Irodov 1.261: The Ball Keeps Two Sevenths – Amit & Deepali Sinha Foundation

Problem Irodov 1.261

A plank of mass $m_1$ lies on a horizontal plane that is perfectly smooth, so the plank slides on it without friction. A uniform solid sphere of mass $m_2$ is placed on top of the plank. A constant horizontal force $F$ is then applied to the plank, along the plank’s length. The sphere never slides on the plank. Find the acceleration of the plank and the acceleration of the center of the sphere.

The sphere of mass m_2 sits on the plank of mass m_1, which lies on a smooth horizontal floor (drawn hatched, as a fixed surface). The constant horizontal force F is applied to the plank and is shown acting on its leading face. — The sphere of mass $m_2$ sits on the plank of mass $m_1$ , which lies on a smooth horizontal floor (drawn hatched, as a fixed surface). The constant horizontal force $\vect F$ is applied to the plank and is shown acting on its leading face.

The sphere alone. The slab beneath it is the plank m_1, not the floor. At the contact the plank exerts the normal force N and the forward friction f; the weight m_2 g acts at the center, and β is the angular acceleration. — The sphere alone. The slab beneath it is the plank $m_1$ , not the floor. At the contact the plank exerts the normal force $\vect N$ and the forward friction $\vect f$ ; the weight $m_2\vect g$ acts at the center, and $\beta$ is the angular acceleration.

The plank alone, horizontal forces only: the applied force F forward and the sphere's friction reaction f' backward, equal in magnitude to f. The vertical forces cancel and are omitted. — The plank alone, horizontal forces only: the applied force $\vect F$ forward and the sphere’s friction reaction $\vect f'$ backward, equal in magnitude to $\vect f$ . The vertical forces cancel and are omitted.

Given

$m_1$
$m_2$
$F$
frictionless floor
no sliding of sphere on plank

Find

the plank’s acceleration

w_1

and the sphere’s center acceleration

w_2

Solution

Step 1Newton’s second law for the sphere’s center

Let $r$ be the sphere’s radius, and measure every acceleration positive in the direction of $\vect F$ . The sphere’s weight and the plank’s normal reaction are both vertical and its center has no vertical acceleration, so they cancel and the friction $f$ is the only horizontal force left to move the center.

m_2 w_2 = f \tag{1}

Step 2The same friction is the only torque about the center

Written about the sphere’s own center, the weight and the normal force have no moment: the weight acts at that center and the normal force is directed straight at it. Friction acts at the contact, a distance $r$ from the center and perpendicular to it, so its moment arm is the full radius. With the moment of inertia of a uniform solid sphere about a central axis,

I = \frac{2}{5}m_2 r^2

the rotational equation of motion reads

\begin{align*}fr &= I\beta \\ &= \frac{2}{5}m_2 r^2\beta\end{align*}

and dividing by $r$ and by the coefficient of $\beta$ ,

\beta = \dfrac{5f}{2m_2 r}

(2)

Step 3No sliding ties the two accelerations together

The point of the sphere touching the plank and the point of the plank under it move together, so their accelerations along the surface are equal. For the rigid sphere, that contact point’s acceleration along the surface is the center’s acceleration plus the rotational contribution $\beta r$ , which is why the spin enters the plank’s motion at all.

w_1 = w_2 + \beta r

Substituting the angular acceleration (2) and then the friction (1),

\begin{align*}w_1 &= w_2 + \frac{5f}{2m_2} \\ &= w_2 + \frac{5}{2}w_2\end{align*}

w_1 = \dfrac{7}{2}\,w_2

(3)

Step 4The plank’s own equation closes the system

The floor is frictionless, so the plank feels only the applied force forward and the sphere’s friction reaction backward. Writing its equation of motion and replacing the friction first by (1) and then by the constraint (3),

\begin{align*}m_1 w_1 &= F - f \\ &= F - m_2 w_2 \\ &= F - \frac{2}{7}m_2 w_1\end{align*}

Collecting the two terms in $w_1$ ,

w_1\left(m_1 + \frac{2}{7}m_2\right) = F \tag{4}

w_1 = \dfrac{F}{m_1 + \dfrac{2}{7}m_2}

(5)

Step 5The sphere’s center lags behind

The constraint (3) now delivers the second acceleration directly.

w_2 = \frac{2}{7}\cdot\frac{F}{m_1 + \dfrac{2}{7}m_2}

w_2 = \dfrac{2}{7}\,w_1 = \dfrac{2F}{7m_1 + 2m_2}

(6)

Both accelerations are horizontal and point the way $\vect F$ does.

Check

The floor is smooth, so the only external horizontal force on the plank and sphere together is $F$ , and the center-of-mass theorem demands $F = m_1w_1 + m_2w_2$ . The two boxed results give $m_1w_1 + m_2(2/7)w_1 = w_1(m_1 + 2m_2/7)$ , which is $F$ by (4). As a second test, let the plank become massless: $m_1 \to 0$ leaves $w_1 = 7F/2m_2$ and $w_2 = F/m_2$ , and the second of these is what (1) requires, since a massless plank can carry no net force and must pass the whole of $F$ on to the sphere as friction.

Remark

Of the plank’s acceleration the sphere’s center takes only $2/7$ ; the remaining $5/7$ shows up as $\beta r$ , the rate at which the ball spins up rather than moves along. That split is fixed entirely by the shape factor in $I = km_2r^2$ , here $k = 2/5$ : the center picks up the fraction $1/(1+1/k)$ of the surface’s acceleration, which is $2/7$ for a solid sphere but $1/2$ for a hoop and $1/3$ for a uniform disc. The radius itself never survives, because friction’s moment arm and the contact point’s rotational term are both proportional to $r$ and the two cancel; a marble and a bowling ball on the same plank would trail it in the same proportion, each center taking two sevenths of whatever acceleration the plank has.