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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:

$$\sin { \left( \Delta \theta /2 \right) =\cfrac { \Delta V }{ 2V }  } \\ \Delta V=2V\sin { \left( \Delta \theta /2 \right)  } \\ \cfrac { \Delta V }{ \Delta \theta  } =\cfrac { 2V\sin { \left( \Delta \theta /2 \right)  }  }{ \Delta \theta  } \\ \Rightarrow \cfrac { \Delta V }{ \Delta \theta  } =\cfrac { V\sin { \left( \Delta \theta /2 \right)  }  }{ \left( \Delta \theta /2 \right)  } \\ \Rightarrow \cfrac { dV }{ d\theta  } =\lim _{ \Delta \theta \longrightarrow 0 }{ \cfrac { V\sin { \left( \Delta \theta /2 \right)  }  }{ \left( \Delta \theta /2 \right)  } =V } $$

 $$\Rightarrow \cfrac { dV }{ d\theta  } =\cfrac { dv }{ d\theta  } \cfrac { d\theta  }{ dt } =V\times \cfrac { V }{ r } =\cfrac { { V }^{ 2 } }{ r } $$


1032003_1025325_ans_e44f70e5dc1c44d5b646f59773707c11.png

Physics

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