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?

Length of the rod is 0.043m

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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it