The unit of angular acceleration is rad s−2. Note:
If the angular acceleration, α is positive, positive then the angular velocity, ω is increasing. increasing
If the angular acceleration, α is negative, negative then the angular velocity, ω is decreasing. decreasing Direction of the angular acceleration
If the rotation is speeding up, up α and ω in the same direction as shown in Figure 8.4.
ω Figure 8.4
α
15
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If the rotation is slowing down, down α and ω have the opposite direction as shown in Figure 8.5.
α
ω Figure 8.5
Example 8.3 : The instantaneous angular velocity, ω of the flywheel is given by 3 2
ω=8t −t
where ω in radian per second and t in seconds. Determine a. the average angular acceleration between 2.2 s and 4.8 s, b. the instantaneous angular acceleration at time, 3.0 s. 16
PHYSICS
CHAPTER 8
Solution : a. At time, t1 =2.2 s :
3
2
ω 1 =8 2 . 2 − 2 . 2 −1
ω 1 =80 . 3 rad s
At time, t2 =4.8 s :
3
2
ω 2 =8 4 . 8 − 4 . 8
Therefore the average angular acceleration is
α av =
ω 2 −ω 1 t 2 −t 1
862−80 .3 α av = 4 . 8−2. 2 17
PHYSICS
CHAPTER 8
Solution : b. The instantaneous angular acceleration as a function of time is
dω α= dt d 3 2 α= 8t −t dt
At time, t =3.0 s :
2
α=24 3 . 0 −2 3 . 0
18
PHYSICS
CHAPTER 8
Exercise 8.1 : 1. If a disc 30 cm in diameter rolls 65 m along a straight line without slipping, calculate a. the number of revolutions would it makes in the process, b. the angular displacement would be through by a speck of gum on its rim. ANS. : 69 rev; 138π rad 2. During a certain period of time, the angular displacement of a swinging door is described by
θ=5.0010 .0t2.00 t
2
where θ is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration a. at time, t =0, b. at time, t =3.00 s. ANS. : 5.00 rad, 10.0 rad s−1, 4.00 rad s−2; 53.0 rad, 22.0 rad s−1, 4.00 rad s−2 19
PHYSICS
CHAPTER 8
8.1.2 Relationship between linear and rotational motion 8.1.2 Relationship between linear velocity, v and angular velocity, ω
When a rigid body is rotates about rotation axis O , every particle in the body moves in a circle as shown in the Figure 8.6.
y
v P r O Figure 8.6
θ
s x
20
PHYSICS
CHAPTER 8
Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by
ds v= dt
and
s=rθ
dθ v=r dt v=rω
The direction of the linear (tangential) velocity always tangent to the circular path. path Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle, r is changing depend on the position of the particle. particle Simulation 7.1 21
PHYSICS
CHAPTER 8
8.1.2 Relationship between tangential acceleration, at and angular acceleration, α
If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 8.7.
y
at a O
P
ac x
Figure 8.7 22
PHYSICS
CHAPTER 8
The components are tangential acceleration, at and centripetal acceleration, ac given by
but
dv and v=rω at= dt dω a t =rα a t =r dt 2 v 2 a c = =rω =vω r
The vector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, a given by Vector form t c
a =a a
and its magnitude,
∣a∣= a 2t a 2c
23
PHYSICS
CHAPTER 8
8.1.3 Rotational motion with uniform angular acceleration
Table 8.1 shows the symbols used in linear and rotational kinematics. Linear motion
Quantity
Rotational motion
s
Displacement
u
Initial velocity
θ ω0
v
Final velocity
ω
a
Acceleration
α
t
Time
t
Table 8.1 24
PHYSICS
CHAPTER 8
Table 8.2 shows the comparison of linear and rotational motion with constant acceleration. Linear motion
Rotational motion
a=constant
α =constant
v=uat
ω=ω0 αt
1 2 s=ut at 2
1 2 θ=ω 0 t αt 2
2
2
2
v =u 2 as 1 s= vu t 2 where θ in radian.
ω
2 =ω 0 2 αθ
1 θ= ωω 0 t 2 Table 8.2 25
PHYSICS
CHAPTER 8
Example 8.4 : A car is travelling with a velocity of 17.0 m s−1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s−2 for 5.00 s, calculate a. the number of revolutions of the wheels during this period, b. the angular speed of the wheels after 5.00 s. −1 −2 Solution : u=17.0 m s ,r=0.48 m ,a=2.00 m s ,t=5.00 s a. The initial angular velocity is
u=rω0 17.0=0.48 ω0
and the angular acceleration of the wheels is given by
a=rα
2 . 00=0 . 48 α 26
PHYSICS
CHAPTER 8 −1 −2 Solution : u=17.0 m s ,r=0.48 m ,a=2.00 m s ,t=5.00 s a. By applying the equation of rotational motion with constant angular acceleration, thus 1 2
θ=ω 0 t αt 2 1 2 θ= 35 . 4 5 . 00 4 . 17 5 . 00 2 θ=229 rad
therefore
b. The angular speed of the wheels after 5.00 s is
ω=ω0 αt
ω=35. 4 4.17 5.00 27
PHYSICS
CHAPTER 8
Example 8.5 : The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. The wheels have a diameter of 70 cm. a. Calculate the angular acceleration. b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. 0 .70 Solution : θ=30×2π=60π rad, r= =0 . 35 m ,
2
3
50. 0 km 10 m 1 h u= =13. 9 m s−1 , 1h 1 km 3600 s 3
35.0 km 10 m 1 h −1 v= =9. 72 m s 1h 1 km 3600 s 28
PHYSICS
CHAPTER 8
Solution : a. The initial angular speed of the wheels is
u=rω0 13.9=0.35 ω 0
and the final angular speed of the wheels is
v=rω
9 . 72=0 . 35 ω therefore 2 2 ω =ω 0 2 αθ 2 2 27 . 8 =39 . 7 2α 60π b. The car stops thus ω=0 and ω=ω0 αt Hence
ω 0 =27 . 8 rad s
0=27 .8 −2.13 t 29
−1
PHYSICS
CHAPTER 8
Example 8.6 : A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s -1. The angular acceleration of the blade is 0.750 rev s-2. Determine a. the angular velocity after 4.00 s, b. the number of revolutions for the blade turns in this time interval, c. the tangential speed of a point on the tip of the blade at time,
