A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M = 1.61 kg and length L = 0.153 m that is pivoted freely about one end, with a solid disk of mass 3M and a radius of L4𝐿4 attached to the free end of the rod. What the moment of inertia of the pendulum about its pivot point?
Question
A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M = 1.61 kg and length L = 0.153 m that is pivoted freely about one end, with a solid disk of mass 3M and a radius of L4𝐿4 attached to the free end of the rod. What the moment of inertia of the pendulum about its pivot point?
Solution
The moment of inertia of the pendulum about its pivot point can be calculated by adding the moment of inertia of the rod and the moment of inertia of the disk.
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The moment of inertia of a rod pivoted about one end is given by the formula I_rod = 1/3 * M * L^2. Substituting the given values, we get:
I_rod = 1/3 * 1.61 kg * (0.153 m)^2 = 0.012 kg*m^2
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The moment of inertia of a solid disk about an axis through its center is given by the formula I_disk = 1/2 * m * r^2. However, since the disk is not pivoted about its center, we need to use the parallel axis theorem to find the moment of inertia about the pivot point. The parallel axis theorem states that I = I_cm + m*d^2, where I_cm is the moment of inertia about the center of mass, m is the mass, and d is the distance from the center of mass to the new axis (which is the length of the rod in this case).
Substituting the given values, we get:
I_disk = 1/2 * 31.61 kg * (0.153 m/4)^2 + 31.61 kg * (0.153 m)^2 = 0.020 kg*m^2
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Adding the two moments of inertia together, we get:
I_total = I_rod + I_disk = 0.012 kgm^2 + 0.020 kgm^2 = 0.032 kg*m^2
So, the moment of inertia of the pendulum about its pivot point is 0.032 kg*m^2.
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