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. groundv-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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