Physics: AP Physics C Mechanics: Center of Mass and Moment of Inertia Worksheet

Question 1

Two particles lie on the x-axis. A 2.0 kg particle is at x = 1.0 m, and a 6.0 kg particle is at x = 5.0 m. Find the x-coordinate of the center of mass.

Accepted Answer

The center of mass is x_cm = (2.0 kg)(1.0 m) + (6.0 kg)(5.0 m) divided by 8.0 kg = 4.0 m. The center of mass is at x = 4.0 m.

Question 2

Three point masses form a triangle in the xy-plane: 1.0 kg at (0, 0), 2.0 kg at (3.0 m, 0), and 3.0 kg at (0, 4.0 m). Find the coordinates of the center of mass.

Accepted Answer

The total mass is 6.0 kg. The x-coordinate is x_cm = [(1.0)(0) + (2.0)(3.0) + (3.0)(0)]/6.0 = 1.0 m. The y-coordinate is y_cm = [(1.0)(0) + (2.0)(0) + (3.0)(4.0)]/6.0 = 2.0 m. The center of mass is at (1.0 m, 2.0 m).

Question 3

A uniform thin rod of length L lies along the x-axis from x = 0 to x = L. Use integration to find the x-coordinate of its center of mass.

Accepted Answer

For a uniform rod, dm = (M/L) dx. The center of mass is x_cm = (1/M) integral from 0 to L of x dm = (1/M) integral from 0 to L of x(M/L) dx = (1/L)(L^2/2) = L/2. The center of mass is at the midpoint of the rod.

Question 4

A nonuniform thin rod extends from x = 0 to x = L with linear density lambda(x) = kx, where k is a constant. Find the center of mass in terms of L.

Accepted Answer

The total mass is M = integral from 0 to L of kx dx = kL^2/2. The numerator is integral from 0 to L of x dm = integral from 0 to L of x(kx dx) = kL^3/3. Therefore x_cm = (kL^3/3)/(kL^2/2) = 2L/3. The center of mass is located at x = 2L/3.

Question 5

A 4.0 kg object moving at 3.0 m/s to the right sticks to a 2.0 kg object initially at rest on a frictionless track. Find the velocity of the center of mass before and after the collision.

Accepted Answer

Before the collision, v_cm = [(4.0 kg)(3.0 m/s) + (2.0 kg)(0)]/(6.0 kg) = 2.0 m/s to the right. With no external horizontal force, the center of mass velocity remains constant, so after the collision v_cm is also 2.0 m/s to the right.

Question 6

Four equal masses m are placed at the corners of a square of side length a. Find the moment of inertia of the system about an axis perpendicular to the square through its center.

Accepted Answer

Each mass is a distance r = a/sqrt(2) from the center of the square. The moment of inertia is I = sum mr^2 = 4m(a^2/2) = 2ma^2. The system has rotational inertia I = 2ma^2 about the central perpendicular axis.

Question 7

A thin hoop of mass M and radius R rotates about its central symmetry axis. Explain why its moment of inertia is MR^2.

Accepted Answer

For a thin hoop, every small mass element dm is the same distance R from the rotation axis. Therefore I = integral r^2 dm = integral R^2 dm = R^2 integral dm = MR^2. The moment of inertia is MR^2 because all of the mass is located at radius R.

Question 8

A uniform solid disk has mass M and radius R. Its moment of inertia about its central axis is I_cm = (1/2)MR^2. Use the parallel-axis theorem to find its moment of inertia about an axis perpendicular to the disk through a point on its rim.

Accepted Answer

The parallel-axis theorem gives I = I_cm + Md^2, where d is the distance between parallel axes. Here d = R, so I = (1/2)MR^2 + MR^2 = (3/2)MR^2. The moment of inertia about the rim axis is (3/2)MR^2.

Question 9

A uniform thin rod of mass M and length L rotates about an axis perpendicular to the rod through one end. Derive its moment of inertia using integration.

Accepted Answer

Let the rod extend from x = 0 to x = L with the axis at x = 0. For a uniform rod, dm = (M/L) dx. Then I = integral x^2 dm = integral from 0 to L of x^2(M/L) dx = (M/L)(L^3/3) = (1/3)ML^2. The moment of inertia is I = (1/3)ML^2.

Question 10

A uniform thin rod of mass 2.0 kg and length 1.5 m rotates about an axis perpendicular to the rod through its center. Find its moment of inertia.

Accepted Answer

For a uniform thin rod about its center, I = (1/12)ML^2. Substituting values gives I = (1/12)(2.0 kg)(1.5 m)^2 = 0.375 kg m^2. The moment of inertia is 0.38 kg m^2 to two significant figures.

Question 11

A composite object consists of a uniform thin rod of mass M and length L with a point mass M attached to one end. The object rotates about an axis perpendicular to the rod through the other end. Find the total moment of inertia.

Accepted Answer

The rod contributes I_rod = (1/3)ML^2 about an end. The point mass is a distance L from the axis, so it contributes I_point = ML^2. The total moment of inertia is I_total = (1/3)ML^2 + ML^2 = (4/3)ML^2.

Question 12

A rigid body consists of two point masses connected by a massless rod: 3.0 kg at x = 0 and 1.0 kg at x = 2.0 m. The body rotates about an axis perpendicular to the rod through its center of mass. Find the moment of inertia about this center-of-mass axis.

Accepted Answer

The center of mass is x_cm = [(3.0)(0) + (1.0)(2.0)]/4.0 = 0.50 m. The 3.0 kg mass is 0.50 m from the center of mass, and the 1.0 kg mass is 1.50 m from the center of mass. The moment of inertia is I = (3.0)(0.50)^2 + (1.0)(1.50)^2 = 0.75 + 2.25 = 3.0 kg m^2. The moment of inertia is 3.0 kg m^2.

Physics: AP Physics C Mechanics: Center of Mass and Moment of Inertia

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