Question

The tangential velocity of a particle performing a U.C.M. of radius π m makes 'p' revolutions in
time t,is
(a)2π^2p/t m/s
(b)2π^2/pt m/s
(c)π^2p/t. m/s
(d)2pπ/t. m/s​

Answer 1

Answer:

Given:

Particle performing Uniform Circular Motion of radius π metres and p revolutions in time t.

To find:

Tangential velocity of that particle.

Calculation :

Circumference of the circular track :

C = 2πR

=> C = 2π (π)

=> C = 2π² metres.

Now p revolutions corresponding to circumference as :

C" = C × p

=> C " = 2π² × p

=> C " = 2pπ² metres.

This distance is traversed in time period t ,

So velocity = distance/time

=> Velocity = 2π²p/t

So final answer :

$\boxed{ \large{ \red{velocity = 2 {\pi}^{2} ( \dfrac{p}{t})}}}$

Additional information :

1. Since the Particle is undergoing UCM , it means that the particle doesn't experience any tangential acceleration. It has only Centripetal acceleration.

2. As the result , the angle between net acceleration and tangential Velocity is 90°.

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Given:

Particle performing Uniform Circular Motion of radius π metres and p revolutions in time t.

To find:

Tangential velocity of that particle.

Calculation :

Circumference of the circular track :

C = 2πR

=> C = 2π (π)

=> C = 2π² metres.

Now p revolutions corresponding to circumference as :

C" = C × p

=> C " = 2π² × p

=> C " = 2pπ² metres.

This distance is traversed in time period t ,

So velocity = distance/time

=> Velocity = 2π²p/t

So final answer :

\boxed{ \large{ \red{velocity = 2 {\pi}^{2} ( \dfrac{p}{t})}}}

Additional information :

1. Since the Particle is undergoing UCM , it means that the particle doesn't experience any tangential acceleration. It has only Centripetal acceleration.

2. As the result , the angle between net acceleration and tangential Velocity is 90°.