Kevin S. Huang

2012: Problem 24

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

Solution:

When the assembly is rotated, the spring forces on the mass provide the centripetal acceleration. Since the springs are stretched by distance $l$ , the spring forces have magnitude $kl$ .

Applying Newton’s 2nd law,

2kl\cos\theta=m\omega^2r

From the figure, we have $r\cos\theta=l$ so

2kl\cos\theta=m\omega^2\left(\frac{l}{\cos\theta}\right)

k=\frac{m\omega^2}{2\cos^2\theta}

Since we have an equilateral triangle, $\theta=\pi/6$ and $\cos\theta=\sqrt{3}/2$ . Finally,

k=\frac{2m\omega^2}{3}

so the answer is C.

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Kevin S. Huang

2012: Problem 24

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2012: Problem 24

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