Question

A large bell is hung from a wooden beam so it can swing back andforth with negligible friction. The center of mass of the bell is0.60 m below the pivot, the bell has mass 34.0 kg, and the momentof inertia of the bell about an axis at the pivot is 18.0 kgm2 . The clapper is a small, 1.8-kg mass attached to oneend of a slender rod that has length L and negligible mass. theother end of the rod is attached to the inside of the bell so itcan swing freely about the same axis as the bell. What should bethe length L of the clapper rod for the bell to ring silently, thatis, for the period of oscillation for the bell to equal that forthe clapper?

Answer 1

Answer:

 L = 0.882 m

Explanation:

This problem can approximate the bell and the clapper as two pendulums, one a simple pendulum and the other a physical pendulum.

Let's start by analyzing the keystone that is a simple pendulum, with angular velocity

      w₁ = √ g / L

The angular velocity is related to the frequency and period

      w = 2π f = 2π / T

       T₁ = 2π √ L / g

We approach the bell to a physical pendulum, the angular velocity is

      w₂ = √ mg d / I

Where m is the mass of the field, d the distance to the pivot point and I the moment of inertia

      T₂ = 2π √ I / Mgd

In this case, they indicate that the two bodies have the same period

       T₁ = T₂

       2π √ L / g = 2π √ I / M g d

       L / g = I / M g d

       L = I / M d

Let's calculate

     L = 18.0 / (34.0 0.60)

     L = 0.882 m