Question

A large, 36.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell’s center of mass is 0.55 m below the pivot. The bell’s moment of inertia about an axis at the pivot is 36.0 kg ·m2 . The clapper is a small, 2.8 kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. 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 of the clapper?

Answer 1

Length of the rod is 0.043m

Step-by-step explanation:

Mass of the bell (M) = 36kg

Distance of the center of mass from pivot = 0.55m

Moment of inertia (I) = 36.0kgm²

Mass of clipper (m) = 2.8kg

Length of the bell to ring = L

The period of the pendulum with small amplitude of oscillation is

T = 2π √(I / mgd)

Where g = acceleration due to gravity = 9.8 m/s²

T = 2π √(36 / 36*9.8*0.55)

T = √(0.1855)

T = 0.43s

The period of the pendulum is 0.43s

To find the length of the Clapper rod for which the bell ring slightly,

T = 2π√(L / g)

T² = 4π²L / g

T² *g = 4π²L

L = (T² * g) / 4π²

L = (9.8* 0.43²) / 4π²

L = 1.7287 / 39.478

L = 0.043m

The length of the Clapper rod for the bell to ring slightly is 0.043m