"Rotational Motion" — Practice

Question 1

Give the moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through one end, perpendicular to the rod.

Accepted Answer

Central value: $I_{cm}=\dfrac{ML^{2}}{12}$, and the end is a distance $d=L/2$ away.

Parallel axis: $I=\dfrac{ML^{2}}{12}+M\left(\dfrac{L}{2}\right)^{2}=\dfrac{ML^{2}}{12}+\dfrac{ML^{2}}{4}=\mathbf{\dfrac{ML^{2}}{3}}$

Four times the central value — because moving the axis to the end puts far more mass at large $r$, and $r$ is squared.

Question 2

A wheel rolls without slipping at $5\ 	ext{m/s}$. Give the speed of its topmost point and of its contact point.

Accepted Answer

**Contact point: $0$** — that is the definition of rolling without slipping.

**Top point: $\mathbf{10\ 	ext{m/s}}$** — it is a distance $2R$ from the instantaneous axis, twice as far as the centre, so it moves twice as fast.

This is why the top of a rolling wheel blurs in a photograph while the bottom is sharp.

Question 3

Why does a skater spin faster when pulling their arms in?

Accepted Answer

The ice is nearly frictionless, so there is **no external torque** about the vertical axis and $L=I\omega$ is conserved.

Pulling the arms in moves mass closer to the axis, reducing $I$. With $L$ fixed, $\omega$ must rise.

Note the kinetic energy **increases**, since $K=L^{2}/2I$ and $I$ fell. That energy is not free — it is the muscular work done pulling the arms inward against the outward pull.

Question 4

A disc of radius $R$ has a hole of radius $R/2$ cut out, centred at $R/2$ from the centre. Find the new centre of mass.

Accepted Answer

Let the full disc have mass $M$. The hole has a quarter of the area, so mass $M/4$, located at $R/2$.

Remaining mass $=M-\dfrac{M}{4}=\dfrac{3M}{4}$.

$x_{cm}=\dfrac{M(0)-\frac{M}{4}\cdot\frac{R}{2}}{\frac{3M}{4}}=\mathbf{-\dfrac{R}{6}}$

It shifts $R/6$ **away** from the hole, as it must. Treating the hole as negative mass turned an integration into two terms.

Question 5

Find the moment of inertia of a disc about a tangent lying in its own plane.

Accepted Answer

**Step 1 — perpendicular axis.** A disc is planar, so $I_z=I_x+I_y$. By symmetry the two in-plane diameters are equal, and $I_z=	frac12MR^{2}$, so each diameter gives

$I_{diam}=\dfrac{MR^{2}}{4}$

**Step 2 — parallel axis.** A tangent in the plane is a distance $R$ from that diameter:

$I=\dfrac{MR^{2}}{4}+MR^{2}=\mathbf{\dfrac{5MR^{2}}{4}}$

The order matters: perpendicular axis first (it needs a central axis), then parallel axis to move out.

Question 6

A ring, a disc and a solid sphere roll from rest down the same incline. Give the order of arrival and justify it.

Accepted Answer

$a=\dfrac{g\sin	heta}{1+k^{2}/R^{2}}$ — both $M$ and $R$ cancel.

| Body | $k^{2}/R^{2}$ | $a$ |
|---|---|---|
| Solid sphere | $2/5$ | $0.71\,g\sin	heta$ |
| Disc | $1/2$ | $0.67\,g\sin	heta$ |
| Ring | $1$ | $0.50\,g\sin	heta$ |

**Sphere, disc, ring.** The ring has all its mass at the rim, so the largest share of the released energy goes into spin rather than translation.

Question 7

A solid sphere rolls without slipping down a $30°$ incline. Find the minimum coefficient of static friction.

Accepted Answer

$\mu_{min}=\dfrac{	an	heta}{1+R^{2}/k^{2}}$

With $k^{2}/R^{2}=	frac25$, we get $R^{2}/k^{2}=2.5$, so the denominator is $3.5$.

