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The linear speed of a particle moving in...

The linear speed of a particle moving in a circle of radius `R` varies with time as `v = v_0 - kt`, where `k` is a positive constant. At what time the magnitudes of angular velocity and angular acceleration will be equal ?

Text Solution

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We have `v = v_0 - kt` (given). Then, the tangential acceleration of the particle is
`a_t = (d v)/(d t) = k`
The angular acceleration of the particle is
`omega = (v)/( R) = (v_0 - kt)/( R)`
If `prop = omega`, we have `(k)/( R) = (v_0 - kt)/( R)`
Then, `t = (v_0 - k)/(k)`.
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