AP Physics 1: Algebra-Based 5.4 Rotational Inertia Questions

Analyse how mass distribution and rotation axis determine rotational inertia, then connect that inertia to measured torque and angular acceleration.

Exam points

explain how moving mass farther from the rotation axis increases inertia and changes angular speed
calculate rotational inertia about an off-centre pivot from torque and angular acceleration
decide whether rotational inertia changes by checking the mass distribution and rotation axis

Covered syllabus points

5.4.A—Describe the rotational inertia of a rigid system relative to a given axis of rotation 5.4.B—Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system’s…

[Maximum number: 1]

Figure for Question 1 — AP Physics 1: Algebra-Based

8 points; suggested time: 15-20 minutes A mass m is initially resting at the center of a rotating disk with rotational inertia $(1 / 2) M R^{2}$ . The entire system is freely rotating with an initial angular speed of $w_{0}$ . As time goes by, the mass is carefully moved at a constant speed outward from the center. This is done in such a way that no torque on the system is introduced.

If the mass m is about half the mass of the spinning disk, estimate the approximate angular speed of the two-mass system after the mass m has been pushed all the way to the edge in terms of $w_{0}$ .

Justify your estimate using qualitative reasoning beyond referencing equations.

4. (a) 1 point. Pushing the mass out to the edge will result in an increase

in the rotational inertia of the system of $m r^{2}$ or $(M / 2) r^{2}$ , which is

same as the original rotational inertia of the disk alone.

1 point. Since this is effectively doubling the rotational inertia, the

angular speed should be reduced to 1/2 of its original value.

Figure for Question 2 — AP Physics 1: Algebra-Based

A static, nonuniform wedge-shaped mass, m (with center of mass x as indicated), with a base length of L (shown below), is supported at two points. The first support point is at the bottom left corner of the wedge.

The second support point is located at a point $\frac{3}{4}$ of the length of the wedge away from that corner.

(a)

Support \#2 is removed. The wedge begins to fall by rotating around support \#1 with an angular acceleration of $2 \mathrm{rad} / \mathrm{s}^{2}$ . If the mass of the wedge is 3.8 kg and $L=75 \mathrm{~cm}$ , determine the rotational inertia of the nonuniform wedge about the top of support \#1.

(b) Torque $=I \alpha$

Since $F_{2}$ is gone, the torque is supplied only by the mass of the

object:

\begin{aligned} & \text { Torque }=F R \sin \theta \\ & \text { Torque }=m g(L / 3)(1) \end{aligned}

Substituting into the above equation:

\begin{aligned} m g L / 3 & =I \alpha=I(2) \\ I & =m g L / 6=(3.8)(9.8)(0.75) / 6=4.7 \mathrm{~kg} \cdot \mathrm{~m}^{2} \end{aligned}

Note that $I=m r^{2}$ cannot be used, as the object has mass spread out

over many different r values.

(b)

As the wedge continues to rotate clockwise but remains in contact with support \#1:

(i)

Figure for Question (i) — AP Physics 1: Algebra-Based

Does the rotational inertia of the wedge increase, decrease, or remain the same? Justify your response.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (i)