Irodov 1.253: When Unwinding Sets the Fall – Amit & Deepali Sinha Foundation

Problem Irodov 1.253

A uniform cylinder of mass $m = 8.0$ kg and radius $R = 1.3$ cm hangs from two threads, one wound around each end of the cylinder, with the free upper end of each thread fixed to the ceiling directly above; the mass of the threads is negligible. Released from rest at time $t = 0$ , the cylinder begins to descend by unwinding both threads together. Find:

the tension in each thread and the angular acceleration of the cylinder;
how the instantaneous power delivered by gravity depends on time.

The cylinder hangs on two threads wound on its ends, each fixed to the ceiling above its own end.

Given

$m = 8.0$ kg
$R = 1.3$ cm
threads massless
released from rest at $t = 0$

Find

Find the tension in each thread, the angular acceleration

\beta

, and the power

P(t)

of gravity

Solution

Step 1What the two threads do, and what they cannot do

The two threads are wound at the two ends of the cylinder, at equal distances from its mid plane, so mirror symmetry about that plane fixes the two thread tensions to be equal:

T_{1} = T_{2} \equiv T.

The same symmetry gives their pulls no net moment about any axis through the center of mass at right angles to the cylinder’s own axis, which is what stops the cylinder tilting. About the cylinder’s own axis the two pulls are quite unbalanced, and they supply exactly the torque that spins it up, taken up in the equation of rotation below. So the cylinder cannot tilt and cannot translate sideways: it descends vertically while turning about its own axis of symmetry, which stays horizontal. This is what reduces a three dimensional apparatus to two scalar equations.

The free body. Each thread pulls up with T, gravity pulls down with mg at the center of mass, and the cylinder descends with acceleration w while spinning up at β. — The free body. Each thread pulls up with $T$ , gravity pulls down with $mg$ at the center of mass, and the cylinder descends with acceleration $w$ while spinning up at $\beta$ .

Step 2The constraint that links the fall to the spin

Take $x$ as the downward displacement of the axis from its position at $t = 0$ , and $\varphi$ as the angle the cylinder has turned through. The threads are inextensible and their upper ends are fixed, so every bit of descent has to be paid for by an equal length of thread unwinding off the rim. The length unwound when the cylinder turns through $\varphi$ is $R\varphi$ , so

x = R\varphi.

Differentiate this twice with respect to time. The radius is a constant, so it comes outside the derivative both times, and with $w = \ddot{x}$ and $\beta = \ddot{\varphi}$ ,

\begin{aligned} \dot{x} &= R\dot{\varphi}, \\ \ddot{x} &= R\ddot{\varphi}. \end{aligned}

w = \beta R

(1)

Step 3Newton’s second law for the center of mass

The external forces on the cylinder are the two tensions, each of magnitude $T$ and directed up, and the weight $mg$ directed down. Writing Newton’s second law for the center of mass along the downward vertical, with $w$ its acceleration,

m w = m g - 2T. \tag{2}

Step 4The equation of rotation about the central axis

Now take moments about the axis of symmetry through the center of mass. The weight acts at that center, so it has zero moment arm and contributes nothing. Each tension contributes the moment $TR$ , and both turn the cylinder the same way, so the two moments add to a net torque about the axis,

\tau = 2TR.

End on view of the cylinder as a free body. Both threads leave the rim at the same point in this projection, so the rim carries a total pull 2T whose moment arm about the axis is R. — End on view of the cylinder as a free body. Both threads leave the rim at the same point in this projection, so the rim carries a total pull $2T$ whose moment arm about the axis is $R$ .

The moment of inertia of a uniform solid cylinder about its own axis follows from the mass per unit radius. A shell of radius $r$ and thickness $dr$ in a cylinder of radius $R$ carries $dm = (2m r/R^2)\,dr$ , so

I = \int r^2\,dm = \int_0^R r^2\,\frac{2m r}{R^2}\,dr = \frac{2m}{R^2}\cdot\frac{R^4}{4} = \frac{mR^2}{2}.

The equation of rotation $I\beta = \tau$ therefore reads, with $\tau = 2TR$ from above,

\frac{mR^2}{2}\,\beta = 2TR. \tag{3}

Step 5Solving the pair

Use the constraint (1) to replace $\beta$ by $w/R$ in the rotational equation, so that both equations speak about the same unknown:

\begin{aligned} \frac{mR^2}{2}\cdot\frac{w}{R} &= 2TR, \\ \frac{mRw}{2} &= 2TR, \end{aligned}

and the radius cancels from both sides, leaving

w = \dfrac{4T}{m}

(4)

Substituting this $w$ into (2) eliminates the acceleration and leaves the tension alone:

\begin{aligned} m\cdot\frac{4T}{m} &= mg - 2T, \\ 4T &= mg - 2T, \\ 6T &= mg. \end{aligned}

Feed $T = mg/6$ back into $w = 4T/m$ for the linear acceleration, then use the constraint (1) to get the angular one:

\begin{aligned} w &= \frac{4T}{m} = \frac{4}{m}\cdot\frac{mg}{6} = \frac{2g}{3}, \\ \beta &= \frac{w}{R} = \frac{2g}{3R}. \end{aligned}

Collecting part (a), with $m = 8.0$ kg, $R = 1.3\times10^{-2}$ m and $g = 9.8\ \text{m/s}^2$ ,

\begin{gathered} T = \frac{mg}{6} = 13\ \text{N}\\ \beta = \frac{2g}{3R} = 5.0\times10^{2}\ \text{rad/s}^{2} \end{gathered}

(5)

Step 6The power developed by gravity

Instantaneous power is the scalar product of the force with the velocity of its point of application. Gravity acts at the center of mass, which moves straight down, and the weight points straight down too, so the two are parallel and

P = \vect{F}\cdot\vect{v} = mg\,v.

The cylinder starts from rest and $w = 2g/3$ is constant, since neither (2) nor (3) contains the time. Integrating $dv/dt = w$ from rest,

v = \int_0^t w\,dt' = w t = \frac{2g}{3}\,t.

Substituting this speed into $P = mgv$ gives the time dependence asked for:

P = mg \cdot \frac{2g}{3}\,t = \frac{2}{3}\,m g^{2} t

(6)

The coefficient evaluates to

\frac{2}{3}mg^{2} = \frac{2}{3}(8.0)(9.8)^{2} = 5.1\times10^{2}\ \text{W/s},

so the power grows by about half a kilowatt per second of fall.

Check

Energy accounts for the same answer independently. The threads do no work, because each acts at a rim point that is instantaneously at rest, so gravity’s power must equal the rate of growth of the kinetic energy. With $v = wt$ and $\omega = v/R$ , that energy is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ , and with $v = 2gt/3$ this is $K = \frac{1}{3}mg^2t^2$ , whose derivative $dK/dt = \frac{2}{3}mg^2 t$ is exactly the boxed $P$ . The dimensions also close: $[mg^2t] = \text{kg}\cdot(\text{m/s}^2)^2\cdot\text{s} = \text{kg}\,\text{m}^2/\text{s}^3 =$ W.

Remark

The fraction of $g$ the body falls with is set entirely by the shape. Repeating (2) and (3) with a thin hoop, $I = mR^2$ , gives $w = 2T/m$ and then $T = mg/4$ with $w = g/2$ : the hoop keeps more of its energy in rotation, so it falls more slowly and pulls its threads harder. A cylinder whose mass were concentrated on the axis, $I \to 0$ , would give $T \to 0$ and $w \to g$ , free fall with the threads slack.