Irodov 1.195: Normal and Weight Torques Quietly Cancel

Problem Irodov 1.195

A uniform sphere of mass $m$ and radius $R$ is released from rest on a plane inclined at angle $\alpha$ to the horizontal and rolls down it without slipping. Find the angular momentum $M(t)$ of the sphere, taken about the fixed point on the incline that coincided with the contact point at the initial moment, as a function of the time $t$ elapsed since release. Then state how this result changes if the incline is perfectly smooth (frictionless) instead.

The sphere of mass m and radius R on the plane inclined at α, drawn after release: the contact point now lies a distance s down the slope from the fixed point O the sphere touched at t=0. — The sphere of mass $m$ and radius $R$ on the plane inclined at $\alpha$ , drawn after release: the contact point now lies a distance $s$ down the slope from the fixed point $O$ the sphere touched at $t=0$ .

Given

$m$
$R$
$\alpha$

Find

M(t)

about the fixed point

O

; the change for a smooth incline

Solution

Step 1Set up coordinates along and normal to the incline

The incline frame and the three forces, each at its point of action: the weight m g at the center, the normal N and the friction f at the contact. x runs down the slope, y along its outward normal. — The incline frame and the three forces, each at its point of action: the weight $m\vect g$ at the center, the normal $\vect N$ and the friction $\vect f$ at the contact. $x$ runs down the slope, $y$ along its outward normal.

Let $O$ be the point on the incline that coincided with the contact at $t=0$ , and use incline coordinates $(x,y)$ with origin at $O$ : $x$ down the slope, $y$ perpendicular to it, away from the surface. Because the sphere rolls without slipping and its center moves parallel to the incline, at time $t$ the contact point sits at $(s,0)$ and the center at $(s,R)$ , where $s$ is the distance rolled.

Step 2Torque of each force about the fixed point

O

The forces are the weight $m\vect g$ at the center, the normal $\vect N$ ( $N=mg\cos\alpha$ , since the center has no acceleration perpendicular to the incline) and the friction $\vect f$ , both at the contact. Writing each force in incline components and using $\tau_z = x F_y - y F_x$ for a force applied at $(x,y)$ ,

\begin{align*}\tau_{mg} &= s(-mg\cos\alpha) - R(mg\sin\alpha) = -mgs\cos\alpha - mgR\sin\alpha \\ \tau_N &= s(N) - 0 = smg\cos\alpha \\ \tau_f &= s(0) - 0(f) = 0\end{align*}

Step 3Sum the torques: the

s

-dependence cancels

Adding the three contributions, the terms carrying the rolled distance $s$ cancel between the weight and the normal force, leaving a constant net torque.

\begin{aligned} \tau_{\text{ext}} &= \tau_{mg}+\tau_N+\tau_f \\ &= \bigl(-mgs\cos\alpha - mgR\sin\alpha\bigr) + smg\cos\alpha + 0 \\ &= -mgR\sin\alpha + \bigl(smg\cos\alpha - mgs\cos\alpha\bigr) \end{aligned} \tag{1}

The bracketed pair cancels exactly, leaving the $s$ -independent remainder.

\tau_{\text{ext}} = -mgR\sin\alpha

(2)

Step 4Integrate the angular momentum theorem

About the fixed point $O$ , $dM/dt$ equals this net external torque, which (2) shows is constant in magnitude and independent of $s$ ; its sign there only records the sense of rotation, so it is the magnitude $mgR\sin\alpha$ that is carried into the integral. Integrating from $M(0)=0$ , the sphere being released from rest,

\int_0^{M} dM' = \int_0^t mgR\sin\alpha\,dt'

M(t) = mgRt\sin\alpha

(3)

Step 5The frictionless case

Nothing in the cancellation (1) used the value of $f$ : the friction term dropped out identically because of where and along what direction it acts, not because of its magnitude. So a perfectly smooth incline, where the sphere slides without rotating and $f=0$ throughout, is governed by exactly the same torque balance.

The result is unchanged:

M(t)=mgRt\sin\alpha

still holds, friction or none.

Check

Dimensionally $[m][g][R][t] = \text{kg}\cdot\text{m/s}^2\cdot\text{m}\cdot\text{s}=\text{kg}\,\text{m}^2/\text{s}$ , an angular momentum, and the $\sin\alpha$ factor correctly sends $M\to0$ as the incline flattens to $\alpha=0$ , where nothing rolls at all.

Remark

The two forces at the contact, $\vect N$ and $\vect f$ , together contribute a torque about $O$ that depends only on $N$ , never on $f$ : friction’s line of action always passes through the $x$ axis $O$ sits on, so friction is torque-free about every point of that axis, $O$ and the instantaneous contact point alike (a force is always torque-free about its own point of application, and here that point already lies on the axis). That is consistent with, not opposed to, the standard rolling equation $I\dot\omega = fR$ , which is a torque balance about the sphere’s center of mass, not about the contact point: there friction’s moment arm is the full radius $R$ , which is exactly why it appears in that equation at all. Choosing the axis is what decides which force carries the torque in the bookkeeping.