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# The period of physical pendulum

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A large bell is hung from a wooden beam so it can swing back and forth with negligible friction. The center of mass of the bell is 0.50m below the pivot, the bell has mass 38.0kg , and the moment of inertia of the bell about an axis at the pivot is 20.0kg*m^2. The clapper is a small, 1.8kg mass attached to one end of a slender rod that has length L and negligible mass. The other end of the rod is attached to the inside of the bell so it can swing freely about the same axis as the bell.

Question - What should be the length L of the clapper rod for the bell to ring silently-that is, for the period of oscillation for the bell to equal that for the clapper?

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https://brainmass.com/physics/mathematical-physics/period-physical-pendulum-214078

#### Solution Preview

The period of a physical pendulum (the bell) is given by:

T = 2*pi * sqrt (I/Mgl)

Where I is the moment of inertia about the axis, M is the mass of the bell and l is the ...

#### Solution Summary

The solution shows how to apply the physics of a physical pendulum and mathematical pendulum to the problem.

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