A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M = 7.01 kg and length L = 0.15 m that is pivoted freely about one end, with a solid sphere of the same mass, 7.01 kg, and a radius of L/2 centered about the free end of the rod.

Question

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M = 7.01 kg and length L = 0.15 m that is pivoted freely about one end, with a solid sphere of the same mass, 7.01 kg, and a radius of L/2 centered about the free end of the rod.

a) What the moment of inertia of the pendulum about its pivot point?

b) What is the period of the pendulum for small oscillations?

c) Determine the length L that gives a period of T = 2.51 s.
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Answer 1

To find the moment of inertia of the pendulum about its pivot point, you need to take into account the moments of inertia of both the thin rod and the solid sphere. The moment of inertia of a thin rod about one end is given by the equation:

I_rod = (1/3) * M * L^2

where M is the mass of the rod and L is the length of the rod. In this case, M = 7.01 kg and L = 0.15 m, so substituting these values into the equation gives:

I_rod = (1/3) * 7.01 kg * (0.15 m)^2 = 0.0526 kg·m^2

The moment of inertia of a solid sphere about an axis through its center is given by the equation:

I_sphere = (2/5) * M * R^2

where M is the mass of the sphere and R is its radius. In this case, M = 7.01 kg and R = L/2 = 0.15 m / 2 = 0.075 m, so substituting these values into the equation gives:

I_sphere = (2/5) * 7.01 kg * (0.075 m)^2 = 0.021 kg·m^2

To find the total moment of inertia of the pendulum, you add the moments of inertia of the rod and the sphere:

I_total = I_rod + I_sphere = 0.0526 kg·m^2 + 0.021 kg·m^2 = 0.0736 kg·m^2

So the moment of inertia of the pendulum about its pivot point is 0.0736 kg·m^2.

To find the period of the pendulum for small oscillations, you can use the equation for the period of a physical pendulum:

T = 2π * √(I_total / (M * g * L))

where T is the period, I_total is the moment of inertia, M is the mass of the pendulum, g is the acceleration due to gravity, and L is the length of the pendulum.

In this case, M = 7.01 kg, g ≈ 9.8 m/s^2, and I_total = 0.0736 kg·m^2. Let's assume we want to find T in seconds, so substituting these values into the equation gives:

T = 2π * √(0.0736 kg·m^2 / (7.01 kg * 9.8 m/s^2 * L))

Now, to determine the length L that gives a period of T = 2.51 s, we can rearrange the equation and solve for L:

L = (0.0736 kg·m^2 / (7.01 kg * 9.8 m/s^2 * T)^2) * (2π)^2

Substituting T = 2.51 s into this equation will give you the length L that corresponds to the desired period of the pendulum.