"Rotational Motion" — Practice

Question 1

A uniform ring of mass $M$ and radius $R$ rolls without slipping. Find its angular momentum about a point on the ground.

Accepted Answer

**Split:** $L=I_{cm}\omega+Mv_{cm}R=MR^{2}\omega+M(\omega R)R=\mathbf{2MR^{2}\omega}$

**Instantaneous axis:** the contact point is momentarily at rest, so

$I_{contact}=MR^{2}+MR^{2}=2MR^{2} \Rightarrow L=2MR^{2}\omega$

The two agree.

*The error to avoid:* quoting $MR^{2}\omega$, which is only the spin term. The orbital term is equally large here, so the answer would be exactly half.

Question 2

At what height above its centre must a ring be struck horizontally so that it begins rolling immediately, with no slipping?

Accepted Answer

For a ring about its central perpendicular axis, $I=MR^{2}$, so $k^{2}=R^{2}$.

$$h=\frac{k^{2}}{R}=\frac{R^{2}}{R}=\mathbf{R}$$

The ring must be struck at its **topmost point** — a full radius above the centre.

*Why this is the extreme case.* Among common bodies the ring has all its mass at the rim, so it has the largest $k^{2}$ and needs the highest strike. A solid sphere needs only $0.4R$. This is why a hoop is awkward to start rolling cleanly: anywhere below the very top and it skids.

Question 3

A block of width $b$ and height $h$ rests on an incline with friction coefficient $\mu$. State the test for whether it slides or topples first as the incline is tilted.

Accepted Answer

**Sliding** begins when friction is exhausted: $	an	heta=\mu$

**Toppling** begins when the weight's line of action falls outside the base: $	an	heta=\dfrac{b}{h}$

Whichever critical angle is **smaller** is reached first.

$$\mu<\frac{b}{h}\ \Rightarrow\ 	ext{slides first};\qquad \mu>\frac{b}{h}\ \Rightarrow\ 	ext{topples first}$$

*The physical picture:* a tall narrow block on a grippy surface tips over, while a short wide block on a slippery one slides. Both failure modes are always present; only one of them wins.

Question 4

A rod of length $L$ has linear density $\lambda=\lambda_0\left(1+\dfrac{x}{L}\right)$, measured from one end. Find its mass and its moment of inertia about an axis through that end, perpendicular to the rod.

Accepted Answer

**Mass:**

$M=\displaystyle\int_0^L\lambda_0\left(1+\frac{x}{L}\right)dx=\lambda_0\left(L+\frac{L}{2}\right)=\frac{3\lambda_0L}{2}$

**Moment of inertia:**

$I=\displaystyle\int_0^L x^{2}\lambda_0\left(1+\frac{x}{L}\right)dx=\lambda_0\left(\frac{L^{3}}{3}+\frac{L^{3}}{4}\right)=\frac{7\lambda_0L^{3}}{12}$

Substituting $\lambda_0=\dfrac{2M}{3L}$:

$$I=\frac{2M}{3L}\cdot\frac{7L^{3}}{12}=\mathbf{\frac{7ML^{2}}{18}}$$

*Sanity check:* $\dfrac{7}{18}=0.389$, against $\dfrac13=0.333$ for a uniform rod about its end. Larger, as it must be — this rod is heavier at the far end, so more of its mass sits at large $x$.

Question 5

A hollow sphere rolls without slipping down a $30°$ incline. Take $g=10$. Find its acceleration and the minimum coefficient of friction required.

Accepted Answer

$a_{cm}=\dfrac{g\sin	heta}{1+k^{2}/R^{2}}=\dfrac{10(0.5)}{1+2/3}=\dfrac{5}{5/3}=\mathbf{3.0\ 	ext{m/s}^{2}}$

$\mu_{min}=\dfrac{	an	heta}{1+R^{2}/k^{2}}=\dfrac{0.577}{1+3/2}=\dfrac{0.577}{2.5}=\mathbf{0.23}$

*Compare with a solid sphere:* $k^{2}/R^{2}=2/5$ gives $a=3.57\ 	ext{m/s}^{2}$ and $\mu_{min}=0.165$.

The hollow sphere is both **slower** and **needs more friction**, and both follow from the same large $k^{2}/R^{2}$: more of its mass sits far from the axis, so more torque is needed to spin it up at the required rate, and only friction can supply it.

