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Question

A solid body rotates about a stationary axis according to the relation θ=24t2t3. Find the angular acceleration at the moment when body stops. (Assume SI units)
  1. 12 rad/s2
  2. 12 rad/s2 
  3. 24 rad/s2
  4. 24 rad/s2 


Solution

The correct option is C 24 rad/s2
It is given that,   θ=24t2t3

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=246t2=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=012t=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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