t =4.00 s, d. the magnitude of the resultant acceleration of a point on the tip −1 of the blade at t =4.00 s. r=0 . 400 m , ω 0 =0. 150×2π=0 .300π rad s , Solution : α=0.750×2π =1. 50π rad s−2
a. Given tω=ω =4.00 0s,αt thus
ω= 0.300π −11.50π 4 .00 ω=19.8 rad s 30
PHYSICS
CHAPTER 8
Solution : b. The number of revolutions of the blade is
1 2 θ=ω 0 t αt 2 1 2 θ= 0 . 300 π 4 . 00 1 .50 π 4 . 00 2 θ=41. 5 rad
c. The tangential speed of a point is given by
v=rω
v= 0.400 19.8 31
PHYSICS
CHAPTER 8
Solution : d. The magnitude of the resultant acceleration is
c
a= a 2 a 2 t
a=
2 2
v r
2
rα 2 2
a=
7 . 92
0 . 400
2
0 . 400×1. 50π
32
PHYSICS
CHAPTER 8
Example 8.7 : Calculate the angular velocity of a. the second-hand, b. the minute-hand and c. the hour-hand, of a clock. State in rad s-1. d. What is the angular acceleration in each case? Solution : a. The period of second-hand of the clock is
2π ω= T
T = 60 s, hence
2π ω= 60
33
PHYSICS
CHAPTER 8
Solution : b. The period of minute-hand of the clock is T = 60 min = 3600 s, hence
2π ω= 3600
c. The period of hour-hand of the clock is hence
2π ω= 4 . 32×10 4
T = 12 h = 4.32 ×104 s,
d. The angular acceleration in each cases is 34
PHYSICS
CHAPTER 8
Example 8.8 : A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s−1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s−2, calculate the distance travelled by the coin before coming to rest. −1 Solution : ω =18 rad s−1
ω=0 rad s
0
−2
α=−1.90 rad s
−2
d=2.40×10 m The radius of the coin is
s
d −2 r= =1 .20×10 m 2 35
PHYSICS
CHAPTER 8
Solution : The initial speed of the point at the edge the coin is
u=rω0 −2 u= 1.20×10 18
v=0 m s
−1
and the final speed is The linear acceleration of the point at the edge the coin is given by
a=rα
−2 a= 1.20×10 −1.90
Therefore the distance by the coin is 2 travelled 2
v =u 2 as 2 0= 0 .216 2 −2 . 28×10−2 s 36
PHYSICS
CHAPTER 8
Exercise 8.2 : 1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev min-1 about its central axis. Determine a. its angular speed, b. the tangential speed at a point 3.00 cm from its centre, c. the radial acceleration of a point on the rim, d. the total distance a point on the rim moves in 2.00 s. ANS. : 126 rad s−1; 3.77 m s−1; 1.26 × 103 m s−2; 20.1 m 2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculate a. its angular velocity in rad s−1, b. the linear speed and the radial acceleration of a point on the edge of the grinding wheel. ANS. : 262 rad s−1; 46 m s−1, 1.2 × 104 m s−2 37
PHYSICS
CHAPTER 8
Exercise 8.2 : 3. A rotating wheel required 3.00 s to rotate through 37.0 revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel. ANS. : 13.6 rad s−2 4. A wheel rotates with a constant angular acceleration of 3.50 rad s−2. a. If the angular speed of the wheel is 2.00 rad s−1 at t =0, through what angular displacement does the wheel rotate in 2.00 s. b. Through how many revolutions has the wheel turned during this time interval? c. What is the angular speed of the wheel at t = 2.00 s? ANS. : 11.0 rad; 1.75 rev; 9.00 rad s−1 38
PHYSICS
CHAPTER 8
Exercise 8.2 : 5. A bicycle wheel is being tested at a repair shop. The angular velocity of the wheel is 4.00 rad s-1 at time t = 0 , and its angular acceleration is constant and equal −1.20 rad s-2. A spoke OP on the wheel coincides with the +x-axis at time t = 0 as shown in y Figure 8.8.
O
P
x
Figure 8.8
a. What is the wheel’s angular velocity at t = 3.00 s? b. What angle in degree does the spoke OP make with the positive x-axis at this time? ANS. : 0.40 rad s−1; 18° 39
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PHYSICS CHAPTER 8 Learning Outcome: 8.2 Equilibrium of a uniform rigid body (2 hour) At the end of this chapter, students should be able to: Define and use torque.
State and use conditions for equilibrium of rigid body:
∑
Fx = 0 ,
∑
Fy = 0 ,
∑
τ= 0
40
PHYSICS 8.2.1
Torque (moment of a force),τ
CHAPTER 8
The magnitude of the torque is defined as the product of a force and its perpendicular distance from the line of action of the force to the point (rotation axis). axis) OR
τ = Fd
where
Because of
τ : magnitude of the torque F : magnitude of the force d : perpendicular distance (moment arm)
d = r sin θ where r : distance between the pivot point (rotation axis) and the point of application of force.
Thus
τ = Fr sin θ OR τ = r× F where θ : angle between F and r 41
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The dimension of torque is
[τ ] = [ F ][ d ] =
ML2T − 2
The unit of torque is N m (newton metre), a vector product unlike the joule (unit of work), work) also equal to a newton metre, which is scalar product. product Torque is occurred because of turning (twisting) effects of the forces on a body. Sign convention of torque: Positive - turning tendency of the force is anticlockwise. anticlockwise Negative - turning tendency of the force is clockwise. clockwise The value of torque depends on the rotation axis and the magnitude of applied force. force
42
PHYSICS
CHAPTER 8
Case 1 : Consider a force is applied to a metre rule which is pivoted at one end as shown in Figures 8.9a and 8.9b.
