Irodov 1.265: When A Rides over the Top – Amit & Deepali Sinha Foundation

Problem Irodov 1.265

A thin rigid hoop of radius $R$ and mass $m$ carries a small body $A$ fixed to it at the rim, and the body’s mass is also $m$ . The hoop rolls without slipping along a horizontal plane. Each time the body $A$ passes through its lowest position, the center of the hoop is moving with speed $v_0$ . For what values of $v_0$ does the hoop keep rolling without ever bouncing, that is, without leaving the plane?

The hoop of mass m rolls to the right along the plane, turning at ω with its center at v_0. The body A, also of mass m, is fixed to the rim, drawn in its lowest position at the contact point. — The hoop of mass $m$ rolls to the right along the plane, turning at $\omega$ with its center at $\vect v_0$ . The body $A$ , also of mass $m$ , is fixed to the rim, drawn in its lowest position at the contact point.

Given

$R$
hoop mass $m$
mass of $A$ also $m$
rolling without slipping
center speed $v_0$ when $A$ is lowest

Find

the values of

v_0

for which the hoop never leaves the plane

Solution

Step 1The hoop’s own kinetic energy

Rolling without slipping locks the hoop’s angular velocity to the speed $v$ of its center, and all the mass of a thin hoop sits at the rim, so its moment of inertia about the central axis is the full $mR^2$ .

v = \omega R, \qquad I = mR^2

Adding the translational and rotational parts,

\begin{align*}K_{\mathrm{hoop}} &= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \\ &= \frac{1}{2}mv^2 + \frac{1}{2}mR^2\cdot\frac{v^2}{R^2}\end{align*}

K_{\mathrm{hoop}} = mv^2

(1)

Half a turn later, with A at the top. The center, a distance R from the stationary contact point, moves with v; A, at distance 2R straight above it, moves with 2 v. — Half a turn later, with $A$ at the top. The center, a distance $R$ from the stationary contact point, moves with $\vect v$ ; $A$ , at distance $2R$ straight above it, moves with $2\vect v$ .

Step 2Energy conservation between the two extreme positions

When $A$ is lowest it sits exactly at the contact point, which rolling without slipping makes instantaneously stationary, so $A$ then contributes no kinetic energy at all. Half a turn later $A$ is at the top, a height $2R$ higher. The contact point is instantaneously at rest, so every point’s speed is proportional to its distance from it, and $A$ , twice as far out as the center, moves at twice the center’s speed. The hoop’s own center stays at height $R$ throughout, so it neither gains nor loses potential energy. Writing the two total energies with $v$ for the center’s speed in the upper configuration, and taking the hoop’s own share from (1) at each end,

\begin{align*}mv_0^2 &= mv^2 + \frac{1}{2}m(2v)^2 + mg(2R) \\ &= 3mv^2 + 2mgR\end{align*}

and solving for $v^2$ ,

v^2 = \dfrac{v_0^2 - 2gR}{3}

(2)

Step 3What the ground has to push with at that instant

The plane acts on the system only through the normal force $N$ and a horizontal friction, so vertically the external forces are $N$ upward and the total weight $2mg$ downward. Newton’s second law for the two bodies together equates their sum to the total rate of change of vertical momentum.

N - 2mg = m\,(a_{\mathrm{C}})_y + m\,(a_A)_y \tag{3}

The hoop’s center $\mathrm{C}$ rides at the constant height $R$ , so its vertical acceleration vanishes identically. The body $A$ turns about $\mathrm{C}$ on a circle of radius $R$ at angular velocity $\omega$ , and at the moment it is at the top its centripetal acceleration points straight down while both its tangential acceleration and the acceleration of $\mathrm{C}$ itself are horizontal.

(a_{\mathrm{C}})_y = 0, \qquad (a_A)_y = -\omega^2R = -\frac{v^2}{R}

Substituting these into (3),

N = 2mg - \dfrac{mv^2}{R}

(4)

Step 4The condition for staying on the ground

A surface can push but not pull, so the hoop stays down only while $N$ remains non-negative. The instant with $A$ at the top is the one to test: it is the only configuration in which the whole of $A$ ’s centripetal acceleration points straight down and its tangential acceleration drops out of the vertical balance completely, and it is there that $N$ first reaches zero as $v_0$ is raised. Imposing that condition on (4),

\begin{align*}2mg - \frac{mv^2}{R} &\ge 0 \\ v^2 &\le 2gR\end{align*}

Now bringing in the speed (2) that the launch condition actually delivers there,

\begin{align*}\frac{v_0^2 - 2gR}{3} &\le 2gR \\ v_0^2 - 2gR &\le 6gR\end{align*}

v_0 \le \sqrt{8gR} = 2\sqrt{2gR}

(5)

Check

Take the critical case $v_0^2 = 8gR$ and look at the opposite configuration, $A$ at the bottom. There $A$ is at the contact point and its acceleration is purely centripetal, of magnitude $v_0^2/R = 8g$ , directed upward toward the center. The vertical balance (3) then gives $N = 2mg + 8mg$ , that is $10mg$ , five times the total weight: the hoop presses hardest exactly when $A$ swings through the bottom, and hardest of all in the very case that leaves it on the point of lifting off half a turn later. The two answers also bracket sensibly, since $N$ runs from $10mg$ to $0$ over that half turn. Dimensionally $\sqrt{gR}$ is a speed, as required.

Remark

The hoop’s own center never accelerates vertically at all, so the whole bouncing question belongs to the little body $A$ . What $A$ traces out as the hoop rolls is a cycloid, and at the top of each arch it is being swung round the hoop’s center on a circle of radius $R$ , at the speed $v$ it has relative to that center, which demands a downward force $mv^2/R$ . The ground frame reaches the same number another way: there $A$ moves at $2v$ , but along a cycloid whose radius of curvature at the top of the arch is $4R$ . The only downward force available is the combined weight of hoop and body, and when the demand exceeds that supply the ground would have to pull the hoop down to keep it in contact, which it cannot do, so the hoop hops. That is the mechanism by which any wheel with an unbalanced mass on its rim starts to hammer once it is spun fast enough. The equality of the two masses matters to the number, not to the mechanism. Running the same two steps with hoop mass $M$ and body mass $m$ gives the critical launch speed $v_0^2 = gR\left(x + 3 + 4/x\right)$ with $x = M/m$ , the top still being the configuration that binds for any hoop at least as heavy as $A$ . That does climb without limit as $x$ grows, but not from the outset: it falls from $8gR$ at equal masses to $7gR$ at $x=2$ before turning upward, so a hoop twice $A$ ’s mass starts to bounce at a lower launch speed than an equal one does.