AP Physics C Mechanics 5.4: Parallel-Axis Theorem

Use the parallel-axis theorem to relate rotational inertia about a center-of-mass axis to inertia about a parallel shifted axis.

[Maximum number: 3]

A uniform rod of length L and mass m is attached to a pivot on a vertical pole, as shown in Figure 1. There is negligible friction between the rod and the pivot. A horizontal string connects Point Q on the rod to the pole. The rod makes an angle $\theta$ with the pole. A block of mass 3 m hangs from the rod at Point P. The center of mass of the rod is located at Point C.

A nonuniform rod is now attached to the pivot, as shown in Figure 3. There is negligible friction between the nonuniform rod and the pivot. The rod has a length of 1.2 m and a linear mass density $\lambda(x)=A+B x$ , where x is the distance from the pivot, $A=6.0 \mathrm{~kg} / \mathrm{m}$ , and $B=10.0 \mathrm{~kg} / \mathrm{m}^{2}$ .

Calculate the rotational inertia of the rod about the pivot.

(ii) For a correct substitution of $\lambda$ into an integral expression of rotational inertia 1 point
Example Response

I=\int_{0}^{1.2 \mathrm{~m}}(A+B x) x^{2} d x

For a correct integration 1 point
Example Response

I=\left.\left(\frac{A x^{3}}{3}+\frac{B x^{4}}{4}\right)\right|_{0} ^{1.2 \mathrm{~m}}

For a correct numeric answer with correct units 1 point
Example Response

I=8.64 \mathrm{~kg} \cdot \mathrm{~m}^{2}

Example Solution

\begin{aligned} & I=\int r^{2} d m \quad d m=\lambda d r \text { and } r=x \\ & I=\int_{0}^{1.2 \mathrm{~m}} \lambda x^{2} d x \\ & I=\int_{0}^{1.2 \mathrm{~m}}(A+B x) x^{2} d x \\ & I=\left.\left(\frac{A x^{3}}{3}+\frac{B x^{4}}{4}\right)\right|_{0} ^{1.2 \mathrm{~m}} \\ & I=\frac{(6.0 \mathrm{~kg} / \mathrm{m})(1.2 \mathrm{~m})^{3}}{3}+\frac{\left(10.0 \mathrm{~kg} / \mathrm{m}^{2}\right)(1.2 \mathrm{~m})^{4}}{4} \\ & I=8.64 \mathrm{~kg} \cdot \mathrm{~m}^{2} \end{aligned}

Total for part (d) for question 315 points

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5.4.B—Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system’s… question 1