F τ = Fd
(anticlockwise)
d
Figure 8.9a
Line of action of a force Pivot point (rotation axis)
d = r sin θ r
Figure 8.9b
θ
F
Point of action of a force
τ = Fd = Fr sin θ
(anticlockwise) 43
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CHAPTER 8
Case 2 : Consider three forces are applied to the metre rule which is pivoted at one end (point O) as shown in Figures 8.10.
F3 d 1 = r1 sin θ1 r2 O
d 2 = r2 sin θ2
θ2
θ1
F2
r1 Figure 8.10
F1
τ 1 = F1 d 1 = F1 r1 sin θ1 τ 2 = − F2 d 2 = − F2 r2 sin θ 2 τ 3 = F3 d 3 = F3 r3 sin θ3 = 0 Therefore the resultant (nett) torque is
∑ ∑
τ O = τ1 + τ 2 + τ 3 τ O = F1 d 1 − F2 d 2
Caution : If the line of action of a force is through the rotation axis τ = Fr sin θ and θ = 0 then τ= 0 Simulation 5.1 44
PHYSICS
CHAPTER 8
Example 8.9 : Determine a resultant torque of all the forces about rotation axis, O in the following problems. a. F = 10 N 2
F1 = 30 N
5m
5m 3m
6m
O 3m
10 m
F3 = 20 N 45
PHYSICS
CHAPTER 8
Example 8.9 : b.
F1 = 30 N
10 m 3m
β F3 = 20 N
6m
O
3m 5m
5m
F2 = 10 N
α F4 = 25 N
46
PHYSICS
F2 = 10 N
Solution : a. 5m
CHAPTER 8 F1 = 30 N
5m
d1 = 3 m O
6m
d2 = 5 m 10 m Force
F3 = 20 N The resultant torque:
F1 F2 F3
Torque (N m), τo=Fd=Frsinθ
− ( 30 )( 3) = − 90
+ (10)( 5) = + 50
0 47
PHYSICS
CHAPTER 8
Solution : b.
3m
β F3 = 20 N
F1 = 30 N
10 m
d3 d = 3 m 1
β r = 5m 5m
O
6m
sin β = 5m
F2 = 10 N Force
F1 F2 F3 F4
α
3 32 + 5 2
F4 = 25 N
Torque (N m), τo=Fd=Frsinθ
− ( 30 )( 3) = − 90 0
The resultant torque:
F3 r sin β = ( 20 )( 5)( 0.515) = 51.5
0
48
= 0.515
PHYSICS
CHAPTER 8
8.2 Equilibrium of a rigid body 8.2.1.1 Non-concurrent forces
is defined as the forces whose lines of action do not pass through a single common point. The forces cause the rotational motion on the body. The combination of concurrent and non-concurrent forces cause rolling motion on the body. (translational and rotational motion) Figure 8.11 shows an example of non-concurrent forces.
F1
F2
F4 Figure 8.11
49
F3
PHYSICS
CHAPTER 8
8.2.1.2 Equilibrium of a rigid body
Rigid body is defined as a body with definite shape that doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it. it If the rigid body is in equilibrium, equilibrium means the body is translational and rotational equilibrium. equilibrium There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must be zero.
∑
F = Fnett = 0 OR
∑
Fx = 0 ,
∑
Fy = 0 ,
∑
Fz = 0 50
PHYSICS
CHAPTER 8
The vector sum of all external torques acting on a rigid body must be zero about any rotation axis. axis
∑
τ = τ nett = 0
This ensures rotational equilibrium. equilibrium This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
∑
τx = 0 ,
∑
τy = 0,
∑
τz = 0
Centre of gravity, CG is defined as the point at which the whole weight of a body may be considered to act. act A force that exerts on the centre of gravity of an object will cause a translational motion. motion 51
PHYSICS
CHAPTER 8
Figures 8.14 and 8.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere rod – refer to the midway point between its end. end
l l 2
CG
Figure 8.12
l 2
sphere – refer to geometric centre. centre
CG Figure 8.13 52
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CHAPTER 8
8.2.4 Problem solving strategies for equilibrium of a rigid body
The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components and to determine the torque by each force. Apply the condition for equilibrium of a rigid body :
∑
Fx = 0 ;
∑
Fy = 0
and
∑
τ= 0
Solve the equations for the unknowns. 53
PHYSICS
CHAPTER 8
Example 8.10 : A 35 cm O
B
75 cm
W1
W2 Figure 8.14
A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in Figure 8.14. If the mass of the beam is 3.0 kg, calculate a. the weight, W1 needed, b. the force exerted on the beam at point O. (Given g =9.81 m s−2)
54
PHYSICS
CHAPTER 8 N
Solution : m = 3 kg; W2 = 23 N The free body diagram of the beam :
A W2
0.20 m
0.35 m
O
Let point O as the rotation axis. Force
W1 W2 mg N
y-comp. (N)
− W1 − 23
