AP Physics 1 5.4.B: Rotational Inertia About an Off-Centre Axis

Practise AP Physics 1 5.4.B by calculating rotational inertia about an off-centre axis and explaining when it stays constant as a rigid body swings.

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)

(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)

(c) As the wedge continues to rotate clockwise but remains in contact

with support \#1:

(i)

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

remain the same? Justify your response.

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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