Kevin S. Huang

2016: Problem 9

Topic: Energy
Concepts: Circular motion, Conservation of energy, Forces in mechanics

Solution:

If the bead leaves the sphere at angle $\theta$ , we can find its speed by conservation of energy. Setting the potential energy reference at the center of the sphere, we have

E_i=mgR

E_f=\frac{1}{2}mv^2+mgR\cos\theta

Since $E_i=E_f$ ,

mgR=\frac{1}{2}mv^2+mgR\cos\theta

v^2=2gR(1-\cos\theta)

To find $\theta$ , we note the bead is undergoing circular motion and apply Newton’s 2nd law,

mg\cos\theta-N=\frac{mv^2}{R}

At the instant the bead leaves the sphere, $N \rightarrow 0$ so

mg\cos\theta=\frac{mv^2}{R}

\cos\theta=\frac{v^2}{gR}

Plugging this into our earlier equation,

v^2=2gR(1-v^2/gR)=2gR-2v^2

3v^2=2gR

v=\sqrt{\frac{2gR}{3}}=\sqrt{\frac{gD}{3}}

using the fact that $D=2R$ . Thus, the answer is E.

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Kevin S. Huang

2016: Problem 9

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2016: Problem 9

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