Question

The angular position of a point on the rim of a rotating wheel is given by $\theta =4t-3t^2+t^3$, where $\theta$ is in radians and $t$ is in seconds. What is the instantaneous angular acceleration at $t=2\text{sec}$?

• A. $4{\text{rad/s}}^2$
• B. $2{\text{rad/s}}^2$
• C. $6{\text{rad/s}}^2$
• D. $3{\text{rad/s}}^2$

Answer 1

The correct option is C $6{\text{rad/s}}^2$

Given, $\theta =4t-3t^2+t^3$
$\omega =\frac{d\theta }{dt}$ & $\alpha =\frac{d\omega }{dt}$
$\Rightarrow \omega =\frac{d\theta }{dt}=4-6t+3t^2$
& $\alpha =\frac{d\omega }{dt}=\frac{d}{dt}(4-6t+3t^2)$
$=-6+6t$
At $t=2\text{s}$, $\alpha =-6+6\times 2$
$=6{\text{rad/s}}^2$