2023 A/L Physics Past Paper · Question 50 (English Medium)

Answer: (4) $\dfrac{g}{4r}$

What the question asks

We want: the angular acceleration $\alpha$ of the pulley.

What each letter means:

$M$ ($\text{kg}$): the mass of each block, and of the pulley.

$r$, $\dfrac{r}{2}$ ($\text{m}$): the radii of the two parts. In the figure the right string comes off the big part at $r$, and the left string off the small part at $\dfrac{r}{2}$.

$I$ ($\text{kg m}^2$): the moment of inertia of the pulley, $\dfrac{3}{4}Mr^2$.

$T_1$, $T_2$ ($\text{N}$): the tensions in the right and the left string.

$\alpha$ ($\text{rad s}^{-2}$): the angular acceleration of the pulley.

Relevant theory

In translatory motion, force is defined as the rate of change of linear momentum. Likewise in rotational motion, torque or moment of the force is defined as the rate of change of angular momentum.

Equation (2) is similar to Newton's 2nd law, F = m a

Newton's first law states that a resultant force can change the state of motion of an object. Newton's second law states about this resultant force.

The first quote says that in rotation, torque does what force does in straight-line motion. A force $F$ at a radius $r$ gives the torque $\tau = Fr$.

The second quote says that $\tau = I\alpha$ is like Newton's second law: torque, moment of inertia and angular acceleration take the places of force, mass and acceleration.

The third quote gives Newton's second law, $F = ma$, for each block.

For example, two children of the same weight sit on a seesaw. One sits at the far end, the other halfway in. The far one goes down, because the same weight turns more at a longer distance.

Required equations

$Mg - T_1 = M r\alpha$

$T_2 - Mg = M \dfrac{r}{2}\alpha$

$T_1 r - T_2 \dfrac{r}{2} = \dfrac{3}{4}M r^2 \alpha$

$\dfrac{Mgr}{2} = 2 M r^2 \alpha$

$\alpha = \dfrac{g}{4r}$

Reasoning and calculation

Which way it turns:

The right block pulls at $r$, the left block at $\dfrac{r}{2}$. The same weight at a larger radius gives a larger torque, so the pulley turns clockwise. The right block goes down, the left block goes up.

The strings do not stretch, so each block moves with the rim of its part: the right block with $a_1 = r\alpha$, the left block with $a_2 = \dfrac{r}{2}\alpha$.

Use $F = ma$ for each block to find the tensions.

Right block: $Mg - T_1 = Mr\alpha$, so $T_1 = Mg - Mr\alpha$.

Left block: $T_2 - Mg = M\dfrac{r}{2}\alpha$, so $T_2 = Mg + \dfrac{1}{2}Mr\alpha$.

$Mgr - Mr^2\alpha$

$T_1 r = Mgr - Mr^2\alpha$

$\dfrac{Mgr}{2} + \dfrac{1}{4}Mr^2\alpha$

$T_2 \dfrac{r}{2} = \dfrac{Mgr}{2} + \dfrac{1}{4}Mr^2\alpha$

$\dfrac{Mgr}{2} = 2Mr^2\alpha$

$\dfrac{Mgr}{2} - \dfrac{5}{4}Mr^2\alpha = \dfrac{3}{4}Mr^2\alpha$

$\dfrac{g}{4r}$

$\alpha = \dfrac{Mgr}{2 \times 2Mr^2}$

$2.5$

$\alpha = \dfrac{10}{4 \times 1}$

$7.5$

$T_1 = 10 - 2.5$

$11.25$

$T_2 = 10 + 1.25$

$1.875$

$7.5 \times 1 - 11.25 \times 0.5$

$0.75 \times 2.5$

$(M * g * r / 2) / (0.75 * M * r^2 + M * r^2 + M * (r / 2)^2)$

Reading the options

Now compare $\dfrac{g}{4r}$ with the five options.

Option (4) is $\dfrac{g}{4r}$. It matches.

Option (1) says the pulley stays still. Options (2), (3) and (5) come from a wrong total inertia.

Checks

Check the units. $\dfrac{g}{r}$ is $\dfrac{\text{m s}^{-2}}{\text{m}} = \text{s}^{-2}$, which is $\text{rad s}^{-2}$, the unit of an angular acceleration.

Does it make sense?

If the two parts had the same radius, the torques would balance and $\alpha$ would be $0$. Our radii differ, so $\alpha$ is not zero.

If the pulley had no inertia, the total would be $\dfrac{5}{4}Mr^2$ and $\alpha = \dfrac{2g}{5r}$. The real pulley adds $\dfrac{3}{4}Mr^2$, so $\alpha$ is smaller: $\dfrac{g}{4r}$ is less than $\dfrac{2g}{5r}$, as it should be.

Conclusion

$\alpha = \dfrac{g}{4r}$

The net torque $\dfrac{Mgr}{2}$ turns a total inertia of $2Mr^2$, so the angular acceleration of the pulley is $\alpha = \dfrac{g}{4r}$.

The answer is (4).

The quotes give torque, $\tau = I\alpha$ and Newton's second law. That the right string is on the part of radius $r$, and the block accelerations $r\alpha$ and $\dfrac{r}{2}\alpha$, are our reading of the figure and our reasoning.

Teacher's note

For a pulley with hanging blocks, a quick route: $$angular\,acceleration = net\,torque$$ $\div$ total inertia. Each block at radius $R$ adds $MR^2$ to the inertia of the pulley.

Exam tip