Kevin S. Huang

2009: Problem 25

Topic: Rigid Bodies
Concepts: Angular kinematics, Rotational Newton’s 2nd law

Solution:

Since both discs come to rest, they go to rest at the same time. From angular kinematics,

\omega-\alpha t=0

t=\frac{\omega}{\alpha}

so the quantity $\omega/\alpha$ is the same for both disks:

\frac{\omega_1}{\alpha_1}=\frac{\omega_2}{\alpha_2}

Let’s rewrite $\alpha$ in terms of $r$ . We have

\alpha=\frac{\tau}{I} \propto \frac{r}{mr^2} \propto \frac{r}{r^4}=\frac{1}{r^3}

using the fact that $\tau \propto r$ as the same force (friction) acts on both disks and also $m \propto r^2$ as both disks have the same density and thickness. Substituting into the earlier equation,

\omega_1r_1^3=\omega_2r_2^3

so the answer is D.

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Kevin S. Huang

2009: Problem 25

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2009: Problem 25

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