− ( 3)( 9.81) = − 29.4 N
CG
mg
0.55 m
0.55 m
Torque (N m), τo=Fd=Frsinθ
− W1 ( 0.75) = − 0.75W1
+ ( 23)( 0.35) = 8.05
− ( 29.4 )( 0.20 ) = − 5.88 0
B
0.75 m
55
W1
PHYSICS
CHAPTER 8
Solution : Since the beam remains at rest thus the system in equilibrium. a. Hence τ = 0
∑
O
− 0.75W1 + 8.05 − 5.88 = 0 b. and
∑
Fy = 0
− W1 − 23 − 29.4 + N = 0
− ( 2.89 ) − 23 − 29.4 + N = 0
56
PHYSICS
CHAPTER 8
Example 8.11 : A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 8.15. The height of the A end A of the ladder is 8.0 m from the rough floor. a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B. b. If the ladder is just on the verge of smooth slipping when the firefighter is 7.0 m wall B up the ladder , Calculate the coefficient rough floor of static friction between ladder and Figure 8.15 floor. (Given g =9.81 m s−2)
57
PHYSICS
Force
ml g mf g N1 N2 fs
CHAPTER 8
Solution : ml = 5.0 kg; m f = 60 kg a. The free body diagram of the ladder : Let point B as the rotation axis. A Torque (N m), x-comp. y-comp. (N) (N) τB=Fd=Frsinθ
0
− 49.1
0
− 589
N1
0
0
N2
− fs
0
( 49.1)( 5.0) sin β
α
N1
β
8 sin α = = 0.8 10 6 sin β = = 0.6 10
= 147 ( 589)( 3.0) sin β 8.0 m CG 10 m = 1060 ml g β − N 1 (10) sin α = − 8N1 mf g β α 5.0 m 0
0
6.0 m 58
fs
3.0 m
N2 B
PHYSICS
CHAPTER 8
Solution : Since the ladder in equilibrium thus
∑
τB = 0
147 + 1060 − 8 N 1 = 0
∑
N 1 = 151 N
Fx = 0
N1 − f s = 0 Horizontal force:
∑
Fy = 0
− 49.1 − 589 + N 2 = 0 Vertical force:
59
PHYSICS
Force
ml g mf g N1 N2 fs
CHAPTER 8
Solution : sin α = 0.8; sin β = 0.6 b. The free body diagram of the ladder : Let point B as the rotation axis. A Torque (N m), x-comp. y-comp. (N) (N) τB=Fd=Frsinθ β
0 0 N1
− 49.1 − 589
0
( 49.1)( 5.0) sin β
= 147 ( 589)( 7.0) sin β = 2474 − N 1 (10) sin α = − 8N1
0
N2
0
− μs N 2
0
0
α
mf g 8.0 m
N1
β
10 m
ml g β
5.0 m 6.0 m 60
7.0 m
N2 α fs
B
PHYSICS
CHAPTER 8
Solution : Consider the ladder stills in equilibrium thus
∑
τB = 0
∑
N 1 = 328 N Fy = 0
∑
N 2 = 638 N Fx = 0
147 + 2474 − 8 N 1 = 0
− 49.1 − 589 + N 2 = 0
N 1 − μs N 2 = 0 ( 328) − μs ( 638) = 0
61
PHYSICS
CHAPTER 8
Example 8.12 : A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 8.16. A cable at an angle 30° with the beam helps to support the light. a. Sketch a free body diagram of the beam. b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole.
Figure 8.16
(Given g =9.81 m s ) −2
62
PHYSICS
CHAPTER 8
Solution : m f = 20.0 kg; mb = 10.0 kg a. The free body diagram of the beam :
T
S
O
30 mb g
CG
0.5l
l
mf g
b. Let point O as the rotation axis. Force x-comp. (N) y-comp. (N)
mf g 0 mb g 0 T − T cos 30 Sx S
− (196 ) l
− 196
− ( 98.1)( 0.5l ) = − 49.1l
− 98.1 T sin 30
Sy
Torque (N m), τo=Fd=Frsinθ
Tl sin 30 = 0.5Tl 0 63
PHYSICS
CHAPTER 8
Solution : b. The floodlight and beam remain at rest thus i. τ = 0
∑
O
− 196l − 49.1l + 0.5Tl = 0 ii.
∑
Fx = 0
− T cos 30 + S x = 0
∑
S x = 424 N
Fy = 0
− 196 − 98.1 + T sin 30 + S y = 0
S y = 49.1 N 64
PHYSICS
CHAPTER 8
Solution : b. ii. Therefore the magnitude of the force is 2
2
S=
Sx + S y
S=
( 424) 2 + ( 49.1) 2
and its direction is given by − 1
Sy θ = tan Sx − 1 49.1 θ = tan 424 65
PHYSICS
CHAPTER 8
Exercise 8.3 : Use gravitational acceleration, g = 9.81 m s−2 F1 1. a B A
F2
b D
C Figure 8.17
γ
F3
Figure 8.17 shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle γ is 30°. Calculate the resultant torque about point D. ANS. : -3.7 N m 66
PHYSICS
CHAPTER 8
Exercise 8.3 : 2.
Figure 8.18
A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 8.18. The fulcrum is under the centre of gravity of the board. Determine a. the magnitude of the force exerted by the fulcrum on the board, b. where the father should sit from the fulcrum to balance the system. 67 ANS. : 1128 N; 1.31 m
PHYSICS
CHAPTER 8
Exercise 8.3 : 3.
Figure 8.19
A traffic light hangs from a structure as show in Figure 8.19. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. 68 ANS. : 248 N; 197 N, 248 N
PHYSICS
CHAPTER 8
Exercise 8.3 : 4.
