Dynamics MCB2043 Revision Questions
Dereje Engida Woldemichael (PhD, CEng MIMechE) MIMechE)
[email protected]
September 2014 Semester
EXAMPLE # 1 Ball bearings leave the horizontal trough with a velocity of m agnitude u and fall through the 70-mm-diameter holes as shown. Calculate the permis sible range of u which will enable the balls to enter the hole. Take Take the dotted positions to represent the limiting conditions. Note the origin of the coordinate system is attached to the point leaving the trough with y-direction positive downwards .
x
s = = ut
Horizontal motion Vertical motion
B
A
h
1
=
2
gt
u
2
s
=
2 h
y
g
C
To fall through the hole, the centre of the ball bearing must be within the range of points C and D with the limited trajectories between BC and BD. D d 70 20 For trajectory BC s BC = 120 − 2 + 2 = 120 − 2 + 2 = 95 mm
120 −
=
u BC
=
95 × 10 −
−3
2 × 80 × 10 −
=
For trajectory BD s BD
=
(120
+
2
)−
d 2
2
2
s BD
0.744 m / s
120
9.81 D
D
s BC
−3
umin
D
d
d +
D
2
2
−3
=
145 mm
umax
=
u BD
=
145 × 10 −
−3
2 × 80 × 10 −
= 1.135 m / s
9.81
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EXAMPLE # 2
At the instant shown, the bicyclist at A is traveling at 7 m/s around the curve on the race track while increasing his speed at 0.5 m/s2. The bicyclist at B is traveling at 8.5 m/s along the straight-a-way and increasing his speed at 0.7 m/s2. Determine the relative velocity and relative acceleration of A with respect to B at the instant.
EXAMPLE # 3 A child twirls a small 50-g ball attached to the end of a 1-m string so that the ball traces a circle in a vertical plane as shown. What is the minimum speed v that the ball must have when in position 1? If this speed is maintained throughout the circle, calculate the tension T in the string when the ball is in position 2. Neglect any small motion of the child’s hand. FBD in position 1
From FBD of the ball in position 1, in order to have the minimum speed, there should be no string tension and only the weight of ball is applied in the n direction. Thus, the equation of motion yields
∑ F
n
= mg = ma n = m
mg
t
2 vmin
r
n
Hence, the minimum speed at position 1 v min
=
gr
=
9.81 × 1
=
3.13 m / s
FBD in position 2
Using the FBD at position 2, the equation of motion is given by
∑ F
n =
2
T − mg = m
vmin r
���
=
T
( gr ) = m r
Thus the tension in position 2
T = 2 mg
n
2
t
mg
2 × 0.05 × 9.81 = 0.981 N
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EXAMPLE # 4 The 100 kg crate is carefully placed with zero velocity on an incline as shown. Describe what happens if (a) θ =15º and (b) θ =20º. Calculate the distance x travelled by the crate when reaching v=4 m/s with θ =20º. 100 kg
(a) Assume no slip for the crate (θ=15º) With reference to the FBD of crate, the equations of motion in x and y directions
∑ F
∑ F
mg sin 15 − f
= 0
x
y
=
=
N − mg cos 15
0
0
f = 100 × 9.81 × sin 15 = 253.9 N
0
N = 100× 9.81× cos15 = 947.6 N
=
f
If no slip, f ≤ µ s N should hold. Since
=
253 .9 ≤ µ µ µ s N = 0 .3 × 947 .6
µ s=0.3 µ k=0.25
x
θ
FBD =
y
284 .3
N = 100 × 9.81 × cos 20 = 921.8 N
Assumption of no slip is true (Crate is static )
f
x
(b) Assume no slip for the crate again (θ=20º)
θ
N
Using the similar approach, one can find
f = 100 × 9.81 × sin 20 = 335.5 N
f
=
m g
335 .5 > µ µ µ s N = 0 .3 × 921 . 8 = 276 .5
Assumption of no slip is false (Crate starts to slip )
Thus, the equation of motion in x and y directions can be changed to
∑ F
x
∑ F
=
ma x
y =
mg sin 20 − µ µ µ k N = ma x N − mg cos 20 = 0
0 a x
= g (sin
Using kinematic equation
x
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=
20 − µ µ k cos 20 ) = 9 .81 × (sin 20 − 0 .25 × cos 20 ) = 1 .05 m / s µ µ
v
v2 2 a x
2
−
=
0
=
2
2 a x x
42 2 × 1 . 05
=
7 . 62 m
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Exercise A 2 kg box slides on a smooth ramp from A to B and subsequently undergoes a projectile motion until it strikes the ground at C as shown. The speed of the box at A is v A = 2 m/s. a. Determine the velocity of the box at B using energy method. b. Determine the horizontal distance from B to C .
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