Angular Momentum: notes and previous year questions

Answer: Option B

Nothing outside acts, so $I\omega$ is kept. Twice the radius with the same mass makes $I = \tfrac{2}{5}MR^{2}$ four times as large, so $\omega$ becomes a quarter as large and the period four times as long: $4 \times 27 = 108$ days.

The trap is to double the period along with the radius, $54$ days, which is not even offered; the radius enters $I$ squared. The mass has not changed, only how far out it sits.

Answer: 60 kg m²/s

Two ways. Components: $(0,\ 4)$ is a point on the line, and a line of slope $1$ has a direction with equal $x$ and $y$ parts, so a speed of $3\sqrt{2}$ along it is the velocity $(3,\ 3)$, taking the body to be moving towards $+x$. Then $L_{z} = m(xv_{y} - yv_{x}) = 5(0 \times 3 - 4 \times 3) = -60$: a size of $60\ \mathrm{kg\,m^{2}/s}$, clockwise.

Perpendicular distance: the line cuts the axes at $(0,\ 4)$ and $(-4,\ 0)$, and the perpendicular from O meets it halfway between them, at $(-2,\ 2)$, which is $2\sqrt{2}\ \mathrm{m}$ from O. So $L = mv\,r_{\perp} = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 60$.

The trap is taking the intercept, $4\ \mathrm{m}$, as $r_{\perp}$, which gives $60\sqrt{2} \approx 85$: the $4\ \mathrm{m}$ is measured along the $y$-axis, not at right angles to the line. The body is moving at a steady speed along a straight line, so $60$ holds at every point of its path.

Answer: Option D

At the top the velocity is horizontal, $u\cos 30^{\circ} = \dfrac{\sqrt{3}}{2}u$, and its line is the height $h$ above the point of projection: $h = \dfrac{u^{2}\sin^{2}30^{\circ}}{2g} = \dfrac{u^{2}}{8g}$. So $L = m \times \dfrac{\sqrt{3}}{2}u \times \dfrac{u^{2}}{8g} = \dfrac{\sqrt{3}}{16}\dfrac{mu^{3}}{g}$.

B is the trap of the last question: at the top the vertical velocity is zero, not the angular momentum. C cannot be right, since $\dfrac{mu^{2}}{g}$ is a mass times a length, not a unit of angular momentum. A is far too big: it would need the height to be $\dfrac{\sqrt{2}u^{2}}{\sqrt{3}g}$, more than the $\dfrac{u^{2}}{2g}$ that even a straight-up throw reaches.

Answer: Option D

$\Delta\vec{L} = \vec{\tau}\,\Delta t = 0.02\hat{j}$, so $\vec{L} = 2\hat{i} + 0.02\hat{j}$. Its size is $\sqrt{4 + 0.0004} \approx 2.0001$: practically unchanged, while its direction has swung through about $\dfrac{0.02}{2} = 0.01\ \mathrm{rad}$ towards $\hat{j}$.

A and B add or subtract the sizes as if the torque lay along $\vec{L}$. C forgets that the torque adds to the angular momentum already there instead of replacing it.

Answer: Option C

Linear momentum: $mv = 4m\,v_{\mathrm{cm}}$, so $v_{\mathrm{cm}} = \dfrac{v}{4}$. Angular momentum about the rod's centre: the ball's line is $\dfrac{L}{2}$ from it, so $mv\dfrac{L}{2} = \dfrac{4mL^{2}}{12}\omega$, and $\omega = \dfrac{3v}{2L}$.

A takes the rod as turning about its far end, with $\dfrac{4mL^{2}}{3}$ and an arm of $L$. B assumes the far end stays still, $\omega = \dfrac{v_{\mathrm{cm}}}{L/2}$. D forgets the rod's mass is $4m$.

Check the energy: after, $\tfrac{1}{2}(4m)\left(\dfrac{v}{4}\right)^{2} + \tfrac{1}{2}\dfrac{mL^{2}}{3}\left(\dfrac{3v}{2L}\right)^{2} = \dfrac{mv^{2}}{8} + \dfrac{3mv^{2}}{8} = \tfrac{1}{2}mv^{2}$, all the ball had. So this is an elastic collision; with a lighter rod the ball could not stop dead.