The drive chain in a bicycle is applying a torque of $0.850 \mathrm {~N} \cdot \mathrm { m }$ to the wheel of the bicycle. You can treat the wheel as a thin uniform hoop (or ring) with a mass of 0.750 kg and a radius of 33.0 cm. What is the angular acceleration of the wheel?
A) $20.8 \mathrm { rad } / \mathrm { s } 2$
B) $5.20 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $10.4 \mathrm { rad } / \mathrm { s } ^ { 2 }$
D) $3.43 \mathrm { rad } / \mathrm { s } ^ { 2 }$
E) $1.06 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Question

The drive chain in a bicycle is applying a torque of $0.850 \mathrm {~N} \cdot \mathrm { m }$ to the wheel of the bicycle. You can treat the wheel as a thin uniform hoop (or ring) with a mass of  0.750 kg and a radius of  33.0 cm. What is the angular acceleration of the wheel?
A) $20.8 \mathrm { rad } / \mathrm { s } 2$
B) $5.20 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $10.4 \mathrm { rad } / \mathrm { s } ^ { 2 }$
D) $3.43 \mathrm { rad } / \mathrm { s } ^ { 2 }$
E) $1.06 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Quizplus · Accepted Answer

The moment of inertia $I$ for a thin hoop is $I = m r^2$. With $m = 0.750 \, \text{kg}$ and $r = 0.33 \, \text{m}$, $I = 0.750 \times (0.33)^2 = 0.081675 \, \text{kg} \cdot \text{m}^2$. The angular acceleration $\alpha$ is given by $\alpha = \frac{\tau}{I} = \frac{0.850}{0.081675} \approx 10.4 \, \text{rad/s}^2$.

The Drive Chain in a Bicycle Is Applying a Torque 0.850 N⋅m0.850 \mathrm {~N} \cdot \mathrm { m }0.850 N⋅m

The Drive Chain in a Bicycle Is Applying a Torque $0.850 \mathrm {~N} \cdot \mathrm { m }$