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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 0.50 m below the pivot, the bell has mass 32.0 kg , and the moment of inertia of the bell about an axis at the pivot is 20.0 kg⋅m2 . The clapper is a small, 1.8 kg 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 bel

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Answer:

Tp = 2.244 s

L = 1.25 m

Step-by-step explanation:

Given :

mb = 32 kg , mc = 1.8 kg , d = 0.50 m , I = 20.0 kg / m²

Now to determine the length can use the equation

L = g * ( Ts / 2 π )

Also knowing that amplitude of the oscillation

Ts = Tp

To determine Tp

Tp = 2 π * √ I / (mb * g * d)

Tp = 2 π * √ 20.0 kg * m² / 32.0 kg * 9.8 m / s² * 0.50 m

Tp = 2.244 s

Finally replacing

L = g * ( Ts / 2 π )²

L = 9.8 m/s² * ( 2.244 s / 2π )²

L = 1.25 m

User JohnN
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