AP Physics 1 5.4 Rotational Inertia

Practise AP Physics 1 5.4 questions by comparing mass distributions, summing point inertias and applying the parallel-axis idea.

Exam points

identify the governing concept or relationship in the problem
justify the result with equations, diagrams or contextual evidence

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 5.4 Rotational Inertia 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.

(a) 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 5.4 Rotational Inertia 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)

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

(b) 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)

with support \#1:

(i)

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

(ii) Does the rotational inertia of the wedge increase, decrease, or

remain the same? Justify your response.

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Covered syllabus points

5.4 Rotational Inertia question 1

5.4 Rotational Inertia question 2

Question (a)

Question (b)

Question (i)