Irodov 1.260: The Thread End Outruns Both Bodies – Amit & Deepali Sinha Foundation

Problem Irodov 1.260

A uniform solid cylinder $A$ of mass $m_1$ turns without friction about a horizontal axle carried by a mount $B$ of mass $m_2$ , so the cylinder’s axis and the mount move together. The mount stands on a horizontal plane and slides along it with no friction at all. A light thread is wound tightly on the cylinder and leaves it horizontally at the top; its free end $K$ is pulled by a constant horizontal force $F$ , and the system starts from rest. Find:

the acceleration of the point $K$
the kinetic energy of the whole system $t$ seconds after the motion begins

The arrangement. The cylinder A of mass m_1 turns on an axle carried by the mount B of mass m_2, which slides on the smooth floor. The thread leaves the rim horizontally, and F pulls its free end K. — The arrangement. The cylinder $A$ of mass $m_1$ turns on an axle carried by the mount $B$ of mass $m_2$ , which slides on the smooth floor. The thread leaves the rim horizontally, and $\vect F$ pulls its free end $K$ .

Given

$m_1$
$m_2$
$F$ , constant and horizontal
no friction at the axle or under the mount
from rest

Find

(a) the acceleration of

K

, (b) the kinetic energy after time

t

Solution

Step 1How fast the axle itself moves off

Take the cylinder and the mount together. The floor is smooth and its reaction is vertical, the weights are vertical, and the only horizontal force from outside is the pull the thread transmits, which is $F$ because the thread is light. The axle ties the cylinder’s center to the mount, so both bodies share one horizontal acceleration $w_0$ .

\begin{align*}(m_1 + m_2)\,w_0 &= F\end{align*}

w_0 = \dfrac{F}{m_1 + m_2}

(1)

Step 2How fast the cylinder spins up

Now take the cylinder alone and its rotation about its own axis. The axle’s reaction acts on that axis and so has no moment about it, and the axle is frictionless, so the thread is the only thing that can turn the cylinder: it pulls tangentially at the rim, a distance $R$ from the axis, with $R$ the cylinder’s radius. For a uniform solid cylinder the moment of inertia is $m_1R^2/2$ , and $\beta$ is the angular acceleration.

\begin{align*}\frac{m_1R^2}{2}\,\beta &= FR\end{align*}

\beta = \dfrac{2F}{m_1 R}

(2)

The kinematics of the thread's end. The cylinder's center accelerates at w_0 while the cylinder spins up at β, and the free end K of the thread accelerates at w_K. — The kinematics of the thread’s end. The cylinder’s center accelerates at $\vect w_0$ while the cylinder spins up at $\beta$ , and the free end $K$ of the thread accelerates at $\vect w_K$ .

Step 3The acceleration of the thread’s end

The point of the rim where the thread leaves moves with the center and, in addition, turns about it, so its acceleration along the thread is $w_0$ plus the tangential contribution $\beta R$ . The straight part of the thread lengthens at exactly the rate the rim pays it out, so $K$ takes that same acceleration. Using (1) and (2),

\begin{align*}w_K &= w_0 + \beta R \\ &= \frac{F}{m_1 + m_2} + \frac{2F}{m_1 R}\,R \\ &= \frac{F}{m_1 + m_2} + \frac{2F}{m_1} \\ &= \frac{F\,\bigl[m_1 + 2(m_1 + m_2)\bigr]}{m_1(m_1 + m_2)}\end{align*}

w_K = \dfrac{F(3m_1 + 2m_2)}{m_1(m_1 + m_2)}

(3)

The radius has canceled: a bigger cylinder is harder to spin up, and pays out more thread per turn, in exactly the same proportion.

Step 4The kinetic energy after

t

seconds

Both accelerations are constant, so after a time $t$ the axle and the mount are moving at $w_0t$ and the cylinder is spinning at $\beta t$ . The energy of the system is then the translation of the two bodies plus the cylinder’s rotation about its own axis.

\begin{align*}T &= \frac{1}{2}(m_1 + m_2)\,(w_0 t)^2 + \frac{1}{2}\cdot\frac{m_1R^2}{2}\,(\beta t)^2 \\ &= \frac{1}{2}(m_1 + m_2)\,\frac{F^2t^2}{(m_1+m_2)^2} + \frac{m_1R^2}{4}\cdot\frac{4F^2t^2}{m_1^2R^2} \\ &= \frac{F^2t^2}{2(m_1+m_2)} + \frac{F^2t^2}{m_1} \\ &= \frac{F^2t^2\bigl[m_1 + 2(m_1+m_2)\bigr]}{2m_1(m_1+m_2)}\end{align*}

T = \dfrac{F^2t^2(3m_1 + 2m_2)}{2m_1(m_1 + m_2)}

(4)

Check

The two answers must agree through the work-energy theorem, and they were reached by different routes. The only force that does any work is $\vect F$ , since the floor’s reaction and both weights are perpendicular to the motion and the axle’s forces on the cylinder and on the mount are equal, opposite and applied at the same moving point. Its point of application travels $w_Kt^2/2$ , so the work is $F\,w_Kt^2/2 = F^2t^2(3m_1+2m_2)/[2m_1(m_1+m_2)]$ , which is the boxed energy exactly. As a limit, letting $m_2\to\infty$ makes the mount immovable and leaves $w_K \to 2F/m_1$ , which is $\beta R$ alone, the pure unwinding of a cylinder on a fixed axle.

Remark

The work done by a force is the force times the displacement of the point it acts on, and here that point is neither body: $K$ runs ahead of the mount by exactly the length of thread that has been unwound. Take the two displacements apart and the energy separates with them. The mount and axle advance $w_0t^2/2$ , and $F$ times that is $F^2t^2/[2(m_1+m_2)]$ , precisely the translational kinetic energy; the extra travel of $K$ , which is $\beta R\,t^2/2$ , accounts for $F^2t^2/m_1$ , precisely the rotational kinetic energy. The unwound thread is the channel through which the spin is paid for.