Irodov 1.248: Two Contacts Brake One Cylinder – Amit & Deepali Sinha Foundation

Problem Irodov 1.248

A uniform cylinder of radius $R$ is spun up about its own axis to angular velocity $\omega_0$ and then set down into a right-angle corner formed by a vertical wall and a horizontal floor, so that it touches both surfaces at once. The coefficient of friction is $k$ at each of the two contacts. Find the number of turns the cylinder makes before friction brings it to rest.

The cylinder seen end on, wedged into the corner between the wall and the floor and spinning counterclockwise with ω_0. — The cylinder seen end on, wedged into the corner between the wall and the floor and spinning counterclockwise with $\vect\omega_0$ .

Given

$R$
$\omega_0$
$k$ at both contacts
uniform cylinder

Find

Find the number of turns

n

before it stops

Solution

Step 1Two contacts, and why both matter

The cylinder touches the floor and the wall at the same time, so there are two normal forces and two friction forces, one pair at each contact. Both frictions oppose the sliding of the rim over the surface it touches, and with the spin counterclockwise as drawn, the rim moves to the right where it meets the floor and downwards where it meets the wall. Friction therefore acts to the left at the floor and upwards at the wall, and both retard the rotation.

The five forces. The two normals act along their own surface normals at the two contact points, the two frictions along the surfaces, and the weight at the center.

Step 2The center does not move, so the forces balance

Wedged in the corner, the cylinder spins but does not translate, so the four contact forces and the weight sum to zero. Take $x$ horizontal, away from the wall, and $y$ vertically upwards. The normal from the wall is $N_2\vect{\hat x}$ and the friction from the floor is $-kN_1\vect{\hat x}$ ; the normal from the floor is $N_1\vect{\hat y}$ and the friction from the wall is $kN_2\vect{\hat y}$ . The two component equations are

\begin{aligned} x:&\quad N_2 - kN_1 = 0,\\ y:&\quad N_1 + kN_2 - mg = 0 . \end{aligned}

The first equation gives the wall’s normal in terms of the floor’s, and putting that into the second eliminates $N_2$ altogether:

\begin{align*}N_2 &= kN_1, \\ N_1 + k\,(kN_1) - mg &= 0, \\ N_1\,(1+k^{2}) &= mg .\end{align*}

Solving for $N_1$ and carrying it back into the first line gives both normals.

N_1 = \frac{mg}{1+k^{2}},\qquad N_2 = \frac{kmg}{1+k^{2}}

(1)

Step 3The retarding moment about the axis

Each friction force is tangent to the rim, so each has the moment arm $R$ about the axis, and both moments turn the cylinder the same way, against the spin. Their sum is

N_{\mathrm{fr}} = kN_1R + kN_2R = kR\,(N_1+N_2),

and inserting (1) collects the two contacts into one factor:

N_{\mathrm{fr}} = \frac{kmgR\,(1+k)}{1+k^{2}}

(2)

Step 4From the moment to the number of turns

The moment is constant, so the deceleration is uniform. With $I = \frac{1}{2}mR^{2}$ for a uniform cylinder about its axis, the equation of rotation gives the magnitude of the angular acceleration,

\beta = \frac{N_{\mathrm{fr}}}{I} = \frac{kmgR(1+k)}{1+k^{2}}\cdot\frac{2}{mR^{2}} = \frac{2kg(1+k)}{R(1+k^{2})} .

The mass has canceled, as it must, since both the friction and the inertia are proportional to it. The deceleration being uniform, the total angle $\Phi$ swept in coming from $\omega_0$ to rest follows from the rotational analogue of the uniform-acceleration relation between speed and distance,

\omega_0^{2} = 2\beta\Phi ,

which gives the angle once the deceleration just found is inserted:

\begin{align*}\Phi &= \frac{\omega_0^{2}}{2\beta} \\ &= \frac{\omega_0^{2}}{2}\cdot\frac{R\,(1+k^{2})}{2kg\,(1+k)} \\ &= \frac{\omega_0^{2}R\,(1+k^{2})}{4kg\,(1+k)} .\end{align*}

The number of turns is that angle divided by $2\pi$ :

n = \frac{\Phi}{2\pi} = \frac{(1+k^{2})\,\omega_0^{2}R}{8\pi k\,(k+1)\,g}

(3)

Check

The expression is dimensionless, since $\omega_0^{2}R/g$ is, and it behaves correctly at both ends of the friction range. As $k \to 0$ the answer diverges: with no friction there is no retarding moment and the cylinder never stops. As $k \to \infty$ the normals themselves are throttled, $N_1 \to mg/k^{2}$ , but the retarding moment does not fade away with them: (2) saturates at $mgR$ , so the turn count settles on the finite value $\omega_0^{2}R/(8\pi g)$ rather than dropping to zero. No corner is sticky enough to stop the cylinder in appreciably fewer turns than that, and the count is smallest not at large $k$ but at $k = 1+\sqrt{2}$ , where it is $0.83$ of that value.

Remark

Only the leading power of $k$ comes from the floor. Expanding the answer for small $k$ gives $n \approx \omega_0^{2}R/(8\pi kg)$ , which is what the floor alone would give; the wall’s contribution enters at the next order, through the factors $1+k^{2}$ and $1+k$ .