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# Suppose the Platform with the Kid in the Previous Problem is Rotating in Anticlockwise Direction at an Angular Speed ω. - Physics

Sum

Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed ω. The kid starts walking along the rim with a speed $\nu$ relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.

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#### Solution

Let ω' be the angular velocity of platform after the kid starts walking.

Let ω be the angular velocity of the wheel before walking.

When we see the (kid-wheel) system from the initial frame of reference, we can find that the wheel moves with a speed of $\omega$ and the kid with a speed of $\left( \omega' + \frac{\nu}{R} \right),$ after the kid has started walking.

Initial angular momentum of the system = (I + MR2

Angular velocity of the kid after it starts walking = $\left( \omega' + \frac{v}{R} \right)$

External torque is zero; therefore, angular momentum is conserved.

$\Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)$

$\Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)$

$\Rightarrow \left( I + M R^2 \right)\omega - MvR = \left( I + M R^2 \right) \omega'$

$\Rightarrow \omega' = \frac{\left( I + M R^2 \right)\omega - MvR}{\left( I + M R^2 \right)}$

$\Rightarrow \omega' = \omega - \frac{MvR}{I + M R^2}$

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 58 | Page 198
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