Irodov 1.254: The Car Only Changes Effective Gravity

Problem Irodov 1.254

Thin threads are wound tightly on the two ends of a uniform solid cylinder of mass $m$ , and the free end of each thread is fastened to the ceiling of an elevator car, so that the cylinder hangs inside the car and can unwind. The car then starts moving upward with the acceleration $\vect w_0$ . Find the acceleration $\vect w'$ of the cylinder relative to the car, and the force $\vect F$ that the cylinder exerts on the ceiling through the two threads.

The cylinder hanging from the ceiling of the car. Seen along the axis, the two threads lie one behind the other and are drawn as a single vertical strand leaving the rim tangentially. The car accelerates upward with w_0. — The cylinder hanging from the ceiling of the car. Seen along the axis, the two threads lie one behind the other and are drawn as a single vertical strand leaving the rim tangentially. The car accelerates upward with $\vect w_0$ .

Given

$m$
$\vect w_0$ , upward

Find

\vect w'

, the cylinder’s acceleration relative to the car ·

\vect F

, its pull on the ceiling

Solution

Step 1The bodies, the forces and the one moment of inertia needed

Everything will be written in the ground frame, where Newton’s laws hold as they stand, so the car’s motion enters only through the thread. Let $R$ be the cylinder’s radius and let $F$ be the combined pull of the two threads; by Newton’s third law that same $F$ is what the cylinder exerts on the ceiling. The cylinder is uniform and solid, so its moment of inertia about its own axis is

I = \frac{mR^2}{2} \tag{1}

The cylinder as a free body in the ground frame. The two threads pull upward with F at the point where they leave the rim, the weight m g acts at the center, and unwinding gives the angular acceleration β. — The cylinder as a free body in the ground frame. The two threads pull upward with $\vect F$ at the point where they leave the rim, the weight $m\vect g$ acts at the center, and unwinding gives the angular acceleration $\vect\beta$ .

Step 2What the inextensible thread demands

The thread is vertical and cannot stretch, and its upper end is fixed to the ceiling, so the material point of the cylinder at which the thread leaves the rim can gain no height on the ceiling and lose none to it: its vertical acceleration is $w_0$ upward, the same as the ceiling’s. Only that one component is tied down. The same point also carries a centripetal acceleration toward the axis, which is horizontal here, and about which a vertical thread says nothing; the equation below is the vertical projection alone. Taking upward as positive, writing $a$ for the cylinder’s own (absolute) acceleration and $\beta$ for its angular acceleration in the sense the figure shows, the departure point sits a distance $R$ from the axis and is carried upward by the rotation.

\begin{align*}a + \beta R &= w_0 \\ \beta R &= w_0 - a\end{align*}

The cylinder’s acceleration relative to the car is the difference of the two absolute accelerations, and the line above shows at once which way it points: $\beta R$ is positive while the cylinder unwinds, so the difference is negative, that is, directed downward. Write $w'$ for its magnitude:

w' = w_0 - a, \qquad\text{so}\qquad a = w_0 - w'

(2)

With that definition the constraint above reads

w' = \beta R

(3)

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

The center of mass responds to the vector sum of the external forces, which are the thread’s pull upward and the weight downward. Taking upward as positive again,

F - mg = m a ,

and substituting the absolute acceleration from (2) puts the equation in terms of the relative one.

F - mg = m\,(w_0 - w')

(4)

Step 4The moment equation about the cylinder’s own axis

The weight acts at the center of mass and so has no moment about the axis through it; only the thread’s pull does, at the arm $R$ . Substituting the moment of inertia (1) and the constraint (3), which supplies $\beta = w'/R$ , eliminates both $\beta$ and $R$ .

\begin{align*}F R &= I\beta \\ &= \frac{mR^2}{2}\cdot\frac{w'}{R}\end{align*}

F = \dfrac{mw'}{2}

(5)

Step 5The acceleration relative to the car

Equations (4) and (5) are two equations in the two unknowns $F$ and $w'$ . Putting the second into the first leaves the relative acceleration alone.

\begin{align*}\frac{mw'}{2} - mg &= m(w_0 - w') \\ \frac{3}{2}\,w' &= g + w_0\end{align*}

w' = \dfrac{2}{3}(g + w_0)

(6)

Directed vertically downward relative to the car.

Step 6The pull on the ceiling

The force follows from (5) once $w'$ is known.

F = \frac{mw'}{2} = \frac{m}{2}\cdot\frac{2}{3}(g+w_0)

F = \dfrac{1}{3}\,m(g + w_0)

(7)

Directed vertically downward on the ceiling.

Check

The two results must satisfy (4), and they do: the cylinder’s absolute acceleration is $a = w_0 - w' = (w_0 - 2g)/3$ , and $mg + ma = m(3g + w_0 - 2g)/3$ , which is exactly the boxed $F$ . Two special cases are also right. With $w_0 = 0$ the car stands still and the formulas give $w' = 2g/3$ and $F = mg/3$ , the familiar falling spool that descends at two thirds of $g$ while the threads carry only a third of its weight. And if the car were cut loose into free fall, its acceleration being $g$ downward, then $w_0 = -g$ and both $w'$ and $F$ vanish: inside a freely falling car the cylinder is weightless, nothing unwinds and the threads go slack.

Remark

The car never appears in the answers on its own, only ever in the combination $g + w_0$ , which is the effective gravity an observer riding in the car would measure. That is the whole content of the result, and it is what lets the two answers be written compactly as $\vect w' = \frac{2}{3}(\vect g - \vect w_0)$ and $\vect F = \frac{1}{3}m(\vect g - \vect w_0)$ , forms that stay correct whichever way the car accelerates, since $\vect g$ and $\vect w_0$ point opposite ways when the car rises and the same way when it descends. The numbers $2/3$ and $1/3$ are the cylinder’s shape and nothing else: redoing the moment equation and the elimination with a general $I$ in place of (1) gives $w' = (g+w_0)/(1 + I/mR^2)$ , so a thin hoop wound the same way, with $I = mR^2$ , would descend at only half the effective gravity and hang on the ceiling with half its effective weight.