30.0 cm
50.0
15.0 cm
F Figure 8.20
A uniform 10.0 N picture frame is supported by two light string as shown in Figure 8.20. The horizontal force, F is applied for holding the frame in the position shown. a. Sketch the free body diagram of the picture frame. b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F . ANS. : 1.42 N, 11.2 N; 7.20 N 69
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PHYSICS CHAPTER 8 Learning Outcome: 8.3 Rotational dynamics (1 hour) At the end of this chapter, students should be able to: Define the moment of inertia of a rigid body about an axis,
I=
n
∑
mi ri
2
i= 1
State and use torque,
τ = Iα 70
PHYSICS
CHAPTER 8
8.3.1 Centre of mass, moment of inertia and torque 8.3.1.1 Centre of mass (CM)
is defined as the point at which the whole mass of a body may be considered to be concentrated. concentrated
Its coordinate (xCM, yCM) is given the expression below:
n
n
∑ mi y i
∑ mi x i
x CM =
i =1 n
; y CM =
∑ mi
∑ mi
i=1
i=1
where
i=1 n
th
mi : mass of the i particle th x i : x coordinate of the i th particle y i : y coordinate of the i particle 71
PHYSICS
CHAPTER 8
Example 8.13 :
5m
Two masses, 3 kg and 5 kg are located on the y-axis at y =1 m and y =5 m respectively. Determine the centre of mass of this system. Solution : m 1 =3 kg; m 2 =5 kg
m2
3.5 m CM
2
∑ mi y i
y CM = i=12
=
m 1 y 1 m 2 y 2
∑ mi
m1 m 2
i=1
3 1 5 5 y CM = 35 1m
y= 0
m1 72
PHYSICS
CHAPTER 8
Example 8.14 : A system consists of three spheres have the following masses and coordinates : (1) 1 kg, (3,2) ; (2) 2 kg, (4,5) and (3) 3 kg, (3,0). Determine the coordinate of the centre of mass of the system. Solution : m1 =1 kg; m 2 =2 kg; m 3 =3 kg The x coordinate of the CM is 3
∑ mi x i
i=1 x CM= 3
∑ mi
=
m1 x 1 m 2 x 2m3 x 3 m1 m 2m3
i=1
1 3 2 4 3 3 x CM= 123 73
PHYSICS
CHAPTER 8
Solution : The y coordinate of the CM is 3
∑ mi y i
y CM = i=13
∑ yi
=
m 1 y 1 m 2 y 2 m3 y 3 m 1 m2 m3
i=1
1 2 2 5 3 0 y CM = 123 Therefore the coordinate of the CM is
74
PHYSICS
CHAPTER 8
8.3.1.2 Moment of inertia, I
Figure 8.21 shows a rigid body about a fixed axis O with angular velocity ω.
m1 mn
rn
r1 Or
3
ω
r 2 m2 m3
Figure 8.21
is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis. axis 75
PHYSICS
CHAPTER 8 n
OR
I=m1 r 21 m 2 r 22 m3 r 23 . .. mn r 2n =∑ mi r 2i i=1
where
I : moment of inertia of a rigid body about rotation axis m : mass of particle r : distance from the particle to the rotation axis
It is a scalar quantity. quantity
Moment of inertia, I in the rotational kinematics is analogous to the mass, m in linear kinematics. The dimension of the moment of inertia is M L2. The S.I. unit of moment of inertia is kg m2.
The factors which affect the moment of inertia, I of a rigid body: a. the mass of the body, b. the shape of the body, c. the position of the rotation axis. axis 76
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape
Diagram
Hoop or ring or thin cylindrical shell
CM
I CM=MR
CM
1 2 I CM= MR 2
Solid cylinder or disk
Equation
77
2
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Uniform rod or long thin rod with rotation axis through the centre of mass.
Solid Sphere
Diagram
Equation
CM
1 2 I CM = ML 12
CM
2 2 I CM= MR 5 78
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape
Hollow Sphere or thin spherical shell
Diagram
Equation
CM
2 2 I CM= MR 3
Table 8.3
79
PHYSICS
CHAPTER 8
Example 8.15 : Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 8.22.
2 kg
3 kg 60 cm
A
O
5 kg
250 cm
B
60 cm
4 kg
Figure 8.22
The spheres are connected by light rods . Determine the moment of inertia of the system about an axis a. through point O, b. along the line AB. 80
PHYSICS
CHAPTER 8
Solution : m1 =2 kg; m 2 =3 kg; a. rotation axis about point O,
m3 =4 kg; m 4 =5 kg m2
m1
r2
r1
r4
O
1.25 m
r3 m3
m4 Since r1= r2=
r3= r4= r thus
0.6 m
2
2
r= 0.6 1.25 =1.39 m
and the connecting rods 2 2 are light 2 therefore 2 I O=m1 r 1 m2 r 2 m3 r 3 m4 r 4
2
2
I O=r m 1 m2 m3 m 4 = 1 .39 2345 81
PHYSICS
CHAPTER 8
Solution : m1 =2 kg; m 2 =3 kg; b. rotation axis along the line AB,
m3 =4 kg; m 4 =5 kg m2
m1
r2
r1 A
r4
r3 m3
m4 r1= r2= r3= r4= r=0.6 m therefore
2 2 2 2 I AB =m1 r 1 m2 r 2m3 r 3 m4 r 4 2 I AB =r m1 m 2 m 3 m 4
2
I AB = 0 . 6 234 5 82
B
PHYSICS
CHAPTER 8
8.3.2 Torque,τ Relationship between torque,τ and angular acceleration, α
Consider a force, F acts on a rigid body freely pivoted on an axis through point O as shown in Figure 8.23.
mn
r1
an rn
Or
2
a 1 m1
m2
F
a2
Figure 8.23
The body rotates in the anticlockwise direction and a nett torque is produced. 83
PHYSICS
CHAPTER 8
A particle of mass, m1 of distance r1 from the rotation axis O will experience a nett force F1 . The nett force on this particle is F 1=m1 a 1 and a 1 =r 1 α
F 1=m1 r 1 α
° The torque on the mass m1 is τ 1=r 1 F 1 sin 90
2 τ 1=m1 r 1 α
The total (nett) torque on 2 the rigid2body is given by2
∑ τ=m1nr1 αm2 r2 α .. .m nrnα n
∑ τ=
2 m r ∑ ii α i =1
∑ τ=Iα
and
2 m r ∑ i i =I i=1
84
PHYSICS
CHAPTER 8
From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration. acceleration
Note :
Nett torque , ∑ τ =Iα is analogous to the
Nett force, ∑ F =ma
85
PHYSICS
CHAPTER 8
Example 8.16 : Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 8.24.