$\mu_{min}=\dfrac{	an 30°}{3.5}=\dfrac{0.577}{3.5}\approx\mathbf{0.165}$

Below this the sphere slips, the rolling condition fails, and friction becomes kinetic — and then it *does* dissipate energy.

Question 8

A rotating stool system has $I=6\ 	ext{kg m}^2$ at $2\ 	ext{rad/s}$. Pulling masses inward reduces $I$ to $2\ 	ext{kg m}^2$. Find the new angular speed and the change in kinetic energy, and account for it.

Accepted Answer

No external torque $\Rightarrow$ $L$ conserved: $6(2)=2\omega\Rightarrow\omega=\mathbf{6\ 	ext{rad/s}}$

$K_i=	frac12(6)(4)=12\ 	ext{J}$, $K_f=	frac12(2)(36)=36\ 	ext{J}$

Kinetic energy **rose by 24 J**. With $L$ fixed, $K=L^{2}/2I$, so halving $I$ triples... more precisely, cutting $I$ to a third triples $K$.

The 24 J is the **muscular work** done pulling the masses inward against their outward tendency. Angular momentum conservation says nothing about energy.

Question 9

Masses of 3 kg and 5 kg hang over a uniform disc pulley of mass 4 kg and radius $0.5\ 	ext{m}$, $g=10$. Find the acceleration.

Accepted Answer

Disc pulley: $I=	frac12MR^{2}$, so its effective inertia is $\dfrac{I}{R^{2}}=\dfrac{M}{2}=2\ 	ext{kg}$.

Driving force $=(5-3)g=20\ 	ext{N}$; total effective mass $=5+3+2=10\ 	ext{kg}$.

$a=\dfrac{20}{10}=\mathbf{2\ 	ext{m/s}^2}$

*Compare:* a massless pulley would give $2.5\ 	ext{m/s}^2$. Note also that the tensions on the two sides are now **different** — that is what supplies the pulley's torque.

Question 10

Explain why friction on a body rolling down an incline acts up the slope, yet does no work.

Accepted Answer

**Why up the slope.** Gravity acts at the centre of mass, so it exerts **no torque about the centre of mass**. Something else must supply the torque that spins the body up, and friction at the rim is the only candidate. To produce a torque in the rolling sense, it must point up the slope.

**Why no work.** The contact point is instantaneously at rest, so there is no relative sliding and no displacement for the friction force to act through. Work requires motion at the point of application.

That combination — a force essential to the motion doing zero work — is unique to rolling, and is why energy conservation still applies.

Question 11

A 60 kg man walks from one end of a 120 kg boat to the other, a distance of 4 m along the boat. The water is frictionless. How far does the boat move?

Accepted Answer

With no external horizontal force, $\vec{a}_{cm}=0$ and the centre of mass stays fixed.

Let the boat move $X$ backward. The man walks 4 m **relative to the boat**, so his ground displacement is $4-X$ forward.

$\sum m_i\Delta x_i=0$:

$60(4-X)+120(-X)=0$

$240=180X \Rightarrow X=\mathbf{1.33\ 	ext{m}}$

*Sanity check:* the man is lighter, so he should move further — he covers $4-1.33=2.67$ m to the boat's 1.33 m, exactly the inverse ratio $120:60$ of the masses.

Question 12

A uniform rod of mass $M$ and length $L$ is hinged at one end and released from rest in the horizontal position. Find its initial angular acceleration and the initial acceleration of its free end.

Accepted Answer

About the hinge, $I=	frac13ML^{2}$ and the weight acts at $L/2$:

$	au=Mg\dfrac{L}{2}=\dfrac{1}{3}ML^{2}\alpha \Rightarrow \alpha=\mathbf{\dfrac{3g}{2L}}$

The free end is at distance $L$, so

$a_{end}=\alpha L=\mathbf{\dfrac{3g}{2}}=1.5g$

**The tip accelerates faster than free fall.** A coin resting near the end of the rod is left behind in mid-air as the rod drops away beneath it.

*Where the extra comes from:* the hinge supplies a downward force on the rod. Only the point at $x=2L/3$ falls at exactly $g$; everything beyond it falls faster.

Practice — "Rotational Motion"

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