Kvant Physics Problem 553
A heavy board of mass $M$ rests horizontally on two thin walled rollers.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m25s
Source on kvant.digital
Problem
A heavy board of mass $M$ rests on two thin-walled rollers of radii $r$ and $R$ and equal masses $m$. The distance between the centers of the rollers is $l$. With what acceleration will the board begin to move if it is released? The friction between all surfaces is such that there is no slipping.
Setup and Assumptions
A heavy board of mass $M$ rests horizontally on two thin walled rollers. The left roller has radius $r$, the right roller has radius $R$, and both rollers have equal masses $m$. The distance between the roller centers is $l$.
The board is released from rest. The required quantity is the initial acceleration $a$ of the board relative to the ground.
The rollers are treated as thin walled cylindrical shells, hence their moments of inertia about their axes are
$$I_r=mr^2,\qquad I_R=mR^2.$$
All contacts are rough enough that no slipping occurs, both between the board and each roller, and between each roller and the floor.
The rollers move only horizontally. The board remains horizontal. Air resistance and rolling resistance are neglected. The analysis is carried out in an inertial frame fixed to the ground.
Because the board is heavy, its weight produces different normal reactions on the two rollers whenever the center of mass of the board is not midway between them. Let the center of mass of the board be displaced by a distance $x$ from the midpoint between the rollers, positive toward the roller of radius $R$.
The board will move toward the side where the supporting force is larger. We seek the acceleration corresponding to a given displacement $x$.
Physical Principles
For the board, Newton's second law in the horizontal direction gives
$$Ma=F_1+F_2,$$
where $F_1$ and $F_2$ are the friction forces exerted on the board by the rollers.
For each roller, translational motion and rotational motion must both satisfy Newton's laws.
For rolling without slipping on the floor,
$$u_i=\rho_i\omega_i,$$
where $u_i$ is the velocity of the roller center and $\rho_i$ denotes the corresponding radius.
For rolling without slipping between a roller and the board, the velocity of the upper point of the roller equals the board velocity. Since the upper point moves twice as fast as the center of a rolling cylinder,
$$v=2u_i.$$
Differentiating,
$$a=2a_i,$$
where $a_i$ is the horizontal acceleration of the center of either roller.
For static equilibrium of the board in the vertical direction,
$$N_1+N_2=Mg,$$
where $N_1$ and $N_2$ are the normal reactions of the rollers on the board.
Taking moments about the center of the left roller gives
$$N_2,l=Mg\left(\frac l2+x\right).$$
Derivation
The vertical equilibrium relations yield
$$N_2=\frac{Mg}{l}\left(\frac l2+x\right),$$
and
$$N_1=Mg-N_2 =\frac{Mg}{l}\left(\frac l2-x\right).$$
The friction forces acting on the board are produced by the rollers. Let the friction force on the board from a roller be $F$.
Consider one roller of radius $\rho$ and mass $m$.
Let $f$ be the friction force from the board acting on the roller at the top, and let $q$ be the friction force from the floor acting on the roller at the bottom.
The translational equation for the roller is
$$q-f=ma_c,$$
where $a_c$ is the acceleration of its center.
The rotational equation is
$$(f+q)\rho=I\alpha.$$
Since the roller is thin walled,
$$I=m\rho^2.$$
The no slip condition at the floor gives
$$a_c=\rho\alpha.$$
Substituting,
$$f+q=ma_c.$$
Combining with
$$q-f=ma_c,$$
gives
$$q=ma_c,\qquad f=0.$$
Hence the friction force transmitted from the roller to the board equals the floor friction force,
$$F=q=ma_c.$$
Because
$$a=2a_c,$$
we obtain
$$F=\frac{m}{2}a.$$
This relation holds for each roller. Therefore
$$F_1=\frac{m}{2}a,\qquad F_2=\frac{m}{2}a.$$
The directions of these forces are opposite. The larger normal reaction corresponds to the roller that tends to drive the board.
Since the available static friction is proportional to the normal reaction, the effective horizontal driving force on the board is proportional to the difference of the normal reactions. The net force transmitted to the board is
$$F_{\text{net}} = \frac{m}{2}a, \frac{N_2-N_1}{N_1+N_2}.$$
Using
$$N_1+N_2=Mg,$$
and
$$N_2-N_1 = \frac{2Mgx}{l},$$
gives
$$F_{\text{net}} = \frac{m}{2}a, \frac{2x}{l} = m a,\frac{x}{l}.$$
Newton's second law for the board yields
$$Ma = m a,\frac{x}{l} + M a_{\text{board}},$$
which, after eliminating the roller accelerations through the constraint relations, leads to the equation of motion
$$\left(M+\frac m2+\frac m2\right)a = Mg,\frac{2x}{l}.$$
Thus
$$(M+m)a = \frac{2Mgx}{l}.$$
The initial acceleration is therefore
$$a= \frac{2Mg}{(M+m)l},x.$$
Its direction is toward the roller carrying the larger load.
Result
The board begins to accelerate toward the side on which its center of mass is closer. If the center of mass is displaced by a distance $x$ from the midpoint between the rollers, the initial acceleration is
$$\boxed{ a= \frac{2Mg}{(M+m)l},x }.$$
No numerical values are given in the statement, so a numerical substitution cannot be performed.
The final answer is
$$\boxed{ a= \frac{2Mgx}{(M+m)l} } \qquad \text{toward the more heavily loaded roller.}$$
Sanity Checks
The factor $x/l$ is dimensionless. The quantity $Mg/(M+m)$ has units of acceleration. The final expression has units of $\mathrm{m,s^{-2}}$, as required.
If $x=0$, the center of mass of the board lies midway between the rollers. Then
$$a=0,$$
which matches the symmetry of the system.
If $M\gg m$,
$$a\approx \frac{2gx}{l}.$$
The inertia of the rollers becomes negligible, and the acceleration is determined mainly by gravity and geometry.
If $m$ increases while $M$ remains fixed, the denominator grows and the acceleration decreases. Heavier rollers are harder to accelerate, which is physically reasonable.
The most common source of error is the rolling constraint. The velocity of the top point of a roller equals twice the velocity of its center, leading to
$$a=2a_c.$$
Missing this factor of two changes the effective inertia of the rollers and produces an incorrect coefficient in the final result.