Irodov 1.70: The Reeling Motor Chases Itself – Amit & Deepali Sinha Foundation

Problem Irodov 1.70

A bar and a motor-and-battery block both sit on the same horizontal plane, coefficient of kinetic friction $k$ with the plane, separated by distance $l$ . A thread runs from the bar to the motor’s shaft, which winds it in. The bar’s mass is twice the block’s mass. Once the motor starts, the bar moves toward the block with constant acceleration $w$ . Find the time $\tau$ until the two bodies meet.

The bar (mass 2m_0) and the motor block (mass m_0), separated by l on the frictional plane, joined by the thread the motor winds in. — The bar (mass $2m_0$ ) and the motor block (mass $m_0$ ), separated by $l$ on the frictional plane, joined by the thread the motor winds in.

Given

$l$
$k$
$m_{\text{bar}} = 2m_0$
$w$ = bar’s acceleration

Find

the time

\tau

until collision

Solution

Step 1Two free bodies, one thread

Bar: normal N_1, weight, thread tension T toward the motor, kinetic friction f_1 opposing its motion. Each panel draws its own body's motion as rightward, so the two T arrows appear parallel. — Bar: normal $\vect N_1$ , weight, thread tension $\vect T$ toward the motor, kinetic friction $\vect f_1$ opposing its motion. Each panel draws its own body’s motion as rightward, so the two $\vect T$ arrows appear parallel.

Motor block: normal N_2, weight, the reaction tension T toward the bar (same thread, same magnitude), kinetic friction f_2 opposing its own motion. — Motor block: normal $\vect N_2$ , weight, the reaction tension $\vect T$ toward the bar (same thread, same magnitude), kinetic friction $\vect f_2$ opposing its own motion.

The thread is a single massless strand, so it pulls the bar and the motor block toward each other with the same tension $T$ : the motor’s shaft winds the thread in, but by Newton’s third law that same winding pulls the shaft, and with it the whole motor block, toward the bar just as hard. Kinetic friction on each body is $k$ times its own weight, opposing that body’s own motion.

Step 2Newton’s second law for the bar

Taking the bar’s direction of motion (toward the motor) as positive, with mass $2m_0$ ,

2m_0 w = T - k(2m_0)g \;\Longrightarrow\; T = 2m_0(w+kg). \tag{1}

Step 3Newton’s second law for the motor block

Taking the block’s direction of motion (toward the bar) as positive, with mass $m_0$ and its own acceleration $w_2$ ,

m_0 w_2 = T - km_0 g \;\Longrightarrow\; w_2 = \frac{T}{m_0} - kg.

Substituting $T$ from (1),

\begin{align*}w_2 &= \frac{2m_0(w+kg)}{m_0} - kg \\ &= 2(w+kg) - kg\end{align*}

w_2 = 2w+kg

(2)

Step 4Closing speed and the time to collide

The two bodies start at rest, a distance $l$ apart, and approach each other with constant accelerations $w$ and $w_2$ from (2); their separation shrinks at the combined rate $w+w_2$ .

l = \frac12(w+w_2)\tau^2 = \frac12\bigl(w+2w+kg\bigr)\tau^2 = \frac12(3w+kg)\tau^2

Solving for $\tau$ ,

\tau = \sqrt{\dfrac{2l}{3w+kg}}

(3)

Check

With $k=0$ (frictionless plane), $\tau=\sqrt{2l/3w}$ , and (2) gives $w_2=2w$ : the tension is the only force on either body, an internal action-reaction pair, so momentum conservation on the two-body system, $2m_0 w = m_0 w_2$ , independently forces $w_2=2w$ , matching the boxed relation with $k$ set to zero.

Remark

The friction correction in (2) is $+kg$ , not $+2kg$ or any other multiple: kinetic friction contributes an acceleration $kg$ to a sliding body regardless of that body’s own mass, since the friction force ( $k$ times the body’s own weight) and the body’s own inertia both scale with the same mass and cancel. That is why the bar and the motor block, despite differing by a factor of two in mass, each lose exactly the same $kg$ from their momentum-conservation values, and the closing acceleration ends up as $3w+kg$ rather than some mass-weighted combination.