Question

A solid body rotates about a stationary axis according to the relation $\theta =24t-2t^3$. Find the angular acceleration at the moment when body stops. (Assume SI units)

• A. $-12{\text{rad/s}}^2$
• B. $12{\text{rad/s}}^2$
• C. $-24{\text{rad/s}}^2$
• D. $24{\text{rad/s}}^2$

Answer 1

The correct option is C $-24{\text{rad/s}}^2$

It is given that, $\theta =24t-2t^3$

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 $(\omega =0)$

$\omega =\frac{d\theta }{dt}=24-6t^2=0$
at $t=2\text{sec}$, the body comes to rest.
now we know that angular acceleration refers to time rate of change of angular velocity

$\alpha =\frac{d\omega }{dt}=0-12t=-12t$

Hence the angular acceleration when the body stops is
$\alpha {|}_{t=2}$ $=-12\times 2=-24{\text{rad/s}}^2$.
as the body is slowing down the negative sign comes and its retardation.