Irodov 1.252: Rolling Sets Its Own Friction – Amit & Deepali Sinha Foundation

Problem Irodov 1.252

A uniform solid sphere of mass $m$ and radius $R$ is released from rest on a fixed plane inclined at an angle $\alpha$ to the horizontal, and it rolls straight down the plane without slipping. The coefficient of friction between the sphere and the plane is $k$ . Find:

the values of $k$ for which the sphere does not slip;
the kinetic energy $T$ of the sphere at a time $t$ after the motion began.

The setup. The sphere rests on the fixed plane inclined at α to the horizontal and rolls straight down it without slipping, so its center moves down the slope with velocity v while the sphere turns with angular velocity ω about its own center. — The setup. The sphere rests on the fixed plane inclined at $\alpha$ to the horizontal and rolls straight down it without slipping, so its center moves down the slope with velocity $\vect v$ while the sphere turns with angular velocity $\omega$ about its own center.

Given

$m$
$R$
$\alpha$
$t$

Find

(a) the values of

k

at which no slipping occurs, (b) the kinetic energy

T

after time

t

Solution

Step 1The forces at the contact, and the weight resolved on the slope

Three forces act on the sphere: its weight at the center, and, at the single contact point, the normal force and the friction. The friction must point up the slope, because that is the only sense in which it can spin the sphere up while the center speeds up on the way down.

The sphere as a free body. The normal force N and the friction f both act at the single contact point, while the weight m g acts at the center. The friction points up the slope. — The sphere as a free body. The normal force $\vect N$ and the friction $\vect f$ both act at the single contact point, while the weight $m\vect g$ acts at the center. The friction points up the slope.

The plane fixes two natural directions, one along the slope and one normal to it, and the weight is the only force with a component in each. The center of the sphere travels parallel to the plane, so it has no acceleration along the normal, and the two normal forces balance there.

The weight resolved in the plane's own directions. The dashed component down the slope has magnitude mgα, the dashed component pressing into the plane has magnitude mgα, and α is marked between the weight and the inward normal. — The weight resolved in the plane’s own directions. The dashed component down the slope has magnitude $mg\sin\alpha$ , the dashed component pressing into the plane has magnitude $mg\cos\alpha$ , and $\alpha$ is marked between the weight and the inward normal.

N = mg\cos\alpha \tag{1}

Step 2Newton’s law along the slope and the moment about the center

Two equations govern the motion: the second law for the center of mass along the slope, and the moment equation for the rotation about the center, where only the friction has a moment arm (the normal force and the weight both pass through the center). Writing $a$ for the center’s acceleration down the slope and $\beta$ for the angular acceleration, and using the moment of inertia of a uniform sphere about a diameter,

\begin{align*} ma &= mg\sin\alpha - f \\ I\beta &= fR, \qquad I = \frac{2}{5}mR^{2} \tag{2}\end{align*}

Rolling without slipping ties the two accelerations together, because the contact point is momentarily at rest and the center advances exactly the arc the rim unrolls.

a = \beta R \tag{3}

Step 3The acceleration and the friction the constraint demands

Eliminating $\beta$ between (2) and (3) expresses the friction through the acceleration alone, and substituting that back into the equation along the slope leaves one unknown.

\begin{align*}f &= \frac{I\beta}{R} = \frac{I a}{R^{2}} = \frac{2}{5}ma \\ ma &= mg\sin\alpha - \frac{2}{5}ma \\ \frac{7}{5}\,a &= g\sin\alpha\end{align*}

a = \frac{5}{7}\,g\sin\alpha

(4)

Putting this acceleration back into the friction just obtained gives the force the contact must supply.

f = \frac{2}{5}ma = \frac{2}{5}m\cdot\frac{5}{7}g\sin\alpha

f = \frac{2}{7}\,mg\sin\alpha

(5)

Step 4The condition that the contact does not slip

Static friction can supply no more than $kN$ , so the sphere rolls only while the demanded friction (5) stays below that ceiling. Combining it with the normal force (1),

\begin{align*}f &\le kN \\ \frac{2}{7}mg\sin\alpha &\le k\,mg\cos\alpha\end{align*}

Dividing by $mg\cos\alpha$ , which is positive for any slope short of vertical,

k \ge \frac{2}{7}\tan\alpha

(6)

Step 5The kinetic energy

t

seconds later

The sphere starts from rest, so after a time $t$ the center has speed $at$ , and the rolling condition fixes the spin from that same speed. The kinetic energy is the translational part plus the rotational part about the center.

\begin{align*}v &= at, \qquad \omega = \frac{v}{R} \\ T &= \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \\ &= \frac{1}{2}mv^{2} + \frac{1}{2}\cdot\frac{2}{5}mR^{2}\cdot\frac{v^{2}}{R^{2}} \\ &= \frac{7}{10}mv^{2}\end{align*}

The radius has canceled, so only the speed is left; substituting the speed from the acceleration (4) gives the answer.

T = \frac{7}{10}m\left(\frac{5}{7}g t\sin\alpha\right)^{2}

T = \frac{5}{14}\,mg^{2}t^{2}\sin^{2}\alpha

(7)

Check

Energy accounts for the same number independently. Static friction does no work, because the point it acts at is instantaneously at rest, so all the kinetic energy comes from the drop in height. In time $t$ the center travels $s = at^{2}/2 = (5/14)gt^{2}\sin\alpha$ along the slope and descends $s\sin\alpha$ , and $mg\,s\sin\alpha = (5/14)mg^{2}t^{2}\sin^{2}\alpha$ , exactly the boxed energy. Dimensionally $mg^{2}t^{2}$ is $\mathrm{kg}\cdot(\mathrm{m/s^{2}})^{2}\cdot\mathrm{s^{2}} = \mathrm{kg}\cdot\mathrm{m^{2}/s^{2}}$ , an energy, and $k \ge (2/7)\tan\alpha$ is dimensionless and vanishes as the plane flattens.

Remark

Neither $m$ nor $R$ survives in the acceleration: every uniform sphere descends a given slope at the same rate, and among rolling bodies the ranking is set by the single number $I/mR^{2}$ alone, $2/5$ for a sphere, $1/2$ for a disc, $1$ for a hoop, since $a = g\sin\alpha/(1+I/mR^{2})$ follows from (2) and (3) with no other property of the body used. The same number fixes the friction demanded, $f = mg\sin\alpha\,(I/mR^{2})/(1+I/mR^{2})$ , so a hoop needs a coefficient $(1/2)\tan\alpha$ where the sphere needs only $(2/7)\tan\alpha$ : the harder a body is to spin up, the more grip it asks of the surface.