IPhO 1970 · Theory problem 1

Motor-driven carriage on a bar sliding on a table

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A long bar with the mass \(M = 1\) kg is placed on a smooth horizontal surface of a table where it can move frictionless. A carriage equipped with a motor can slide along the upper horizontal panel of the bar, the mass of the carriage is \(m = 0.1\) kg. The friction coefficient of the carriage is \(\mu = 0.02\).

The motor is winding a thread around a shaft at a constant speed \(v_0 = 0.1\) m/s. The other end of the thread is tied up to a rather distant stationary support in one case (Fig.1, a), whereas in the other case it is attached to a picket at the edge of the bar (Fig.1, b). While holding the bar fixed one allows the carriage to start moving at the velocity \(V_0\) then the bar is let loose.

Figure 1

Fig. 1

By the moment the bar is released the front edge of the carriage is at the distance \(l = 0.5\) m from the front edge of the bar. For both cases find the laws of movement of both the bar and the carriage and the time during which the carriage will reach the front edge of the bar.

Official English text of the problem.

Answers

Take \(t=0\) at release and the forward direction as positive. Let \(x_b\) and \(x_c\) denote the bar’s and carriage’s displacements from their respective initial positions. Using \(g=9.8\,\mathrm{m\,s^{-2}}\), define \[ a=\frac{\mu mg}{M}=0.0196\,\mathrm{m\,s^{-2}}, \qquad t_*=\frac{v_0}{a}\approx5.1\,\mathrm{s}. \]

  • (a) Thread attached to the stationary support: \[ \boxed{x_c(t)=v_0t,} \qquad \boxed{ x_b(t)= \begin{cases} \dfrac12at^2,&0\le t\le t_*,\\[2mm] v_0t-\dfrac{v_0^2}{2a},&t\ge t_*. \end{cases}} \] After \(t_*\), both bodies move at \(0.1\,\mathrm{m\,s^{-1}}\). The carriage never reaches the front edge: the remaining gap is \[ \boxed{d_{\min}=l-\frac{Mv_0^2}{2\mu mg}\approx0.25\,\mathrm{m}.} \]

  • (b) Thread attached to the bar: \[ \boxed{x_b(t)=0,\qquad x_c(t)=v_0t.} \] The bar remains at rest, and the carriage reaches its front edge after \[ \boxed{t_{\mathrm{edge}}=\frac{l}{v_0}=5.0\,\mathrm{s}.} \]

Solution

Treat the thread as massless and inextensible, and interpret \(v_0\) as its constant linear winding speed. The thread is horizontal, so the normal force on the carriage is \(mg\). During sliding, the friction force has magnitude \[ f=\mu mg. \] The table exerts no horizontal force on the bar.

Before release, the bar and the thread’s attachment point are stationary in both arrangements. Consequently, the carriage’s initial velocity is \[ V_0=v_0. \] Removing the restraint produces no impulse, so immediately after release \[ v_b(0)=0,\qquad v_c(0)=v_0. \]

(a) Thread attached to the stationary support

Because the thread’s far end is fixed, winding it at speed \(v_0\) makes the carriage approach that end at the same speed: \[ v_c=v_0,\qquad x_c=v_0t. \]

Initially the carriage slides forward relative to the bar. Friction acts backward on the carriage and forward on the bar. If \(T\) is the thread tension, Newton’s second law gives \[ m\dot v_c=T-f=0, \qquad M\dot v_b=f. \] Thus \(T=f\), and the bar has constant acceleration \[ a=\frac{f}{M}=\frac{\mu mg}{M}. \] As long as the carriage is sliding forward, \[ v_b=at,\qquad x_b=\frac12at^2. \]

The relative velocity is \[ v_c-v_b=v_0-at. \] It vanishes at \[ t_*=\frac{v_0}{a} =\frac{Mv_0}{\mu mg} \approx5.1\,\mathrm{s}. \]

The carriage’s forward displacement relative to the bar up to this instant is \[ s_{\max}=x_c(t_*)-x_b(t_*) =v_0t_*-\frac12at_*^2 =\frac{v_0^2}{2a} =\frac{Mv_0^2}{2\mu mg}. \] Numerically, \[ s_{\max} =\frac{1(0.1)^2}{2(0.02)(0.1)(9.8)} \approx0.255\,\mathrm{m}. \] Since \(s_{\max}<l=0.5\,\mathrm{m}\), the carriage has not reached the edge when the relative motion stops.

At \(t_*\), both bodies have velocity \(v_0\). They can thereafter continue together without horizontal forces: friction and tension are both zero. The sliding-friction acceleration must therefore cease at this instant. For \(t\ge t_*\), \[ v_b=v_c=v_0, \] and \[ x_b(t)=x_b(t_*)+v_0(t-t_*) =v_0t-\frac{v_0^2}{2a}. \]

The separation between the front edges is \[ d(t)=l+x_b(t)-x_c(t) = \begin{cases} l-v_0t+\dfrac12at^2,&0\le t\le t_*,\\[2mm] l-\dfrac{v_0^2}{2a},&t\ge t_*. \end{cases} \] Its final value is approximately \(0.245\,\mathrm{m}\), so there is no finite arrival time.

(b) Thread attached to the bar

Here the motor shortens the distance between the carriage and a point fixed to the bar. Therefore, \[ v_c-v_b=v_0. \]

All horizontal forces are internal to the bar–carriage system: the thread pulls the carriage forward and the bar backward, while friction supplies an equal-and-opposite pair in the other directions. Hence total horizontal momentum is conserved: \[ Mv_b+mv_c=mv_0, \] where the right-hand side is the momentum at release.

Substituting \(v_c=v_b+v_0\), \[ Mv_b+m(v_b+v_0)=mv_0, \] so \[ v_b=0,\qquad v_c=v_0. \] Thus \[ x_b(t)=0,\qquad x_c(t)=v_0t. \] The tension equals the sliding friction, \(T=\mu mg\), giving zero net force on each body.

The front-edge separation decreases uniformly: \[ d(t)=l-v_0t. \] Setting \(d=0\) gives \[ \boxed{t_{\mathrm{edge}}=\frac{0.5\,\mathrm{m}}{0.1\,\mathrm{m\,s^{-1}}} =5.0\,\mathrm{s}.} \]

The solution was checked against the official solution.