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Question

Initially the platform with the child is rotating with an angular speed $$\omega_{1}$$. If the child starts walking along its periphery in opposite direction with speed $$u$$ relative to the platform, what will be the new angular speed of the platform.


Solution

Let $${ v }_{ 0 }$$-velocity (tangential) of peripheral point w.r.t. ground
v-velocity of boy w.r.t ground
$$\therefore u=v+{ v }_{ 0 }$$
Again using conservation of angular momentum
$$\left( I+m{ R }^{ 2 } \right) { \omega  }_{ 1 }=I{ \omega  }_{ 0 }+mR\left( u-{ v }_{ 0 } \right) $$
$$\left( I+m{ R }^{ 2 } \right) { \omega  }_{ 1 }=I{ \omega  }_{ 0 }+mr\mu -m{ R }^{ 2 }{ \omega  }_{ 0 }$$
$$\left( I+m{ R }^{ 2 } \right) { \omega  }_{ 1 }=\left( I-m{ R }^{ 2 } \right) { \omega  }_{ 0 }+mRv$$
Therefore, $${ \omega  }_{ 0 }=\cfrac { \left( I+m{ R }^{ 2 } \right) { \omega  }_{ 1 }-mRv }{ \left( I-m{ R }^{ 2 } \right)  } $$

Physics

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