Question

A 1.80- kgkg monkey wrench is pivoted 0.250 mm from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s.

A) What is the moment of inertia of the wrench about an axis through the pivot?
B) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

Answer 1

The moment of inertia of the wrench about an axis through the pivot is 1.13 x 10^-6 kg m^2. The angular speed of the wrench as it passes through the equilibrium position is 4.95 rad/s.

Step-by-step explanation:

To calculate the moment of inertia of the wrench about an axis through the pivot, we can use the formula:
I = mL^2, where m is the mass of the object and L is the distance from the axis of rotation. In this case, the mass of the wrench is 1.80 kg and the distance from the axis of rotation is 0.250 mm (or 0.250 x 10^-3 m). Plugging these values into the formula, we get:

I = (1.80 kg)(0.250 x 10^-3 m)^2 = 1.13 x 10^-6 kg m^2

To find the angular speed of the wrench as it passes through the equilibrium position, we can use the formula:
ω = √(g/L), where g is the acceleration due to gravity (9.8 m/s^2) and L is the distance from the axis of rotation. In this case, the distance from the axis of rotation is 0.400 rad. Plugging these values into the formula, we get:

ω = √(9.8 m/s^2 / 0.400 rad) = √24.5 rad/s = 4.95 rad/s