2020 A/L Physics Past Paper · Question 05 (English Medium)

Answer: (1) $My^2$

What the question asks

What is asked: the moment of inertia of rod $AB$ about the axis $PQ$, parallel to it at distance $y$ — an expression in $M$, $L$, $y$.

Relevant theory

If a system of particles of masses $m_1, m_2, m_3, \ldots, m_n$ are situated at perpendicular distances $r_1, r_2, r_3, \ldots, r_n$ from the axis l, Then the moment of inertia of the system of particles $I_l$ is given by

$$I_l = \sum_{i=1}^{n} m_ir_i^2$$

"I" depends on the position of the axis of rotation and how the mass of the body is distributed about it.

Required equations

$I = \sum m_i r_i^2$

$I = \sum m_i y^2$

$I = y^2 \sum m_i$

$I = My^2$

Reasoning and calculation

Symbols:

$m_i$ — the mass of a small particle of the rod.

$r_i$ — the perpendicular distance of that particle from the axis $PQ$.

$M = \sum m_i$ — the total mass of the rod; $L$ — the length of the rod; $y$ — the distance between the rod and $PQ$.

Removing the in-between symbols: since the rod is thin and parallel to $PQ$, the particle at $A$, the one in the middle, and the one at $B$ — all of them are at the same perpendicular distance $y$ from $PQ$.

So every $r_i$ is replaced by $y$; a particle's position along the rod (that is, where it sits within $L$) does not affect this distance.

Rearranging: $I = \sum m_iy^2 = y^2\sum m_i = My^2$.

$L$ does not appear in the answer — the rod's length plays no part in rotation about an axis parallel to it.

Reading the options

Comparing with the options: the expression obtained is $My^2$ — option (1).

Options (2), (4) contain $L^2$, (3) has no $y$, (5) is zero; none of these match $My^2$.

Option (3)'s $\frac{1}{3}ML^2$ belongs to a different axis; the book states it as:

1) Moment of inertia of a uniform rod of length L, mass M about an axis passing normally through the end of the rod is given by

Checks

Dimension check: $[My^2] = ML^2$ — the dimension of moment of inertia (unit $\text{kg m}^2$) ✓.

Every option except (5) has this dimension, so the choice is also confirmed by the limiting-case check.

Limiting case: as $y \to 0$, the axis runs along the thin rod itself through its length; since the particles sit on the axis, $I \to 0$.

$My^2 \to 0$ ✓; but (2) $\to ML^2$, (3) $= \frac{1}{3}ML^2$, (4) $\to \frac{1}{2}ML^2$ — all three fail at this limit.

Conclusion

$I = My^2$

Every particle of the rod is at the same distance $y$ from $PQ$, so $I = \sum m_iy^2 = My^2$ — the answer is (1).

Using $\frac{1}{3}ML^2$ or $\frac{1}{12}ML^2$ for the rod regardless of where the axis is.

Both formulas are only for axes perpendicular to the rod. First check what the perpendicular distance from each particle to the axis actually is.