Question 6

A solid sphere is projected along a rough floor with forward speed $v_0$ and **backspin** $\omega_0=v_0/R$. Find the speed at which pure rolling eventually begins.

Accepted Answer

Friction acts horizontally at a point on the ground line, so its moment about any point $O$ on that line is zero, and $L_O$ is conserved.

**Before.** The orbital term is $+Mv_0R$; the spin term is **negative**, because the backspin opposes rolling:

$L_i=Mv_0R-I_{cm}\omega_0=Mv_0R-	frac25MR^{2}\left(\frac{v_0}{R}ight)=	frac35Mv_0R$

**After**, rolling with $v=\omega R$:

$L_f=	frac25MvR+MvR=	frac75MvR$

$$	frac35Mv_0R=	frac75MvR \Rightarrow v=\mathbf{	frac37 v_0}$$

*Compare with no spin*, which gave $	frac57v_0$. The backspin destroys extra forward momentum, so the ball ends up slower — and a stronger backspin, $\omega_0>	frac52 v_0/R$, would make $L_i$ negative and send the ball rolling **backwards**, which is exactly the trick shot.

Question 7

A man of mass $m$ stands at the rim of a turntable of mass $M$ and radius $R$ rotating freely at $\omega_0$. He walks to the centre. Find the new angular speed.

Accepted Answer

No external torque, so $L=I\omega$ is conserved.

**Initially**, with the man at the rim:

$I_i=	frac12MR^{2}+mR^{2}$

**Finally**, with the man at the axis contributing nothing:

$I_f=	frac12MR^{2}$

$$\omega=\omega_0\frac{	frac12MR^{2}+mR^{2}}{	frac12MR^{2}}=\omega_0\left(1+\frac{2m}{M}ight)$$

*Where the energy comes from.* $K=L^{2}/2I$, and $I$ has fallen, so the kinetic energy has **risen**. The extra energy is supplied by the man's muscles, working against the outward tendency as he walks inward. Conserving $L$ says nothing about energy in either direction.

Question 8

A uniform rod of mass $M$ and length $L$ lies at rest on frictionless ice. It is struck by a horizontal impulse perpendicular to the rod at a distance $d$ from its centre. Find $d$ such that one end of the rod is instantaneously at rest immediately afterwards.

Accepted Answer

$J=Mv_{cm}$ and $Jd=I_{cm}\omega=\dfrac{ML^{2}}{12}\omega$

For the far end (at $L/2$ from the centre, on the opposite side) to be at rest:

$v_{cm}-\omega\dfrac{L}{2}=0 \Rightarrow v_{cm}=\omega\dfrac{L}{2}$

Substituting $J=M\omega L/2$ into the second equation:

$\dfrac{M\omega L}{2}\,d=\dfrac{ML^{2}\omega}{12} \Rightarrow d=\mathbf{\dfrac{L}{6}}$

*The same point as before.* Measured from the stationary end this is $\dfrac{L}{2}+\dfrac{L}{6}=\dfrac{2L}{3}$ — identical to the centre of percussion of a rod hinged at that end, and for the same reason: a hinge that supplies no impulse is indistinguishable from no hinge at all.

Question 9

A solid sphere is released on a $30°$ incline with $\mu=0.10$ and $g=10$. Determine whether it rolls or slips, and find its linear acceleration.

Accepted Answer

**Required for rolling:**

$\mu_{min}=\dfrac{	an	heta}{1+R^{2}/k^{2}}=\dfrac{0.577}{1+5/2}=\dfrac{0.577}{3.5}=0.165$

Available is only $0.10$, so **the sphere slips**. Friction is kinetic and equal to $\mu Mg\cos	heta$ in a known direction, up the slope.

$a_{cm}=g\sin	heta-\mu g\cos	heta=5-0.10(10)(0.866)=\mathbf{4.13\ 	ext{m/s}^{2}}$

The angular acceleration is now independent of that:

$\alpha=\dfrac{\mu Mg\cos	heta\,R}{	frac25MR^{2}}=\dfrac{5\mu g\cos	heta}{2R}=\dfrac{4.33}{R}$

*Check the consistency:* $\alpha R=4.33 < a_{cm}=4.13$? No — $4.33>4.13$, so $\omega R$ grows faster than $v_{cm}$ and the two converge. The sphere is slipping now but will reach rolling further down the slope, exactly as the attractor argument predicts.

