Test Index

Systems of Particles and Rotational Motion

© examsnet.com
Question : 13 of 33
Marks: +1, -0
(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/ min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution:  
(a) Given that, initial angular speed, ω1=40ω_1 = 40 rev/min, ω2=?ω_2 = ?
Suppose that initial moment of inertia of the child is I1I_1. Then, final moment of inertia of the child
I2=2/5I1I_{2}= {2}/{5} I_{1}
As no external torque acts in the process, therefore, according to the principle of conservation of angular momentum,
I1ω1=I2ω2I_{1} \omega_{1}=I_{2} \omega_{2}
⇒ω2=I1I2ω1\Rightarrow \omega_{2}= \frac{I_{1}}{I_{2}} \omega_{1} =5/2×40=100rpm= {5}/{2} ×40=100 {rpm}
(b) final kinetic energy of rotation K2initial kinetic energy of rotation K1=12I2ω2212I1ω12\frac{\text{final kinetic energy of rotation } K_{2}}{\text{initial kinetic energy of rotation } K_{1}} = \frac{\frac{1}{2} I_{2} \omega_{2}^{2}}{\frac{1}{2} I_{1} \omega_{1}^{2}}
K2/K1=I2ω22/I1ω12=(I2/I1)(ω2/ω1)2{K_{2}}/{K_{1}}= {I_{2} \omega_{2}^{2}}/{I_{1} \omega_{1}^{2}}= ( {I_{2}}/{I_{1}} ) ( {\omega_{2}}/{\omega_{1}} )^{2}
=2/5×(100/40)2=5/2=2.5= {2}/{5} ×( {100}/{40} )^{2}= {5}/{2}=2.5
K2/K1=2.5⇒K2=2.5K1{K_{2}}/{K_{1}}=2.5 ⇒ K_{2}=2.5 K_{1}
The new kinetic energy is 2.5 times initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.
© examsnet.com
Go to Question:
Prev Question
Next Question