IB Physics HL A.4.5 Angular motion equations Question Bank

Practise IB Physics HL A.4.5 by applying angular motion equations concepts to exam-style physics questions.

Exam points

identify the relevant physical quantity, model or diagram
apply the correct equation and units to the evidence
justify the result using conservation laws, forces or energy relationships

[Maximum number: 3]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

Figure for Question A.4.5 (HL)—Angular motion equations question 1 — IB Physics HL

A torque of 0.010 Nm brings the system to rest after a number of revolutions. For this case $R=0.50 \mathrm{~m}, M=0.70 \mathrm{~kg}$ and $v=2.1 \mathrm{~ms}^{-1}$ .

Calculate the number of revolutions made by the system before it comes to rest.

$\begin{aligned} \theta=\frac{\omega_{i}^{2}}{2 \alpha}<\text { from } \omega_{f}^{2}=\omega_{i}^{2}+2 \alpha \theta>\text { » } \theta<=\frac{v^{2}}{32 R^{2} \alpha}=\frac{2.1^{2}}{32 \times 0.5^{2} \times 0.043} \text { » }=12.8 \text { OR } 12.9 \text { «rad» } \text { number of rotations }<=\frac{12.9}{2 \pi}>=2.0 \text { revolutions } \end{aligned}$

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A.4.5 (HL)—Angular motion equations question 1