Kevin S. Huang

2014: Problem 5

Topic: Dynamics
Concepts: Circular motion, Newton’s laws, Statics

Solution:

Applying Newton’s 2nd law, we have in the vertical direction

N=mg

Since friction provides the centripetal acceleration,

f=\frac{mv^2}{r}

We choose the CM as the pivot point and balance torques. The weight of the unicyclist acts at the CM so doesn’t contribute to the torque. If the unicyclist has length $l$ then

N\left(\frac{l}{2}\sin\theta\right)=f\left(\frac{l}{2}\cos\theta\right)

Solving for $\theta$ ,

\tan\theta=\frac{f}{N}=\frac{mv^2/r}{mg}=\frac{v^2}{gr}=\frac{(10\,\mathrm{m/s})^2}{(10\,\mathrm{m/s^2})(30\,\mathrm{m})}=\frac{1}{3}

\theta=\arctan\left(\frac{1}{3}\right)=18.4^{\circ}

so the answer is C.

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Kevin S. Huang

2014: Problem 5

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2014: Problem 5

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