UNIT 6: Circular Motion
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6.1 Uniform Circular Motion
Definition - is motion in a circle circle (circular arc) arc) at a constant speed. Uniform circular motion of a particle about the axis of rotation, O can be described by the angular position (θ ) of a reference line with respect to the x-axis as shown below. y
θ = r θ
O
s r
where θ : angular position in radian
s
s : the length of arc. (arc length) r : the radius of the circular motion
x
y ∆θ t 1
t 2
If the particle changes its angular position of the reference line from θ 1 to θ 2 , the angular displacement ∆θ is given by
θ 1
O
θ 2
x
∆θ = θ 2 − θ 1
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∆θ is positive if the motion is anticlockwise. ∆θ is ne ative if the motion is clockwise.
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6.1.1 Angular velocity (angular frequency),
Average angular velocity, ω av Definition – is the rate of change of angular displacement.
ω av =
θ 2 − θ 1 t 2 − t 1
∆t →0
=
∆θ ∆t
Instantaneous angular velocity, ω Definition – is the instantaneous rate of change of angular displacement.
ω = limit
r
∆θ d θ = ∆t dt
Vector quantity. The unit of angular velocity is radian per second (rad s-1). Notes : For uniform circular motion, the angular velocity is constant therefore the angular displacement in time, t, θ =ω t t Unit conversion :
π rad = 180
o
2π rad = 360
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o
The direction of angular velocity will be learnt in unit 7.
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r
6.1.2 Tangential velocity (linear velocity),
y
v
From the definition of instantaneous linear velocity, we get
r θ
s x
ds
where s = r θ θ
dt d θ d θ = ω v = r where dt dt
v = r ω
v
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v=
v
O
v
The relationship between angular velocity and linear velocity
Vector quantity.
The unit of the tang ential (linear) velocity is m s-1.
It is directed tangentially to the circular path.
The magnitude of the linear velocity velocity (speed) of an object is constant in uniform circular motion but the direction is continually changing.
The linear velocity is difficult to measure but we can measure the period, T of an object in circular motion. 4
The period, T
Definition – is the time taken for one complete revolution (cycle/rotation).
The S.I. unit of the period is second.
If the object makes one revolution (rotation), the angular displacement, ∆θ = 2 π radian and the time interval, ∆t = T hence ∆θ
ω =
ω =
∆t
2π T
or ω = 2π f where f =
1 T
The frequency, f
Definition – is the number of revolutions (cycles/rotations) completed in one second.
The S.I. unit of the frequency is hertz (Hz) or s-1.
Therefore we can determine the linear velocity by using equ ation below.
v=
2π r T
= 2π rf
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r
6.1.3 Centripetal (radial) acceleration,
r
a c or a r
The figure below shows a particle moving with constant speed in a circular path of radius, r with centre at O. The particle moves from A to B in r a time, ∆t .
v0
r
v1
Consider sector OAB of the circle. The arc length AB is given by ∆ s = r ∆θ ∆ s ∆θ =
r
(1)
The velocities of the particle at A and B are v 0 and v 1 respectively where r r
v0 = v1 = v
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Let PQ and PR represent the velocity vectors v 0 and v 1 respectively, as shown in figure below.r
v0
P
Q r
∆ v
r
v1
r
r
= v1 − v0
R Then QR represent the change in velocity vector v of the particle in time interval ∆t . Since the angle between PQ and PR is small hence
(QR ) = (PQ )∆θ ∆v = v ∆θ ∆v ∆θ =
v
(2)
Equation (1) is equal to equation (2) then
∆ s
r
=
∆v
Dividing by time, ∆t , we get
v
1 ∆ s
1 ∆v = r ∆t v ∆t
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As ∆t → 0, we have
1 ∆ s 1 ∆v = limit limit r ∆t →0 ∆t v ∆t →0 ∆t 1 ds
1 dv = r dt v dt
v r
a
=
ac =
v v2 r
where ac : the centripetal acceleration v : the linear(tangential) velocity
or
r : the radius of circular motion
ac = r ω 2 = vω
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Definition – is defined as the acceleration of an object moving in circular path whose direction is towards the centre of the circular path and whose magnitude is equal to the square of the speed divided by the radius. 8
