Irodov 1.243: Half the Cylinder Joins the Load – Amit & Deepali Sinha Foundation

Problem Irodov 1.243

A uniform solid cylinder of mass $M$ and radius $R$ turns freely about a fixed horizontal axle through its own axis, carried by a bracket; the friction in the axle is negligible. A light inextensible thread is wound on the cylinder’s lateral surface and hangs vertically from the rim, and a body of mass $m$ is tied to its free end. The thread does not slip on the cylinder. The system is at rest until the moment $t = 0$ , when it is released and begins to move. Find, as functions of the elapsed time $t$ :

the angular velocity $\omega$ of the cylinder;
the kinetic energy $K$ of the whole system, cylinder and hanging body together.

The cylinder of mass M turns on a fixed horizontal axle held by a hatched bracket. The light thread, wound on its lateral surface, leaves the rim vertically and carries the body of mass m. The cylinder turns clockwise with angular velocity ω. — The cylinder of mass $M$ turns on a fixed horizontal axle held by a hatched bracket. The light thread, wound on its lateral surface, leaves the rim vertically and carries the body of mass $m$ . The cylinder turns clockwise with angular velocity $\omega$ .

The cylinder alone. The thread pulls tangentially at the rim with tension T, and the gray leader from the axle to that rim point is the moment arm, the full radius R. The resulting angular acceleration β is clockwise. — The cylinder alone. The thread pulls tangentially at the rim with tension $\vect T$ , and the gray leader from the axle to that rim point is the moment arm, the full radius $R$ . The resulting angular acceleration $\beta$ is clockwise.

The hanging body alone. It does not rotate, so its weight $m\vect g$ and the thread tension $\vect T$ are drawn concurrent at its center of gravity. The weight is the longer arrow because the body accelerates downward.

Given

$M$
$R$
$m$
axle friction negligible
at rest at $t=0$

Find

(a)

\omega(t)

, (b) the total kinetic energy

K(t)

Solution

Step 1Newton’s second law for the hanging body

The body is acted on by its weight and by the thread tension, both vertical, and it accelerates downward with some magnitude $a$ . Taking the downward direction as positive gives one equation in the two unknowns $a$ and $T$ .

ma = mg - T \tag{1}

Step 2The torque equation supplies the second relation

The cylinder turns about a fixed frictionless axle, so the only torque about that axle is the one the tension exerts at the rim, with moment arm $R$ . Its moment of inertia about its own axis is that of a uniform solid cylinder, and the thread does not slip, so the rim’s tangential acceleration equals the body’s acceleration $a$ , which ties $\beta$ to $a$ .

I = \frac{MR^2}{2}, \qquad \beta = \frac{a}{R}

Writing the rotational equation of motion and substituting both of these,

\begin{align*}TR &= I\beta \\ &= \frac{MR^2}{2}\cdot\frac{a}{R} \\ &= \frac{MRa}{2}\end{align*}

and canceling the common factor $R$ ,

T = \dfrac{Ma}{2}

(2)

Step 3The acceleration of the load

Eliminating the tension between (1) and (2) leaves $a$ alone.

\begin{align*}ma &= mg - \frac{Ma}{2} \\ ma + \frac{Ma}{2} &= mg \\ a\left(m + \frac{M}{2}\right) &= mg\end{align*}

a = \dfrac{mg}{m + \dfrac{M}{2}} = \dfrac{g}{1 + \dfrac{M}{2m}}

(3)

Step 4Angular velocity as a function of time

Every quantity on the right of (3) is constant, so $a$ is constant and with it the angular acceleration $\beta = a/R$ . Starting from rest, the angular velocity therefore grows linearly.

\omega(t) = \beta t = \frac{a t}{R}

Substituting the acceleration (3) gives the answer to part (a).

\omega(t) = \dfrac{gt}{R\left(1 + \dfrac{M}{2m}\right)}

(4)

Step 5Kinetic energy of the two bodies together

The system’s energy is the body’s translational part plus the cylinder’s rotational part. At time $t$ the body has fallen with speed $v = at$ and the cylinder turns at $\omega = at/R$ , so both parts can be written in terms of $a$ alone.

\begin{aligned} K &= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \\ &= \frac{1}{2}m(at)^2 + \frac{1}{2}\cdot\frac{MR^2}{2}\cdot\frac{a^2t^2}{R^2} \\ &= \frac{1}{2}a^2t^2\left(m + \frac{M}{2}\right) \end{aligned} \tag{5}

Now substituting the acceleration (3),

\begin{align*}K &= \frac{1}{2}t^2\left(\frac{mg}{m + \dfrac{M}{2}}\right)^{2}\left(m + \frac{M}{2}\right) \\ &= \frac{m^2g^2t^2}{2\left(m + \dfrac{M}{2}\right)}\end{align*}

Dividing numerator and denominator by $m$ puts part (b) in the same shape as part (a).

K(t) = \dfrac{mg^2t^2}{2\left(1 + \dfrac{M}{2m}\right)}

(6)

Check

Energy bookkeeping. In time $t$ the body has fallen a distance $at^2/2$ , so gravity has done work $mg\,at^2/2$ on it. Rearranged, (3) says the weight is $mg = a(m+M/2)$ , and putting that into the work makes it $a^2t^2(m+M/2)/2$ , which is exactly the kinetic energy (5), as it must be with a frictionless axle and an inextensible thread. The limit $M \to 0$ removes the cylinder from both answers: the acceleration becomes $g$ , the angular velocity becomes $gt/R$ , and the energy becomes $mg^2t^2/2$ , the free fall values.

Remark

The cylinder never appears as a moment of inertia in the final answers, only through the combination $I/R^2 = M/2$ , which has the dimensions of mass. The whole system behaves as a single body of mass $m + I/R^2$ pulled by the weight $mg$ of the load alone, and that is why the acceleration comes out smaller than $g$ by exactly the factor $1 + M/2m$ . The same reading fixes what a different drum would do: a thin ring of the same mass $M$ wound at the same radius has $I = MR^2$ , so it would contribute a full $M$ rather than $M/2$ , and the load would accelerate more slowly still.