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Ask experts Expert Question: A Particle is moving with a constant angular acceleration of 4 rad/s^2 in a circular path. At time
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JRK (0)

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A Particle is moving with a constant angular acceleration of 4 rad/s^2 in a circular path. At time t=0, particles was at rest. Find the time at which the magnitudes of centrepetal acceleration and tangential acceleration are equal.
    
Akshay (1466)

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Angular velocity after time t seconds = 4t rad/s

i.e. tangential velocity v = 4rt m/s where r is the radius of the circle .

So, the tangential acceleration = 4r m/s2

now, v2/r = 4r

i.e. 16rt2 = 4r

t = 0.5 second


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Transmigrator (494)

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Sorry for not writing the units.

Given, angular accn (aa) = 4

        So, at any instant, tangential acceleration = Radius * aa

                                                                                   = 4R

              Now after any time t, angular velocity = 0+4t = 4t

                                                   => Linear velocity(v) = Radius*Angular velocity = 4Rt

                                                   So, centripetal accn = v2 /R = 16Rt2

                                                          According to question, 16Rt2 = 4R => t = 1/2 sec

 


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