Physics: AP Physics C Mechanics: Center of Mass and Moment of Inertia
Calculating centers of mass and rotational inertia for systems and continuous bodies
Physics: AP Physics C Mechanics: Center of Mass and Moment of Inertia
Calculating centers of mass and rotational inertia for systems and continuous bodies
Physics - Grade 9-12
- 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.
Use x_cm = sum(m_i x_i) divided by sum(m_i).
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. - 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.
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). - 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.
Write dm in terms of the constant linear density lambda = M/L.
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. - 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.
First find the total mass by integrating lambda(x) over the length of the rod.
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. - 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.
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. - 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.
Find the distance from the center of the square to one corner, then multiply the contribution by four.
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. - 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.
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. - 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.
The new axis is parallel to the central axis and is one radius away from it.
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. - 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.
Each mass element is a distance x from the rotation axis.
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. - 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.
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. - 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.
Add the rotational inertia of the rod and the rotational inertia of the point mass.
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. - 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.
First locate the center of mass, then use each mass's distance from that point.
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.