The direction of centripetal (radial) acceleration is always directed toward the centre of the circle and perpendicular to the linear (tangential) velocity as shown below. r
r
ac
ac
r
ac r
ac
r
ac
Because of
v=
r
ac
2π r T
therefore we can obtain the alternative expression of centripetal acceleration is
ac =
4π 2 r T 2
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Example 1: A sprinter is running at constant speed 9.2 m s-1 in a circular track with centripetal acceleration 3.8 m s-2. Find a. the radius of the circular track. b. the time required by the sprinter to make one revolution. Solution: v = 9.2 m s-1 and ac = 3.8 m s-2 a. Applying equation of centripetal a cceleration:
ac = 3.8 =
v2 r
(9.2 )2
r r = 22.3 m
b. the period of the sprinter,
v=
2π r
T T = 15.2 s SF017
or
ac =
4π 2 r T 2
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Example 2: A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?(HRW.70.51) (Use g = 9.81 m s -2) Solution:
v0
1.5 m
2.0 m
2.0 m 10 m After
Before
The time taken by the stone to strikes the ground,
1
s y = v0 y t − gt 2 where v0 y = 0 2 2h t = = 0.64 s g
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Considering the horizontal distance, x = 10 m
x = v0 x t = v0 t v0 =
10 0.64
= 15.6 m s −1
The initial speed of the stone after the string breaks is equal to the speed of the stone in horizontal circular motion. Therefore
ac =
v2
r
= 162 m s −2
Example 3: (exercise) The astronaut orbiting the Earth is preparing to dock with Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth’s surface, where the free fall acceleration is 8.21 m s-2. Take the radius of the Earth as 6400 km. Determine the speed of the satellite and the time interval required to complete one orbit around the Earth. Ans. : 7581 m s-1, 5802 s No. 32, pg. 104,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.
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r
6.1.4 Uniform circular motion in terms of position vector,
r
For example, an object is in th e circular motion as shown in figure below. The position vector at point P in the circular v path is v r r = r xiˆ + r y jˆ where r x = r cos θ P r
r y r
r y = r sin θ ˆ r = r cos θ i + r sin θ jˆ where θ = ωt r
θ
hencer
r x
O
v
r = r cos ω t iˆ + r sin ω t jˆ r r = r cos ω t iˆ + sin ω t jˆ
(
)
where r : the radius of circular path
By using the equation of instantaneous velocity, the tangential velocity vector is r
r
v=
d r dt d
(r cos ω t iˆ + r sin ω t jˆ ) dt v = r ω (− sin ω t iˆ + cos ω t jˆ ) r
v= r
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By using the equation of instantaneous acceleration, the centripetal (radial) acceleration vector is r r
d v
r
dt d
ac =
(− r ω sin ω t iˆ + r ω cos ω t jˆ ) dt = −r ω (cos ω t iˆ + sin ω t jˆ )
ac = r
ac
2
Negative sign means the direction of centripetal acceleration always r opposite to the direction of po sition vector, r
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6.2 Non-uniform Circular Motion In non-uniform circular motion, the tangential (linear) velocity changes both in direction and in magnitude. Therefore the resultant acceleration can be written as the vector sum of the component vectors :
r
r
r
a = ac + at where ac : the centripetal acceleration at : the tangential acceleration The tangential acceleration component causes the change in the speed of the particle (object). its direction always parallel to the tangential (linear) velocity and given by r
at =
d v
dt
If a particle (object) is speeding up the direction of a t is in the same direction of the tangential velocity, v
If a particle (object) is slowing up the d irection of a t is opposite direction of the tangential velocity, v
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The centripetal (radial) acceleration component
arises from the change in direction of the tangential (linear) velocity.
its direction always toward the centre of the circular path and given by
ac =
v2
r
The example of non-uniform circular motion is a particle moving in a vertical loop, like a roller coaster car with a varying speed as shown in figure r below. Because ac and at are always r B perpendicular to each other, the
v
r
r
a = ac ar c
v
magnitude of resultant acceleration, a at any time is
Ar
at
r
a = ac2 + at 2
r
a
r
ac C
r
at SF017
r
r
a a=a c r
r
v
r
D
v
At point D, the resultant acceleration is maximum.
