Question

A $5Nm$ torque acts on a thin hoop with a mass of $500g$ and a radius of $15cm$ for $10s$.
If the hoop starts from rest, what is the new angular velocity of the disc?

• A. $666\frac{rad}{s}$
• B. $1,300\frac{rad}{s}$
• C. $2,000\frac{rad}{s}$
• D. $4,444\frac{rad}{s}$
• E. $8,888\frac{rad}{s}$

Answer 1

Answer: (D)

$4,444\frac{rad}{s}$

Given : $m=0.5$ kg $r=0.15$ m $t=10$ s $\tau =5$ Nm $w_i=0$ rad/s

Moment of inertia of the hoop $I=m{r}^2=0.5\times (0.15{)}^2=0.01125$ $kg{m}^2$

Thus angular acceleration of the hoop $\alpha =\frac{\tau }{I}=\frac{5}{0.01125}=444.44$ $rad/s^2$

Using $w_f=w_i+\alpha t$

$\therefore$ $w_f=0+444.44\times 10\approx 4444$ $rad/s$