Worked examples · Sheet 1

Moments of inertia and the parallel-axis theorem

Theory used: Moment of inertia Moments of inertia of uniform bodies Parallel-axis theorem (Steiner's theorem)

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Three point masses are fixed to a rigid rod of negligible mass $1.2 m$ long: two masses of $2.0 kg$ at the ends and a mass of $1.0 kg$ at the midpoint. Compute the moment of inertia of the system about the axis perpendicular to the rod through the midpoint $C$ and about the parallel axis through one end $A$ , and verify the second result with the parallel-axis theorem.

I_{C} = i \sum m_{i} R_{i}^{2} = 2.0 \cdot 0.6 0^{2} + 1.0 \cdot 0^{2} + 2.0 \cdot 0.6 0^{2} = 1.44 kg \cdot m^{2}

I_{A} = 2.0 \cdot 0^{2} + 1.0 \cdot 0.6 0^{2} + 2.0 \cdot 1. 2^{2} = 0.36 + 2.88 = 3.24 kg \cdot m^{2}

M = 2.0 + 1.0 + 2.0 = 5.0 kg, I_{C} + M d^{2} = 1.44 + 5.0 \cdot 0.6 0^{2} = 3.24 kg \cdot m^{2}

For point masses the moment of inertia is the sum of the products of each mass and the square of its distance from the axis. About the axis through the midpoint the two outer masses are at distance $0.60 m$ , while the central one lies on the axis and does not contribute. About the axis through the end $A$ the mass at $A$ does not contribute, the central one is at distance $0.60 m$ and the other at $1.2 m$ . The masses are arranged symmetrically with respect to the midpoint, so the center of mass is $C$ and the parallel-axis theorem applies with $d = 0.60 m$ ; the result coincides with the direct calculation. The moment of inertia about the end is more than twice as large, because the farther $2.0 kg$ mass is at twice the distance and contributes with the square of the distance.

Exercise 2

A uniform rod of mass $M = 3.0 kg$ and length $L = 0.80 m$ rotates about an axis perpendicular to the rod through a point $0.20 m$ from one end. Compute the moment of inertia about this axis.

I_{C} = \frac{1}{12} M L^{2} = \frac{1}{12} \cdot 3.0 \cdot 0.8 0^{2} = 0.16 kg \cdot m^{2}

d = \frac{L}{2} - 0.20 m = 0.40 - 0.20 = 0.20 m

I = I_{C} + M d^{2} = 0.16 + 3.0 \cdot 0.2 0^{2} = 0.16 + 0.12 = 0.28 kg \cdot m^{2}

check: \int_{- 0.20}^{0.60} \frac{M}{L} x^{2} d x = 3.75 \cdot \frac{0.6 0 ^{3} + 0.2 0 ^{3}}{3} = 0.28 kg \cdot m^{2}

The center of mass of the uniform rod is the midpoint, $0.40 m$ from each end, and the moment of inertia about the perpendicular axis through the center is read from the table of uniform bodies. The given axis is parallel to this one and is $0.20 m$ from the center, so by the parallel-axis theorem it is enough to add $M d^{2}$ . The check computes the same moment of inertia by integration: with the origin on the axis, the rod occupies the interval from $- 0.20 m$ to $0.60 m$ and the linear density is $\frac{M}{L} = 3.75 kg/m$ . The moment of inertia is $0.28 kg \cdot m^{2}$ , between the value about the center, $0.16 kg \cdot m^{2}$ , and that about the end, $\frac{1}{3} M L^{2} = 0.64 kg \cdot m^{2}$ , as it must be for an intermediate axis.

Exercise 3

A uniform disk of mass $M = 4.0 kg$ and radius $R = 0.25 m$ rotates about an axis perpendicular to the disk through a point of the rim. Compute the moment of inertia and compare it with that of a thin ring with the same mass and the same radius, about an analogous axis.

I_{disk} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} = \frac{3}{2} \cdot 4.0 \cdot 0.2 5^{2} = 0.375 kg \cdot m^{2}

I_{ring} = M R^{2} + M R^{2} = 2 M R^{2} = 2 \cdot 4.0 \cdot 0.2 5^{2} = 0.500 kg \cdot m^{2}

The axis passes through the rim and is parallel to the symmetry axis, which contains the center of mass; the distance between the two axes is the radius $R$ . By the parallel-axis theorem, $M R^{2}$ is added to the moment of inertia about the center, $\frac{1}{2} M R^{2}$ for the disk and $M R^{2}$ for the ring. The ring has the larger moment of inertia, because in the disk part of the mass is close to the center, while in the ring all the mass is at distance $R$ from the center.

Exercise 4

A pendulum consists of a uniform rod of mass $1.2 kg$ and length $L = 0.90 m$ , hinged at one end $O$ , and of a uniform disk of mass $2.0 kg$ and radius $0.10 m$ welded to the other end, with the center of the disk at the end of the rod. The disk lies in the plane of oscillation and the axis of rotation passes through $O$ , perpendicular to that plane. Compute the moment of inertia of the pendulum about the axis.

The pendulum seen along the axis of rotation through

O

: the rod

0.90 m

long and the disk of radius

0.10 m

with its center at the end

I_{rod} = \frac{1}{3} m_{a} L^{2} = \frac{1}{3} \cdot 1.2 \cdot 0.9 0^{2} = 0.324 kg \cdot m^{2}

I_{disk} = \frac{1}{2} m_{d} R^{2} + m_{d} L^{2} = \frac{1}{2} \cdot 2.0 \cdot 0.1 0^{2} + 2.0 \cdot 0.9 0^{2} = 0.010 + 1.62 = 1.63 kg \cdot m^{2}

I_{O} = I_{rod} + I_{disk} = 0.324 + 1.63 = 1.95 kg \cdot m^{2}

The moment of inertia is additive, so the contributions of the rod and of the disk about the same axis through $O$ are computed separately. For the rod the axis passes through one end, and the value $\frac{1}{3} m_{a} L^{2}$ is read from the table of uniform bodies. For the disk the axis is parallel to the symmetry axis and is at distance $L$ from the center, therefore the parallel-axis theorem applies. The moment of inertia of the pendulum is $1.95 kg \cdot m^{2}$ . As a check, treating the disk as a point mass at distance $L$ would give $1.944 kg \cdot m^{2}$ , so the term $\frac{1}{2} m_{d} R^{2}$ changes the result by about $0.5%$ , because the radius of the disk is small compared with the length of the rod.