Irodov 1.247: Half the Pulley Rides Along – Amit & Deepali Sinha Foundation

Problem Irodov 1.247

A body of mass $m_1$ rests on a horizontal plane, the coefficient of friction between the two being $k$ . A thread tied to that body runs horizontally to a pulley mounted at the edge of the plane, passes over the pulley, and then hangs vertically, carrying a second body of mass $m_2$ . The pulley is a uniform disc of mass $m$ and the thread does not slip on its rim; the thread’s own mass and the friction in the pulley’s axle are both negligible. At the instant $t = 0$ the body $m_2$ begins to descend. Find the work done by the friction force acting on the body $m_1$ during the first $t$ seconds of the motion.

The arrangement. The body m_1 slides on the horizontal plane, the thread runs from it horizontally to the pulley of mass m carried on a bracket at the plane's edge, wraps a quarter of the rim, and hangs vertically to the descending body m_2. — The arrangement. The body $m_1$ slides on the horizontal plane, the thread runs from it horizontally to the pulley of mass $m$ carried on a bracket at the plane’s edge, wraps a quarter of the rim, and hangs vertically to the descending body $m_2$ .

Given

$m_1$
$m_2$
$m$
$k$
$t$

Find

A

, the work of the friction force on

m_1

over the first

t

seconds

Solution

Step 1What the inextensible, non-slipping thread ties together

Because the thread does not stretch, the two bodies move with equal speeds and equal accelerations; because it does not slip on the rim, the rim’s material points move with the thread. Write $a$ for that common acceleration, $\beta$ for the pulley’s angular acceleration and $R$ for its radius. The rim’s tangential acceleration is $\beta R$ and it must equal the thread’s:

a = \beta R \tag{1}

The pulley is a uniform disc, so its moment of inertia about its own axis is

I = \frac{mR^2}{2}

The body m_1 as a free body, its four forces drawn concurrent at the center of gravity: the thread's pull T_1, the friction f opposing the sliding, the normal force N from the plane and the weight m_1 g. — The body $m_1$ as a free body, its four forces drawn concurrent at the center of gravity: the thread’s pull $\vect T_1$ , the friction $\vect f$ opposing the sliding, the normal force $\vect N$ from the plane and the weight $m_1\vect g$ .

Step 2The two sliding bodies

Neither body rotates, so each obeys Newton’s second law alone. The body $m_1$ has no vertical acceleration, which fixes the normal force and with it the kinetic friction $f = kN$ ; the thread pulls it forward and the friction retards it. The body $m_2$ falls under its weight against the thread.

\begin{align*}N &= m_1 g \\ T_1 - k m_1 g &= m_1 a \\ m_2 g - T_2 &= m_2 a\end{align*}

The two tensions are written separately on purpose: a pulley with mass cannot pass one on to the other unchanged.

The pulley as a free body. Each strand pulls tangentially at its own departure point, T_1 at the top and T_2 at the side, so both have the same moment arm R about the axle. The angular acceleration β is clockwise. — The pulley as a free body. Each strand pulls tangentially at its own departure point, $\vect T_1$ at the top and $\vect T_2$ at the side, so both have the same moment arm $R$ about the axle. The angular acceleration $\beta$ is clockwise.

Step 3The pulley, which the two tensions have to spin up

Take moments about the pulley’s axle. The axle reaction and the pulley’s weight both act on the axis, so neither has a moment there, and the two tensions act tangentially at arm $R$ in opposite senses. Substituting the disc’s moment of inertia and the constraint (1) turns the moment equation into a relation between the tensions and the linear acceleration.

\begin{align*}(T_2 - T_1)R &= I\beta \\ &= \frac{mR^2}{2}\cdot\frac{a}{R}\end{align*}

T_2 - T_1 = \dfrac{ma}{2}

(2)

Step 4The acceleration of the machine

The three dynamical equations, one per body, contain the two tensions and the acceleration. Adding them all makes both tensions cancel, because each appears once with either sign, and leaves the acceleration alone.

\begin{align*}T_1 - k m_1 g &= m_1 a \\ m_2 g - T_2 &= m_2 a \\ T_2 - T_1 &= \frac{ma}{2}\end{align*}

Adding the three left-hand sides and the three right-hand sides,

m_2 g - k m_1 g = \left(m_1 + m_2 + \frac{m}{2}\right) a

a = \dfrac{2(m_2 - km_1)g}{m + 2(m_1+m_2)}

(3)

Step 5The work the friction does in

t

seconds

The friction force on $m_1$ has the constant magnitude $km_1g$ and points opposite to the displacement throughout, so its work is minus that force times the distance covered. The motion starts from rest with the constant acceleration (3), so the distance follows from the uniform-acceleration formula.

\begin{align*}s &= \frac{a t^2}{2} \\ A &= -k m_1 g\,s \\ &= -k m_1 g\cdot\frac{t^2}{2}\cdot\frac{2(m_2-km_1)g}{m+2(m_1+m_2)}\end{align*}

A = -\dfrac{(m_2 - km_1)\,km_1g^2t^2}{m + 2(m_1+m_2)}

(4)

Check

The bracket $k m_1 g^2 t^2$ carries units $\mathrm{kg}\cdot(\mathrm{m/s^2})^2\cdot\mathrm{s^2} = \mathrm{kg\cdot m^2/s^2}$ , a joule, and the mass ratio beside it is dimensionless, so the result is an energy. Three special cases come out right. With $k = 0$ the friction and its work both vanish. With $m_2 = km_1$ the numerator of the acceleration vanishes, the machine is exactly on the point of moving and no distance is covered, so again $A = 0$ ; and since the statement says $m_2$ does descend, $m_2 > km_1$ and the boxed work is genuinely negative, as the work of a retarding force must be. With $m = 0$ the acceleration collapses to $(m_2-km_1)g/(m_1+m_2)$ , the familiar result for a massless pulley, in which the two tensions become equal by (2).

Remark

The pulley enters the answer only through the combination $I/R^2 = m/2$ , which is why its radius never appears: a uniform disc of any size opposes the thread with an effective inertia of exactly half its mass, and that half-mass simply joins $m_1$ and $m_2$ in the total the driving force has to accelerate. The same trick works for any wheel driven at its rim, with $m/2$ replaced by $I/R^2$ , so a rim-loaded spoked wheel would count for nearly its whole mass and a uniform disc for half of it, whatever either of them weighs. Note also what the pulley’s mass costs structurally: the two strands must carry different tensions, since (2) makes the difference proportional to $ma$ , and the habit of writing one tension throughout a thread is exactly the habit that fails here.