F 2
O
F 1
30.0 cm Figure 8.24
Calculate, a. the nett torque on the disc. b. the magnitude of angular acceleration influence by the disc. 1 2 ( Use the moment of inertia, I CM = MR )
2
86
PHYSICS
CHAPTER 8
Solution : R=0. 30 m ; M=5. 00 a. The nett torque on the disc is
kg
∑ τ=τ 1τ 2
∑ τ=−RF 1RF 2=R −F 1F 2
∑ τ= 0.30 −5.6010.3
b. By applying the relationship between torque and angular acceleration, 1 2
∑ τ=Iα
∑ τ=
MR α
2 1 2 1. 41= 5 . 00 0 . 30 α 2 87
PHYSICS
CHAPTER 8
Example 8.17 : A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is 0.050 kg m2. A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure 8.25. Assume the string does not slip on the wheel.
Figure 8.25
a. Sketch a free body diagram of the wheel and the block. b. Calculate the magnitude of the angular acceleration of the wheel. 88
PHYSICS
CHAPTER 8
Solution : R=0.20 m ; I=0.050 kg m 2 ; a. Free body diagram : for wheel,
T
S
W for block,
P=3.0 N; m=2.0 kg
N
T
a
P
W b 89
PHYSICS
CHAPTER 8
Solution : R=0.20 m ; I=0.050 kg m 2 ; b. For wheel,
∑ τ=Iα
For block,
RT =Iα
∑ F=ma
P=3.0 N; m=2.0 kg
Iα T= R P−T =ma
(1) (2)
By substituting eq. (1) into eq. (2), thus
Iα P− =ma and a=Rα R Iα P− =mRα R 90
PHYSICS
CHAPTER 8
Example 8.18 : An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 8.26. The pulley has a moment of inertia 0.020 kg m2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determine a. the linear acceleration of the object, b. the angular acceleration of the pulley, c. the tension in the string, d. the liner velocity of the object, e. the distance travelled by the object. (Given g = 9.81 m s-2)
R
1.50 kg Figure 8.26
91
PHYSICS
CHAPTER 8
Solution : a. Free body diagram : for pulley,
S
W for block,
T
a
∑ τ=Iα
a RT =Iα and α= R a RT =I R T Ia T= 2 (1) R
∑ F=ma
mg−T =ma
(2)
m g 92
PHYSICS
CHAPTER 8
Solution :
2
R=0.20 m ; I=0.020 kg m ; m=1.50 kg;
u=0; t =0 . 3 s
a. By substituting eq. (1) into eq. (2), thus
Ia mg− 2 =ma R 0 .020 a 1 .50 9 .81 − =1 .50 a 2 0 . 20
b. By using the relationship between a and α, hence
a=Rα 7 . 36=0 . 20 α
93
PHYSICS Solution :
CHAPTER 8 2
R=0.20 m ; I=0.020 kg m ; m=1.50 kg;
u=0; t =0 . 3 s c. From eq. (1), thus
Ia T= 2 R
0 . 020 7 . 36 T= 2 0 .20
d. By applying the equation of liner motion, thus
v=uat
v=0 7.36 0.3 e. The distance travelled by the object in 0.3 s is
1 2 s=ut at 2 1 2 s=0 7 . 36 0 .3 2
94
PHYSICS
CHAPTER 8
Exercise 8.4 : Use gravitational acceleration, g = 9.81 m s−2 1. Three odd-shaped blocks of chocolate have following masses and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m); 0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m). Determine the coordinates of the centre of mass of the system of three chocolate blocks. ANS. : (0.044 m, 0.056 m) 2. Figure 8.27 shows four masses that are held at 70 g 40 cm the corners of a square by a very light 80 cm frame. Calculate the moment of inertia B of the system about an axis perpendicular A to the plane 150 g 150 g a. through point A, and 80 cm b. through point B. Figure 8.27 70 g ANS. : 0.141 kg m2; 0.211 kg m2 95
PHYSICS
CHAPTER 8
Exercise 8.4 : 3. A 5.00 kg object placed on a frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 8.28. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s−2. a. Sketch free body diagrams of both objects and pulley. b. Calculate T1 and T2 the tensions
2.00 m s−2 T1 T2
Figure 8.28
in the string. c. Determine I. ANS. : 10.0 N, 70.3 N; 1.88 kg m2
96
PHYSICS CHAPTER 8 Learning Outcome: http//:kms.matrik.edu.my/
8.4 Work and Energy of Rotational Motion (2 hours) At the end of this chapter, students should be able to: Solve problem related to : kinetic energy,
1 2 K r = Iω 2
work,
power,
W =τθ P=τω
97
PHYSICS
CHAPTER 8
8.4 Rotational kinetic energy and power 8.4.1 Rotational kinetic energy, Kr
Consider a rigid body rotating about the axis OZ as shown in Figure 8.29. Z
vn
mn
rn Or
r1
v 1 m1 r2
v2
m2 v3 3 m3
Figure 8.29
Every particle in the body is in the circular motion about point O. 98
PHYSICS
CHAPTER 8
The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by 1 1 2 1 2 1 2 2 K r = m1 v 1 m2 v 2 m3 v 3 .. . mn v n
2 2 2 2 1 1 2 2 1 2 2 1 2 2 2 2 K r = m1 r 1 ω m2 r 2 ω m3 r 3 ω . .. mn r n ω 2 2 2 2 1 2 2 2 2 2 K r = ω m 1 r 1 m2 r 2m3 r 3 . . .mn r n 2 n n 1 K r = ω 2 ∑ mi r 2i and ∑ mi r 2i =I 2 i=1 i=1
1 2 K r = Iω 2 99
PHYSICS
CHAPTER 8
From the formula for translational kinetic energy,
Ktr
1 2 K tr = mv 2
After comparing both equations thus
ω is analogous to v I is analogous to m
For rolling body without slipping, slipping the total kinetic energy of the body, K is given by
K =K tr K r where
K tr : translational kinetic energy K r : rotational kinetic energy 100
PHYSICS
CHAPTER 8
Example 8.19 : A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25° to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate a. the total kinetic energy of the sphere, b. the linear speed of the sphere, c. the angular speed about the centre of mass. 2 2 (Given the moment of inertia of solid sphere is I CM = mR and 5 the gravitational acceleration, g = 9.81 m s−2)
