Question

3. A roulette wheel with mass 71.0 kg and radius 40.0 cm can be approximated as a disk with a momentum of inertia = ½ mr. A) what is the torque on the wheel if it is spun with a force of 15.0 N on its outer edge? B) what is its angular acceleration? C) If the force is applied for 0.33 s, what is its final angular velocity if it starts from rest? D) What is the final tangential velocity at the outer edge of the wheel?

Answer 1

a)

Torque on the wheel = Force applied on its outer edge x radius of disk

$\tau= F*r=15.0*(0.40)=6.00\,N.m$

b)Torque = moment of inertia x angular acceleration

$\tau=I\alpha=\frac{1}{2}mr^2\alpha$

$\Rightarrow \,\alpha=\frac{2\tau}{mr^2}=\frac{2*6.00}{71.0*0.40^2}=1.06\,rad/s^2$

c)

Initial angular velocity $\omega_0=0\,rad/s$

Final angular velocity $\omega=\omega_0+\alpha t=0+1.06*0.33=0.349\,rad/s$

d)

Final tangential velocity of edge of wheel is $v=r\omega=0.40*0.35=0.139\,m/s$