Question

# Derive the formula for centripetal acceleration of a particle which is moving with a constant speed on a circular path.

Solution

## Consider the distance travelled by the body over a tome $$\Delta t$$ Arc length $$S=V \Delta t$$ also $$S=r \Delta \theta$$ $$\therefore \quad r\Delta \theta =V\Delta t\\ \cfrac { \Delta \theta }{ \Delta t } =\cfrac { V }{ r } \\ \Rightarrow \cfrac { d\theta }{ dt } =\cfrac { V }{ r }$$ angular velocity Now from the second diagram: $$\sin { \left( \Delta \theta /2 \right) =\cfrac { \Delta V }{ 2V } } \\ \Delta V=2V\sin { \left( \Delta \theta /2 \right) } \\ \cfrac { \Delta V }{ \Delta \theta } =\cfrac { 2V\sin { \left( \Delta \theta /2 \right) } }{ \Delta \theta } \\ \Rightarrow \cfrac { \Delta V }{ \Delta \theta } =\cfrac { V\sin { \left( \Delta \theta /2 \right) } }{ \left( \Delta \theta /2 \right) } \\ \Rightarrow \cfrac { dV }{ d\theta } =\lim _{ \Delta \theta \longrightarrow 0 }{ \cfrac { V\sin { \left( \Delta \theta /2 \right) } }{ \left( \Delta \theta /2 \right) } =V }$$  $$\Rightarrow \cfrac { dV }{ d\theta } =\cfrac { dv }{ d\theta } \cfrac { d\theta }{ dt } =V\times \cfrac { V }{ r } =\cfrac { { V }^{ 2 } }{ r }$$Physics

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