Question

The angular velocity of a point on the rim of a rotating wheel is given by $\omega =(4-6t+3t^2)\text{rad/s}$. What is the average angular acceleration for the time interval $t=2\text{s}$ to $t=4\text{s}$ ?

• A. $12{\text{rad/s}}^2$
• B. $6{\text{rad/s}}^2$
• C. $24{\text{rad/s}}^2$
• D. $3{\text{rad/s}}^2$

Answer 1

The correct option is A $12{\text{rad/s}}^2$

As we know,
Average angular acceleration $a_{\text{avg}}=\frac{\Delta \omega }{\Delta t}=\frac{{\omega }_f-{\omega }_i}{t_f-t_i}$
$=\frac{\omega (t=4\text{s})-\omega (t=2\text{s})}{4-2}$
From $\omega =(4-6t+3t^2)\text{rad/s}$
$\omega (t=2\text{s})=(4-6\times 2+3\times 2^2)$
$=4\text{rad/s}$
$\omega (t=4\text{s})=(4-6\times 4+3\times 4^2)$
$=28\text{rad/s}$

$a_{\text{avg}}=\frac{28\text{rad/s}-4\text{rad/s}}{4-2}$
$\therefore a_{\text{avg}}=12{\text{rad/s}}^2$