101
PHYSICS
CHAPTER 8
Solution :
R=0.15 m ; m=10.0 kg s=0 . 75 m
R
h=ssin 25 v
CM
25
°
a. From the principle of conservation of energy,
∑ E i =∑ E f
mgh= K ° K =mgs sin 25 ° K =10.0 9.81 0.75 sin 25 102
°
PHYSICS
CHAPTER 8
Solution : R=0.15 m ; m=10.0 kg b. The linear speed of the sphere is given by 1 1 2 2 K =K tr K r K = mv Iω and
v ω= R
2 2 1 2 1 2 2 v K = mv mR 2 2 5 R 7 2 K = mv 10 7 2 31.1= 10. 0 v 10
2
c. By using the relationship between v and ω, thus
v=Rω
2 .11=0 . 15 ω
103
PHYSICS
CHAPTER 8
Example 8.20 : The pulley in the Figure 8.30 has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s−2)
5.00 kg 7 . 00 m
2.00 kg Figure 8.30
104
PHYSICS
CHAPTER 8
Solution : m1 =5 . 00 kg ;m 2=2. 00 kg; The moment of inertia of the pulley,
R=0 .120 m; h=7 . 00 m
−3 −4 2 10 kg 10 m 2 −9 2 I= 0 . 055 g 1 cm = 5 . 5×10 kg m 1g 1 cm 2
m1 7. 00 m
m2 m2
Initial
∑ E i =U 1
v
m1 Final
v
7 . 00 m
∑ E f = K tr 1105K tr 2K rU 2
PHYSICS
CHAPTER 8
Solution : m 1 =5 . 00 kg ;m 2=2. 00 kg;
−9
R=0 .120 m;
h=7.00 m; I =5.5×10 kg m
2
By using the principle of conservation of energy, thus
∑ E i =∑ E f
U 1 =K tr1 K tr 2 K r U 2 1 2 1 2 1 2 m1 gh= m1 v m2 v Iω m2 gh 2 2 2 2 1 2 1 v m1−m2 gh= 2 v m1m2 2 I R 1 2 1 v −9 5 . 00−2 . 00 9 . 81 7 . 00 = v 5 . 002. 00 5 . 5×10 2 2 0 . 120
106
2
PHYSICS
CHAPTER 8
8.4.2 Work, W
Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 8.31.
R
dθ
ds
R
O
F
Figure 8.31
The work done by the tangential force is given by
dW =Fds and ds=Rdθ dW =FRdθ θ
∫ dW=∫θ τdθ 2 1
θ
W =∫θ τdθ 2 1
107
PHYSICS
CHAPTER 8
If the torque is constant thus θ
W =τ ∫θ dθ 2
1
W =τ θ2 −θ 1
W =τΔθ
is analogous to the
W =Fs
τ : torque Δθ : change in angular displacement
where
W : work done
Work-rotational kinetic energy theorem states
W =ΔK r = K r f − K r i 1 2 1 2 W = Iω − Iω 0 2 2
108
PHYSICS
CHAPTER 8
8.4.3 Power, P
From the definition of instantaneous power,
dW P= dt τdθ P= dt P=τω
and
dW =τdθ
and
dθ =ω dt
is analogous to the
P=Fv
Caution : The unit of kinetic energy, work and power in the rotational kinematics is same as their unit in translational kinematics.
109
PHYSICS
CHAPTER 8
Example 8.21 : A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determine a. the rotational kinetic energy of the merry-go-round, b. the work done by the force on the merry-go-round, c. the average power supplied by the force. (Given g = 9.81 m s−2) Solution :
R=2 . 40 m
F 110
PHYSICS Solution :
CHAPTER 8 2
R=2.40 m ; I=2100 kg m ; F=18.0 N; t=15 . 0 s; ω 0 =0
a. By applying the relationship between nett torque and angular acceleration, thus
∑ τ=Iα
RF=Iα
2.40 18.0 =2100α
Use the equation of rotational motion with uniform angular ω=ω0 αt acceleration,
15.0 ω=0 2.06×10 −1 ω=0.309 rad s −2
Therefore the rotational kinetic energy for 15.0 s is
1 2 K r = Iω 2
1 2 K r = 2100 0 .309 2 111
PHYSICS
CHAPTER 8
Solution :
2
R=2.40 m ; I=2100 kg m ; F=18.0 N; t=15 . 0 s; ω 0 =0
b. The angular displacement, θ for 15.0 s is given by
1 2 θ=ω 0 t αt 2
1 2 −2 θ=0 2. 06×10 15 . 0 2
By applying the formulae of work done in rotational motion, thus
W =τθ
W =RFθ
W =2.40 18.0 2.32 c. The average power supplied by the force is
W P av = t
100 P av = 15.0
112
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PHYSICS CHAPTER 8 Learning Outcome: 8.5 Conservation of angular momentum (1 hour) At the end of this chapter, students should be able to: Define and use angular momentum,
L=Iω
State and use the principle of conservation of angular momentum
113
PHYSICS
CHAPTER 8
8.5 Conservation of angular momentum
L 8.5.1 Angular momentum,
is defined as the product of the angular velocity of a body and its moment of inertia about the rotation axis. axis OR is analogous to the p=mv
L=Iω
where
L : angular momentum
I : moment of inertia of a body ω : angular velocity of a body
It is a vector quantity. Its dimension is M L2 T−1 The S.I. unit of the angular momentum is kg m2 s−1.
114
PHYSICS
CHAPTER 8
The relationship between angular momentum, L with linear momentum, p is given by vector notation : magnitude form :
L =r ×p =r ×m v
L=rpsin θ=mvr sin θ where
r : distance from the particle to the rotation axis θ:the angle between {r with {v ¿ ¿
Newton’s second law of motion in term of linear momentum is
d p ∑ F = F nett = dt
hence we can write the Newton’s second law in angular form as
d L ∑ τ =τ nett = dt
and states that a vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum. momentum 115
PHYSICS
CHAPTER 8
8.5.2 Principle of conservation of angular momentum
states that a total angular momentum of a system about an rotation axis is constant if no external torque acts on the system. system OR
Iω=constant
Therefore
If the
∑ τ =0
d L ∑ τ = dt =0 d L=0 and dL=∑ L f −∑ Li
∑ L i=∑ L f 116
PHYSICS
CHAPTER 8
Example 8.22 : A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s-1 about the rotation axis as shown in Figure 8.32 . A 50 kg bag of sand falls onto the disc at the edge of the wooden disc.
