Question

A solid body rotates about a stationary axis according to the relation θ=24t−2t3. Find the angular acceleration at the moment when body stops. (Assume SI units)−12 rad/s212 rad/s2 −24 rad/s224 rad/s2

Solution

The correct option is C −24 rad/s2It is given that,   θ=24t−2t3 We know that angular velocity is the rate of change of angular position of a rotating body. When the body comes to rest its angular velocity is (ω=0) ω=dθdt=24−6t2=0 at  t=2 sec, the body comes to rest. now we know that angular acceleration refers to time rate of change of angular velocity  α=dωdt=0−12t=−12t Hence the angular acceleration when the body stops is  α|t=2  =−12×2=−24 rad/s2. as the  body is slowing down the negative sign comes and its  retardation.

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