2021 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (5) $mg > T_1 > T_2$

Given information

We know:

The block has mass $M$. The table is frictionless.

The string is light and inextensible. It moves with the block, round the rim of the pulley, and with $m$.

The pulley has a moment of inertia $I$ and radius $r$. It turns as the string moves.

What the question asks

We want to compare $mg$, $T_1$ and $T_2$.

The string cannot stretch. So the block, the mass $m$ and the rim of the pulley all move together.

They share one acceleration $a$, and $a = r\alpha$ for the rim.

Relevant theory

According to Newton's second law resultant force is proportional to the rate of change of momentum.

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.

Required equations

$a = r\alpha$

$T_2 = Ma$

$mg - T_1 = ma$

$(T_1 - T_2)\,r = I\alpha$

Applying it to this question

Apply the law to each body.

The block: only $T_2$ pulls it along the table. So $T_2 = Ma$.

The mass $m$: the weight $mg$ pulls down and $T_1$ pulls up. So $mg - T_1 = ma$.

The pulley: $T_1$ turns it one way and $T_2$ turns it back. So $(T_1 - T_2)\,r = I\alpha$.

$mg = \left(M + m + \dfrac{I}{r^2}\right)a$

$2$

$a = \dfrac{1 \times 10}{2 + 1 + 0.5/0.25} = 2$

$4$

$T_2 = Ma = 2 \times 2 = 4$

$8$

$T_1 = m(g - a) = 1 \times (10 - 2) = 8$

Reasoning and calculation

Signs first. Take the way each body moves as positive.

The block moves toward the pulley, the mass $m$ moves down, and the pulley turns the way $T_1$ pulls it.

Then $T_2$ pulls the block forward, $T_1$ pulls the pulley forward, and $T_2$ holds the pulley back.

Put $\alpha = \dfrac{a}{r}$ in the pulley equation:

$$T_1 - T_2 = \dfrac{I a}{r^2}$$

The pulley accelerates, so $a > 0$. Then $T_1 - T_2 > 0$, and $T_1 > T_2$.

From the hanging mass:

$$mg - T_1 = ma > 0$$

So $mg > T_1$.

Together: $mg > T_1 > T_2$. Option (5).

Checks

Limiting case. If the pulley had no moment of inertia, $I \to 0$.

Then $T_1 - T_2 = 0$ and $T_1 = T_2$. This is option (2), the usual light pulley.

The question says the pulley accelerates, so $I$ is not zero, and $T_1 > T_2$.

$m * g / (M + m + I / (r * r))$

Conclusion

$mg > T_1 > T_2$

The pulley turns faster, so $T_1 > T_2$. The mass $m$ accelerates down, so $mg > T_1$.

The answer is (5).

A pulley with a moment of inertia needs a net torque to turn faster.

So the two tensions on its sides are not equal. The side that pulls the pulley round has the larger tension.

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