A particle moves along a circle of radius $20/π$ m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has began the tangential acceleration is ??

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Answer 1 · 2018-01-13T13:58:26

The tangential acceleration is $=40ms^-2$

Assuming that the particle starts from rest

The initial angular velocity is $omega_0=0rads^-1$

The radius of the circle is $r=20/pim$

The final velocity of the particle is $v=80ms^-1$

The final angular velocity is

$omega=v/r=80/(20/pi)=4pirads^-1$

The angle is $theta=2*2pi=4pi$

Applying the equation

$omega^2=omega_0^2+2alphatheta$

Where $alpha$ is the angular acceleration

$alpha=(omega^2-omega_0^2)/(2theta)$

$=((4pi)^2-0)/(2*4pi)=(16pi^2)/(8pi)=(2pi)rads^-2$

The tangential acceleration is

$a_T=alpha*r=2pi*20/pi=40ms^-2$

A particle moves along a circle of radius 20/π20/π m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has began the tangential acceleration is ??