At point B, the resultant acceleration is minimum. 16
Example 4: A particle revolves in a vertical circle with a radius of 2.50 m. At a particular instant, its total acceleration is 1.20 m s-2 in the direction that makes an angle of 30.0° with the direction of the motion. Find a. its speed at that moment. b. its speed 2.00 seconds later, Assuming constant tangential acceleration. (Csw.CD4.2.6.1.pg7) Solution: a = 1.20 m s-2 and r =2.50 m a. By using centripetal acceleration equation : 2
ac =
v
r
= a sin 30
o
Then the speed is r
v = ra sin 30
v
o
v = 1.22 m s −1
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b.
By using equation of motion: where
v = v0 + at t at = a cos 30 = 1.04 m s −1 o
hence we obtain
v = 1.22 + (1.04 )2.00 = 3.3 m s −1
Example 5: (exercise) A particle starts moving in a circle of radius of 0.20 m with constant tangential acceleration, at = 5 cm s-2. After what time will the value of centripetal acceleration reach twice the value of the tangential acceleration? (Csw.CD4.2.6.1.pg8t3) Ans. : 2.83 s
Example 6: (exercise) A racing car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m s-1 in 11 s, moving on a circular track of radius 500 m. Assuming constant tangential acceleration, find a. the tangential acceleration, and b. the radial acceleration, at the instant when the speed is v =15 m s-1 and again when v =30 m s-1. (Gc.121.ex.5-14) Ans. : 3.2 m s-2, 0.45 m s-2, 1.8 m s-2
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r
r
6.3 Centripetal Force, F c or F r
Any object moving in uniform circular motion has a centripetal acceleration directed towards the centre of the circular path. From Newton’s second law of motion, a force must be associated with the centripetal acceleration. This force is known as the centripetal force and is given by r
r
∑ F = F
net
r
r
r
r
F c = mac where ac = F c =
r
r
= ma where a = ac and ∑ F = F c r
mv 2 r
r
v2 r
= r ω 2 = vω
= mr ω 2 = mvω
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Its direction is in the same direction of centripetal acceleration (directed toward the centre of the circle) as shown in figure below. r
ac
r
r
F c
v
v
r
r
F c
r
ac r
r
ac
F c
r
v
If the centripetal force suddenly ceases to act on a body in circular motion, the body flies off in a straight line with the constant tangential (linear) velocity as show in figure below. r
r
r
v
ac
F c
r
F c
r
ac r
F c SF017
v
r
r
ac r
v
r
F = 0 rc ac r= 0 v
r
F = 0 rc ac =r 0 v
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Definition – is defined as the force acting on a body causing it to move in circular path and its always directed towards the centre of the circular path.
Notes :
In uniform circular motion, the net force on the system is centripetal force.
The work done by the centripetal force is zero but the kinetic energy of the body is not zero and given by
K =
1
2
mv 2 =
1
2
mr 2ω 2
The figure below shows an overhead view of a ball moving in a circular path in a horizontal plane. Wh en the string breaks, the ball moves in the direction tangent to the circle.
r
r
v SF017
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6.4 Examples of the Circular motion 6.4.1 Conical Pendulum
Example 7:
The figure shows an object with mass 40 kg is attached to one end of a string 1.50 m long. The object revolves in a circle of radius, r and makes an angle, θ with the vertical line. The string will break if its tension exceeds 600 N. Find a. the maximum angle, θ b. the maximum speed, c. the minimum period of the object can attain without breaking the string.
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Solution : l= 1.5 m ; m =40 kg ; T ≤ 600 N
T cos θ r
T
T sin θ r
r
m g
m g
The weight of a conical pendulum is supported by the vertical component of the tension in figure above. Hence
T cos θ = mg
Hence
T sin θ = F c =
(1) The centripetal force is contributed by the horizontal component of the tension. 2 Note:
mv
(2)
r
For a conical pendulum, θ cannot be equal to 90°.
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a. By using equation (1),
cos θ max =
mg
T max θ max = 49.2
o
b.
By using equation (2),
l
sin θ =
r
vmax =
l vmax = vmax
c.
ω = T min =
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2π
and v = r ω T 2π r
Tr sin θ m Tl sin 2 θ
m = 3.6 m s −1
then
vmax
T min = 2.00 s
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6.4.2 Motion rounds a curve on a flat (unbanked) track (for car, motorcycle, bicycle, etc…) Example 8: A flat (unbanked) curve on a highway has a radius of 220 m. A car rounds the curve at a speed of 25.0 m s-1. What is the minimum coefficient of friction that will prevent sliding? ( Use g = 9.81 m s-2) No. 5.44, pg. 197, University Physics with Modern Physics,11th edition, Young & r Freedman.