R
ω0
Before
ω
R
Figure 8.32
After
Calculate, a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle) b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same? 1 2 MR ) (Use the moment of inertia of disc is
2
117
PHYSICS
CHAPTER 8 −1
Solution : R=3 . 00 m ;ω 0 =4 . 0 rad s a. The moment of inertia of the disc,
; mw =200 kg; mb =50 kg
1 2 2 1 I w = mw R = 200 3. 00 2 2 2 I w =900 kg m
The moment of inertia of the bag of sand, 2
2
I b =m b R = 50 3 . 00
I b =450 kg m
2
By applying the principle of conservation of angular momentum,
∑ L i=∑ L f
I w ω 0 = I w I b ω
900 4.0 = 900450 ω 118
PHYSICS
CHAPTER 8 −1
Solution : R=3 . 00 m ;ω 0 =4 . 0 rad s b. The initial rotational kinetic energy,
; mw =200 kg; mb =50 kg
1 2 2 1 K r i = 2 I w ω 0= 2 900 4 . 0 K r i =7200 J
The final rotational kinetic energy,
1 2 2 1 K r f = 2 I w I b ω = 2 900450 2 . 67
thus K r i ≠ K r f It is because the energy is lost in the form of heat from the friction between the surface of the disc with the bag of sand.
119
PHYSICS
CHAPTER 8
Example 8.23 : A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer. Solution : The answer is hard-boiled egg. egg
120
PHYSICS
CHAPTER 8
Solution : Reason Raw egg : When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases 2 because of I=mr From the principle of conservation of angular momentum,
Iω=constant
If the I is increases hence its angular velocity, ω will decreases. Hard-boiled egg : The position of the yolk of a hard-boiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.
121
PHYSICS
CHAPTER 8
Example 8.24 : A student on a stool rotates freely with an angular speed of 2.95 rev s−1. The student holds a 1.25 kg mass in each outstretched arm that is 0.759 m from the rotation axis. The moment of inertia for the system of student-stool without the masses is 5.43 kg m2. When the student pulls his arms inward, the angular speed increases to 3.54 rev s−1. a. Determine the new distance of each mass from the rotation axis. b. Calculate the initial and the final rotational kinetic energy of the system. 2 .95 rev 2π rad Solution : ω 0 = =18 . 5 rad s−1
1s
1 rev
3 . 54 rev 2π rad ω= =22. 2 rad s−1 1s 1 rev 122
PHYSICS
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Solution :
2
−1
m=1. 25 kg ; ω0 =18. 5 rad s ; I ss =5. 43 kg m ; −1 r b =0 .759 m ; ω=22. 2 rad s ;
ω0
rb
ω
rb m
m
Before
ra ra
After 123
PHYSICS
CHAPTER 8
Solution :
2
−1
m=1. 25 kg ; ω0 =18. 5 rad s ; I ss =5. 43 kg m ; −1 r b =0 .759 m ; ω=22. 2 rad s ;
a. The moment of inertia of the system initially is
I i =I ss I m
2 mr
I i =I ss mr 2 mr
=I ss
b
b
2
b2
2
I i = 5 . 43 2 1 .25 0 .759 =6 . 87 kg m2
The moment of inertia of the system finally is
I f =I ss 2 mr 2 a = 5 . 43 2 1. 25 r 2 I f =5 . 432 .5r 2 a a
By using the principle of conservation of angular momentum, thus i f
∑ L =∑ L I i ω 0 =I f ω
6 . 87 18 .5 =5 . 432 .5r
a
2
22 .2 124
PHYSICS
CHAPTER 8 2
−1
Solution : m=1. 25 kg ; ω0 =18. 5 rad s
; I ss =5. 43 kg m ; r b =0 .759 m ; ω=22. 2 rad s ; −1
b. The initial rotational kinetic energy is given by
1 K r i = 2 I i ω02 1 2 = 6 .87 18 .5 2
K r i =1 .18×10
3
J
and the final rotational kinetic energy is
1 2 K r f = 2 I f ω
1 2 2 = 5 . 432 .5 0 .344 22. 2 2 125
PHYSICS
CHAPTER 8
Exercise 8.5 : Use gravitational acceleration, g = 9.81 m s−2 1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s−1 relative to the Earth. a. In the what direction and with what value of angular speed does the turntable rotate? b. How much work does the woman do to set herself and the turntable into motion? ANS. : 0.360 rad s−1 ,U think; 99.9 J
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Exercise 8.5 : 2. Determine the angular momentum of the Earth a. about its rotation axis (assume the Earth is a uniform solid sphere), and b. about its orbit around the Sun (treat the Earth as a particle orbiting the Sun). Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m and is 1.5 x 108 km from the Sun. ANS. : 7.1 x 1033 kg m2 s−1; 2.7 x 1040 kg m2 s−1 3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of 1 2 thin rod about the axis through the CM is ML ) 12 ANS. : 4.71 x 10−6 kg m2 s−1 127
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Summary: Linear Motion
ds v= dt dv a= dt m
∑ F=ma p=mv W =Fs P=Fv
Relationship
v=rω a=rα n
Rotational Motion
dθ ω= dt dω α= dt
I=∑ mi r i2 i=1
τ =rF sin θ L=rpsin θ
I
∑ τ=Iα L=Iω
W =τθ P=τω
128
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PHYSICS
CHAPTER 8
THE END… Next Chapter…
CHAPTER 9 : Simple Harmonic Motion (SHM)
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