Kevin S. Huang

2016: Problem 23

Topic: Dynamics
Concepts: Circular motion, Forces in mechanics, Small angle approximation

Solution:

We analyze a small piece of the rubber band with mass $dM$ that subtends an angle $d\theta$ . The spring forces $T$ on its ends provide the centripetal acceleration. We have

2T\sin\left(\frac{d\theta}{2}\right)=dM \omega^2R’

Using the small-angle approximation $\sin\theta \approx \theta$ ,

Td\theta=dM \omega^2R’

T=\frac{dM}{d\theta}\omega^2R’

Since the rubber band has uniform density,

\frac{dM}{d\theta}=\frac{M}{2\pi}

T=\frac{M\omega^2R’}{2\pi}

The change in length of the spring is

\Delta L=2\pi R’-2\pi R

so the tension is

T=k\Delta L=2\pi k(R’-R)

Substituting this into the other equation,

2\pi kR’-2\pi kR=\frac{M\omega^2R’}{2\pi}

4\pi^2 kR’-M\omega^2R’=4\pi^2 kR

R’=\frac{4\pi^2 kR}{4\pi^2 k-M\omega^2}

so the answer is D.

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Kevin S. Huang

2016: Problem 23

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2016: Problem 23

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