A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M = 2.01 kg and length L = 0.19 m that is pivoted freely about one end, with a solid disk of mass 3M and a radius of L/4 attached to the free end of the rod.
a) What the moment of inertia of the pendulum about its pivot point?
0.3711 kg*m*m
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b) What is the period of the pendulum for small oscillations?
0.959 s
c) Determine the length L that gives a period of T = 2.49 s.
1.280 m
CAN YOU EXPLAIN PLEASE
Certainly! Let's break down each part of the problem and explain how to find the answers.
a) To find the moment of inertia of the pendulum about its pivot point, we need to consider the individual moments of inertia of the thin rod and the solid disk. The moment of inertia for a thin rod pivoted about one end is given by the formula: I = (1/3) * m * L^2, where m is the mass of the rod and L is its length.
In this case, the mass of the rod is given as M = 2.01 kg and the length is given as L = 0.19 m. Plugging these values into the formula, we have:
I_rod = (1/3) * 2.01 kg * (0.19 m)^2
= 0.04057 kg*m^2
The moment of inertia for a solid disk rotating about its center is given by the formula: I = (1/2) * m * R^2, where m is the mass of the disk and R is its radius.
In this case, the mass of the disk is given as 3M and the radius is given as L/4. Plugging in the values and simplifying, we get:
I_disk = (1/2) * 3M * [(L/4) * (L/4)]
= (3/8) * M * L^2
To find the total moment of inertia of the pendulum, we simply add the moments of inertia of the rod and the disk:
I_total = I_rod + I_disk
= 0.04057 kg*m^2 + (3/8) * M * L^2
= 0.04057 kg*m^2 + (3/8) * 2.01 kg * L^2
≈ 0.3711 kg*m^2
Therefore, the moment of inertia of the pendulum about its pivot point is approximately 0.3711 kg*m^2.
b) To find the period of the pendulum for small oscillations, we can use the formula:
T = 2π * √(I / (m * g * L))
where T is the period, I is the moment of inertia of the pendulum, 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, the moment of inertia of the pendulum has been calculated as 0.3711 kg*m^2, the mass of the pendulum is M + 3M = 4M = 8.04 kg (since the mass of the disk is 3 times the mass of the rod), and the length of the pendulum is given as L = 0.19 m. The acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in these values, we have:
T = 2π * √((0.3711 kg*m^2) / (8.04 kg * 9.8 m/s^2 * 0.19 m))
≈ 0.959 s
Therefore, the period of the pendulum for small oscillations is approximately 0.959 seconds.
c) To determine the length L that gives a period of T = 2.49 s, we rearrange the formula for the period and solve for L:
L = [(T / (2π))^2 * (m * g * I)]^0.25
Using the given period T = 2.49 s, the mass of the pendulum as 8.04 kg, the acceleration due to gravity as 9.8 m/s^2, and the calculated moment of inertia of the pendulum as 0.3711 kg*m^2, we can plug these values into the formula:
L = [((2.49 s) / (2π))^2 * (8.04 kg * 9.8 m/s^2 * 0.3711 kg*m^2)]^0.25
≈ 1.280 m
Therefore, the length L that gives a period of T = 2.49 s is approximately 1.280 meters.