Question

the same force ( f ) is applied to the edge of two hoops as the drawing below shows. the hoops have the same mass, whereas the radius of the larger hoop is twice the radius of the smaller one. the entire mass of each hoop is concentrated at its rim, so the moment of inertia is equal to ( mr^{2} ), where ( m ) is the mass of the hoop, and ( r ) is the radius. which hoop has the greater angular acceleration, and how many times as great is it compared to the other hoop?

the larger hoop; two times as great.

the larger hoop; four times as great.

the smaller hoop; two times as great.

the smaller hoop; four times as great.

Explanation:

Step1: Recall the formula for torque and angular acceleration

The torque formula is \(\tau = rF\sin\theta\). Here \(\theta = 90^{\circ}\), so \(\tau=rF\). Also, from Newton's second law for rotation \(\tau = I\alpha\), where \(\alpha=\frac{\tau}{I}\) is the angular acceleration, \(I\) is the moment of inertia.

Step2: Calculate the angular acceleration for the smaller hoop

For the smaller hoop with radius \(R\), \(I_{s}=MR^{2}\) (moment of inertia) and \(\tau_{s}=RF\) (torque). Then \(\alpha_{s}=\frac{\tau_{s}}{I_{s}}=\frac{RF}{MR^{2}}=\frac{F}{MR}\)

Step3: Calculate the angular acceleration for the larger hoop

For the larger hoop with radius \(2R\), \(I_{l}=M(2R)^{2} = 4MR^{2}\) (moment of inertia) and \(\tau_{l}=(2R)F\) (torque). Then \(\alpha_{l}=\frac{\tau_{l}}{I_{l}}=\frac{2RF}{4MR^{2}}=\frac{F}{2MR}\)

Step4: Compare the angular accelerations

We find the ratio \(\frac{\alpha_{s}}{\alpha_{l}}=\frac{\frac{F}{MR}}{\frac{F}{2MR}} = 2\)

Answer:

The smaller hoop; two times as great.