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A bob of mass m is attached to a string of length L and is released from rest when the string is horizontal. At what angle alpha (measured from the vertical) during the subsequent swing is the net acceleration of the bob directed purely horizontally?

1. 0
2. 90 deg
3. cos⁻¹(1/sqrt(3))
4. sin⁻¹(1/sqrt(3))

Correct answer: cos⁻¹(1/sqrt(3))

Solution

The vertical component of the total acceleration is 2g*cos²(alpha) - g*sin²(alpha); setting this to zero gives tan²(alpha) = 2, so cos(alpha) = 1/sqrt(3), i.e., alpha = cos⁻¹(1/sqrt(3)).

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