Question

A $250g$ thin hoop with a radius of $25.0$ centimeters is initially rotating at $15.0rev/s$. It takes $30.0$ revolutions for the hoop to roll to a stop under a constant angular acceleration.
If the hoop rolls without slipping, what is the angular acceleration of the hoop in $\frac{rad}{s^2}$?

• A. $-0.25\frac{rad}{s^2}$
• B. $-0.59\frac{rad}{s^2}$
• C. $-3.75\frac{rad}{s^2}$
• D. $-15.0\frac{rad}{s^2}$
• E. $-23.6\frac{rad}{s^2}$

Answer 1

The correct option is E $-23.6\frac{rad}{s^2}$

Initial angular velocity of rotation $w_i=2\pi {\nu }_1=2\pi \times 15=94.2$ rad/s

Final angular velocity $w_f=0$ rad/s

Angular displacement covered before coming to rest $\theta =2\pi \times 30=188.4$ radians

Using $w_f^2-w_i^2=2\alpha \theta$

$\therefore$ $0-(94.2{)}^2=2\alpha \times 188.4$ $⟹\alpha =-23.6$ $rad/s^2$