2022 A/L Physics Past Paper · Question 35 (English Medium)

Answer: (4) $\dfrac{3}{2}g$

Given information

What we know:

The rod is uniform, with mass $M$ and length $L$. It is hinged at $A$ and held level by a string at $B$.

The string is cut. At that first instant the rod is still at rest and starts to turn about $A$.

The moment of inertia about $A$ is $I = \dfrac{1}{3}ML^2$.

The question has no numbers. So we keep the letters and check with sample values $M = 2\ \text{kg}$, $L = 1\ \text{m}$ and $g = 10\ \text{m s}^{-2}$.

What the question asks

We want: the acceleration of the end $B$ at the instant the string is cut.

What each letter means:

$M$ ($\text{kg}$): the mass of the rod.

$L$ ($\text{m}$): the length of the rod.

$I$ ($\text{kg m}^2$): the moment of inertia of the rod about $A$.

$\tau$ ($\text{N m}$): the torque of the weight about $A$.

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

$a_B$ ($\text{m s}^{-2}$): the acceleration of the end $B$.

Relevant theory

What is happening:

While the string holds $B$, the rod stays level.

The string is cut. The end $A$ cannot fall, because the hinge holds it. So the rod starts to swing down about $A$, like a door swinging on its hinge.

Every point of the rod turns through the same angle. A point far from $A$ has to cover a bigger arc than a point near $A$. So the farther a point is from $A$, the faster it speeds up.

The end $B$ is the farthest point, so it speeds up the most. It even starts faster than a stone dropped beside it.

Moment of force F about the point 'O' $= F \times d$ where 'd' is the perpendicular distance from 'O' to the force.

The centre of gravity of a body is the point at which the resultant force of gravity or weight of the body acts.

$\tau = I\alpha$ ........................(2) Equation (2) is similar to Newton's 2nd law, F = m a

$\therefore a_t = r\alpha$ where $a_t$ the acceleration along the tangent.

Which forces turn the rod about $A$? The hinge pushes at $A$ itself. That force is at distance $0$ from $A$, so its moment about $A$ is zero. Only the weight $Mg$ turns the rod.

The rod is uniform, so its weight acts at the middle, $\dfrac{L}{2}$ from $A$.

$$\tau = Mg \times \dfrac{L}{2}$$

In words: the torque is the weight times its distance from $A$.

The torque and the angular acceleration are linked by $\tau = I\alpha$. So:

$$\alpha = \dfrac{\tau}{I} = \dfrac{MgL/2}{ML^2/3}$$

In words: the angular acceleration is the torque divided by the moment of inertia.

The end $B$ is at distance $L$ from $A$. At the first instant the rod is not yet moving, so $B$ starts to move straight down. Its acceleration is $a = r\alpha$ with $r = L$.

For example, hold a ruler level with one end pressed on the edge of a table. Keep that end fixed and let the other end go. The free end swings down, and it drops fastest of all.

The rod is like that ruler, and $B$ is its free end.

Required equations

$\tau = Mg \times \dfrac{L}{2}$

$\alpha = \dfrac{\tau}{I} = \dfrac{3g}{2L}$

$a_B = L\alpha = \dfrac{3}{2}g$

Applying it to this question

$\text{kg} \times \text{m s}^{-2} \times \text{m} = \text{N m}$ is a torque. Dividing by $\text{kg m}^2$ gives $\text{s}^{-2}$, which is $\text{rad s}^{-2}$. Multiplying by $\text{m}$ gives $\text{m s}^{-2}$, an acceleration.

$10$

$\tau = 2 \times 10 \times 0.5$

$0.667$

$I = \dfrac{1}{3} \times 2 \times 1^2$

$15$

$\alpha = \dfrac{10}{0.667}$

$a_B = L\alpha = 1 \times 15$

Reasoning and calculation

With $g = 10\ \text{m s}^{-2}$, $\dfrac{3}{2}g = 15\ \text{m s}^{-2}$. So $a_B = \dfrac{3}{2}g$.

That is option (4).

Checks

$1.5 * g$

Does it make sense?

The middle of the rod has half the distance from $A$, so it starts with $\dfrac{3}{4}g$. The end $B$ has twice that, $\dfrac{3}{2}g$.

A stone in free fall has the acceleration $g$. Here $B$ gets $\dfrac{3}{2}g$, which is more than $g$. That is possible because $B$ is the point of the rod farthest from the hinge and the rod is turning about $A$. The value comes from the torque and the moment of inertia, and it holds only at this first instant. The rod keeps turning, and then the torque and the acceleration change.

Conclusion

$a_B = \dfrac{3}{2}g$

The end $B$ starts with $a_B = \dfrac{3}{2}g$.

The answer is (4).

Take moments about the hinge. The hinge force is at the hinge, so it has no moment about it, and only the weight is left.

Find $\alpha = \dfrac{\tau}{I}$ first, then multiply by the distance to get the linear acceleration of the point you are asked about.

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