N
Solution :
Centre of circle
r
f r
m g From the free body diagram above, x-component : The centripetal force is provided by the frictional force between the wheel (4 tyres) and the road. Hence f = SF017
mv 2 r
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y-component : The normal force is balance by the weight of the car, hence N = mg
µ N =
Therefore
µ mg =
mv 2 r 2 mv
r µ = 0.290
6.4.3 Motion rounds a curve on a banked track (for car, motorcycle, bicycle, etc…)
Example 9: A car rounding a curve on a road banked at an angle, θ to the horizontal with design speed 50 km h-1. If the radius of the curve is 50 m, find the angle, θ at which the car can travel without skidding. Neglect the friction between the car and the road. (use g = 9.81 m s-2) (Gc.120.ex.5-13)
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Solution : v = 50 km h-1 = 14 m s-1 , r = 50 m r
N cos θ
N
θ
r
ac N sin θ
θ r
Centre of circle From the free body diagram above,
m g
x-component : The centripetal force is contributed by the horizontal component of the normal force. Hence
N sin θ =
mv 2
(1)
r
y-component : No vertical motion, hence
N cos θ = mg
(2)
Equation (1) divided by equation (2), hence
tan θ =
v2
rg θ = 22
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o
Notes: Motion rounds a curve on a banked track with friction (for a car, r motorcycle, bicycle, etc…) N cos θ
N
r
θ
ac N sin θ r
f cos θ r
f Centre of circle
m g
θ f sin θ
From the free body diagram above, x-component : The centripetal force is contributed by both the horizontal component of the normal force and frictional force. Hence 2
N sin θ + f cosθ =
y-component : No vertical motion, hence
mv
r
N cos θ = mg + f sin θ SF017
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6.4.4 Motion in a horizontal circle
A figure below shows a ball of mass, m whirled at the end of a thread in a horizontal circle. r
T
r
r
m g
The centripetal force which enables the ball to move in a circle is provided by the tension in the string. Hence
T =
mv 2 r
It is an example of uniform circular motion in which the magnitude of velocity always constant.
Example 10: A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 600 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. Find the maximum speed the stone attain without breaking the string.
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No. 5.43, pg. 197, University Physics with Modern Physics,11th edition, Young & Freedman. 29
Solution : m = 0.80 kg, l=0.90 m=r, T ≤ 600 N r
T
r
r
m g From the figure above, The centripetal force is contributed by the tension of the string. Hence
T = vmax =
mv 2
where l = r r T maxl m
vmax = 26 m s −1
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6.4.5 Motion in a vertical circle
An object of mass, m tied to an inelastic string is moving in a vertical circle of radius, r . The figure shows the mass and the taut string inclined at an angle, θ to the vertical line.
r O
r
T
The forces acting on the mass, m are the tension of the string, T and its weight mg .
In any circular r motion, r
∑ F = F
net
θ
= F c
By resolving forces along the string, we get 2
T − mg cos θ =
mg sin θ
θ mg cos θ
T =
r
m g
mv 2 r
mv
r
+ mg cos θ
The tension of the string changes with the position of the object because of
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r
the angle, θ
the tangential velocity, v
r
For example,
vC
changes
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C
r
Point Equation of Net force
A B C D
T A − mg = T B =
mv B2
mv
r
m g r
r
T D
D
O
r
r
T B
r
r
T C + mg = T D =
T C
2 A
2 D
2 mvC
r
v D
r
r
m g
B r
T A
mv
v B
m g r
v A
r r
m g
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r
vC
C
r
Point
Total Energy
K A = B K B + U B = C K C + U C = D K D + U D =
1 2 1 2 1 2 1 2
T C
mv A2
r
m g r
T D
D
mv B2 + mgr
O
r
v B
r
T B
r
mv + mg (2r )
v D
r
r
2 C
r
T A
m g
m g r
mv + mgr 2 D
v A r
m g Notes: The tension of the string is gre atest at A and smallest at C. The tangential (linear) speed is minimum at C and maximum at A. The vertical circular motion is an example of the non-uniform circular motion. If the mass is fixed to one end of a light rod, it would be possible to rotate the mass in a vertical circle at constant speed f or example Ferris wheel (uniform circular motion) where v A = v B = vC = v D
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Example 11: An object of mass 1.5 kg is tied to a string of length 0.50 m. The object is made to move in a vertical circle, as shown in figure below.
