Q#51
A wheel of moment of inertia 0.500 kg - m² and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.Answer:
The moment of inertia I = 0.500 kg-m²
r = 0.20 m, ⍵ = 20 rad/s.
The mass of the stationary particle m = 200 g = 0.20 kg
Hence the moment of inertia of the system with the particle
I' = I + mr²
= 0.50 + (0.20)(0.20)2 kg.m²
= 0.50 + 0.008 kg.m²
= 0.508 kg.m²
Since there is no external torque on the system, the angular momentum will be conserved. Hence
I'⍵' = I⍵
0.508⍵' = 0.50(20)
⍵' = 10/0.508 rad/s = 19.7 rad/s
Q#52
A diver, having a moment of inertia of 6.0 kg-m² about an axis through its center of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5 kg-m², what will be the new angular speed?
Answer:
The initial moment of inertia I = 6.0 kg-m², ⍵ = 2 rad/s. The angular momentum
L = I⍵ = 6.0(2) = 12 kgm²/s.
The final moment of inertia I' = 5.0 kg.m²
Let the final angular speed = ⍵'.
Final angular momentum = L' = I'⍵' = 5⍵' kg.m²/s
Since there is no external torque on the diver, his angular momentum will be conserved. i.e.
L' = L
5⍵' = 12
⍵' = 12/5 = 2.4 rad/s
Q#53
A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg.m² to 2 kg-m², what will b the new angular speed?
Answer:
The initial moment of inertia I = 6 kgm²,
⍵ = 120 rev/minute
Final M.I. = I' = 2 kg.m²
⍵' =?
Since the angular momentum will be conserved in the absence of a torque,
So, I'⍵' = I⍵
⍵' = I⍵/I' = 6(120/2) = 360 rev/minute.
Q#54
A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of “the platform plus the boy system” is 3.0 x 10⁻³ kg.m² and that of the umbrella is 2.0 x 10⁻³ kg.m². The boy starts spinning the umbrella about the axis at an angular speed of 2.0 rev/s with respect to himself. find the angular velocity imparted to the platform.
Answer:
Let the angular velocity of the platform and boy = ⍵'
The angular velocity of the umbrella with respect to the ground = ⍵ = ⍵'+ 2 rev/minute
The angular momentum given to the umbrella = I⍵
= (2.0 x 10⁻³)(⍵'+ 2) units
The M.I. of the platform+the boy
I' = 3.0 x 10⁻³ kg.m²
The angular momentum of the platform = I'⍵'
Since on the system of (The boy + platform + umbrella) there is no external force or torque, hence its angular momentum will be conserved. Initial angular momentum = 0, hence the final angular momentum I⍵ + I'⍵' = 0
I'⍵' = –I⍵
⍵' = – (⍵' + 2)(2.0 x 10⁻³)/(3.0 x 10⁻³)
3⍵' = –2⍵' – 4
5⍵' = –4
⍵' = –0.80 rev/s
The negative sign denotes the direction of angular velocity which is opposite to the umbrella.
Q#55
A wheel of moment of inertia 0.10 kg-m² is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.
Answer:
The moment of inertia of the first wheel = I = 0.10 kg-m²
Its angular speed ⍵ = 160 rev/minute.
The angular speed of the second wheel ⍵' = 300 rev/minute
Let the moment of inertia of the second wheel = I'
The total angular momentum of both the wheels = I⍵ + I'⍵'
The angular speed of the system after the coupling = ⍵" = 200 rev/minute
The final angular momentum of the system = (I + I')⍵"
The angular momentum will be conserved, hence
I⍵ + I'⍵' = (I + I')⍵"
0.10(160) + I'(300) = (0.10 + I')(200)
16 + 300I' = 20 + 200I'
I' = 0.04 kg.m²
Q#56
A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.
Answer:
The linear momentum of the ball = mv
The angular momentum of the ball about the center of the platform which is transferred to the system = moment of the linear momentum about the center = mvR
The moment of inertia of the system after the ball is caught = I + MR² + mR² = I + (M + m)R²
If the final angular speed of the system = ⍵
its angular momentum = {I + (M + m)R²}⍵
Since the angular momentum will be conserved, hence
{I + (M + m)R²}⍵ = mvR
⍵ = mvR/{I + (M + m)R²}
Q#57
Suppose the platform of the previous problem is brought to rest with the ball in hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed v as seen by his friend. Find the angular velocity with which the platform will start rotating.
Answer:
The moment of inertia of the platform and the kid
= I + MR²
It is initially at rest, so its angular momentum with the ball is zero.
If the angular velocity of the platform after throwing the ball is ⍵, then angular momentum
= (I + MR²)⍵
The angular momentum of the ball after it is thrown = mvR
Since the angular momentum will be conserved, the total angular momentum of the ball and the platform with the boy will also be zero.
(I + MR²)⍵ + mvR = 0
⍵ = –mvR/(I + MR²)
The negative sign denotes the instantaneous speed of the boy will be opposite to the ball.
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