Question

The instantaneous angular position of a particle undergoing a circular motion is given by the equation $\theta \left(t\right)=4t^3-3t^2$. Find the time when the angular acceleration of the particle becomes zero.

• A. $0.50\text{s}$
• B. $0.25\text{s}$
• C. $1\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is B $0.25\text{s}$

As we know that the angular acceleration of the particle is given by $\alpha =\frac{d^2\theta }{d{t}^2}$
We have, $\frac{d\theta }{dt}=\frac{d(4t^3-3t^2)}{dt}=12t^2-6t$
also, $\frac{d^2\theta }{d{t}^2}=\frac{d(12t^2-6t)}{d{t}^2}=24t-6$
Thus, the time when $\alpha$ will be zero is given as
$24t-6=0\Rightarrow t=\frac{6}{24}=0.25\text{s}$