A
B
A
B
∑ E = ∑ E A
0.50 m
r
m g b.
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When it reaches the highest point A, the linear tangential velocity is 4.0 m s-1. Determine a. the angular velocity of the object at B, b. the tension in the string at B. (Use g = 9.81 m s -2)
Solution : m = 1.5 kg, l = r = 0.50 m, v A = 4.0 m s-1 a. By using the principle of conservation of energy, we get
r
T B
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B
K A + U A = K B + U B 1 1 mv A2 + mgh = mr 2ω B2 + mgr 2 2 ω B = 10.2 rad s −1 2 mv T B = B = mr ω B2 r T B = 78.0 N
h = 1.00 m
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Example 12: A small remote control car with mass 1.60 kg moves at a constant speed of v = 12.0 m s-1 in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m as shown in figure below. What is the magnitude of the normal force exerted on the car by the walls of the cylinder at a. point A (at the bottom of the vertical circle)? b. point B (at the top of the vertical circle)? (Use g = 9.81 m s -2) No. 5.118, pg. 205, University Physics with Modern Physics,11th edition, Young & Freedman.
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Solution : m = 1.60 kg, r = 5.00 m, v = 12.0 m s-1
B
r
N r B m g O r
N A A
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r
m g
a. At point A,
∑ F = F = c
N A − mg =
mv 2 r 2 mv
r N A = 61.8 N
b. At point B,
N B + mg =
mv 2
r N B = 30.4 N
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Example 13: A 1000 kg sports car moving at 20 m s-1 crosses the rounded top of a hill (radius = 100 m). Determine a. the normal force on the car, b. the normal force on the 70 kg driver, c. the car speed at which the normal force is zero. (Use g = 9.81 m s -2) (Gc.128.47) Solution : m = 1000 kg, r = 100 m, v = 20 m s-1 r r a. r
∑ F = F m
C
mcar g − N =
v
N
2
car
r
v
r 3 N = 5.81 x 10 N b.
md g − N =
r
md v 2
m g
r N = 4.07 x 10 2 N SF017
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c.
mcar g − N =
mcar v 2 r
where N = 0
v = rg
v = 31.3 m s −1
r
v
Example 14: A rope is attached to a bucket of water and the bucket is then rotated in a vertical circle of 0.70 m radius. Calculate the minimum speed of the bucket of water such that the water will not spill out. (Use g = 9.81 m s -1) (Csw.CD4.2.6.3.pg11t1) Solution : r = 0.70 m The water will spill out when the bucket at the r top of the circle where the equation of the net force is given by N mv 2 r
m g
mg + N =
At this moment, if the water is not falling out from the bucket means the speed of the bucket is minimum and normal force, N =0 therefore
mg = SF017
r
vmin =
2 mvmin
r rg = 2.62 m s -1
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Example 15: (exercise) A cyclist goes around a curve of 50 m radius at a speed of 15 m s-1. The road is banked at an angle θ ; the cyclist travels at the right angle with the surface of the road. The mass of the bicycle and the cyclist together equals 95 kg. Find a. the centripetal acceleration of the cyclist b. the normal force which the road e xerts on the bicycle and the cyclist c. the angle, θ Ans. : 4.5 m s-2, 1.02 kN, 24.6° (Csw.CD4.2.6.3.pg11t4)
Example 16: (exercise) A 4.00 kg object is attached to a vertical rod by two strings, as shown in figure below. The object rotates in a horizontal circle at constant speed 6.00 m s-1. Find the tension in a. the upper string b. the lower string. Ans. : 108 N, 56.2 N No. 11, pg. 172,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.
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Example 17: (exercise) A small mass, m is set on the surface of a sphere as shown in figure below. If the coefficient of static friction is µ s = 0.60, at what angle, φ would the mass start sliding?(Gc.121.83) Ans. : 31°
v
v
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Example 18: (exercise) A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m at constant speed, v as shown in figure above. If the time taken to makes one rotation is 10 s and the mass of the rider is 60 kg, find the normal force exerted on the rider a. at the top of the circle, b. at the bottom of the circle. Ans. : 399 N, 778 N 40