*Note the contrast:* had it rolled, $a$ would have been $3.57\ 	ext{m/s}^{2}$. A slipping sphere accelerates **faster**, because less of the gravitational work goes into spin.

Question 10

A bullet of mass $10\ 	ext{g}$ moving at $200\ 	ext{m/s}$ strikes the end of a $1\ 	ext{kg}$, $1\ 	ext{m}$ rod lying at rest on frictionless ice, perpendicular to the rod, and embeds itself. Find the velocity of the centre of mass and the angular velocity about it.

Accepted Answer

**Centre-of-mass velocity**, from linear momentum:

$v_{cm}=\dfrac{0.01(200)}{1.01}=\mathbf{1.98\ 	ext{m/s}}$

**New centre of mass**, measured from the rod's centre toward the impact end:

$x_c=\dfrac{0.01(0.5)}{1.01}=0.00495\ 	ext{m}$

**Angular momentum about that point, before impact.** Only the bullet contributes, at perpendicular distance $0.5-0.00495=0.49505$ m:

$L_i=0.01(200)(0.49505)=0.990\ 	ext{kg m}^{2}	ext{/s}$

**Moment of inertia after:**

$I=\left[\dfrac{ML^{2}}{12}+Mx_c^{2}ight]+m\left(0.49505ight)^{2}=0.0833+0.0000245+0.00245=0.0858\ 	ext{kg m}^{2}$

$$\omega=\dfrac{0.990}{0.0858}=\mathbf{11.5\ 	ext{rad/s}}$$

*Contrast with the hinged case.* Hinged at one end, the same bullet gave only $5.83$ rad/s, because the hinge absorbed most of the momentum. Free on ice, nothing absorbs it, so the rod both translates and spins faster.

Question 11

A block of width $b$ and height $h$ sits on the flat bed of a truck with friction coefficient $\mu$. The truck brakes with deceleration $a$. Find the condition under which the block topples rather than slides.

Accepted Answer

In the truck's frame a pseudo force $ma$ acts forward at the centre of mass, height $h/2$. **Sliding** begins when the pseudo force exceeds friction: $ma>\mu mg \Rightarrow a>\mu g$ **Toppling** about the leading bottom edge begins when its torque overcomes that of the weight: $ma\dfrac{h}{2}>mg\dfrac{b}{2} \Rightarrow a>\dfrac{gb}{h}$ Toppling happens first if it needs the **smaller** deceleration: $$\frac{gb}{h}<\mu g \qquad\Longleftrightarrow\qquad \frac{b}{h}<\mu$$ *The same condition as on an incline*, and for the same reason — a pseudo force and a component of gravity are indistinguishable to the block. *The practical consequence:* tall narrow cargo on a grippy bed tips forward under braking, which is why loads are strapped and stowed low rather than relying on friction.

Question 12

A solid sphere rolling without slipping at $v_0$ runs up an incline of angle $	heta$. Find the distance travelled along the incline, and compare with a frictionless block sliding up at the same speed.

Accepted Answer

**Rolling friction does no work**, because the contact point is instantaneously at rest and nothing slides. Mechanical energy is therefore conserved.

$	frac12Mv_0^{2}\left(1+	frac25ight)=Mgh \Rightarrow 	frac{7}{10}v_0^{2}=gh$

$h=\dfrac{7v_0^{2}}{10g}, \qquad s=\dfrac{h}{\sin	heta}=\mathbf{\dfrac{7v_0^{2}}{10g\sin	heta}}$

**Frictionless sliding block:** $h'=\dfrac{v_0^{2}}{2g}$, so $s'=\dfrac{v_0^{2}}{2g\sin	heta}$

$$\frac{s}{s'}=\frac{7/10}{1/2}=\mathbf{1.4}$$

**The rolling sphere goes 40% further**, because it arrives carrying rotational kinetic energy as well, and all of it converts to height.

*The apparent paradox worth resolving:* friction is essential for rolling yet does no work. It supplies the torque that converts translational energy into rotational energy, but since the contact point does not move, it transfers no energy into or out of the sphere.

Practice — "Rotational Motion"

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