Classical-Mechanics · Unit 25 · Video 5 · Interactive Practice

Three Routes to the Same Roll: Forces, Energy, and Torque About the Contact Line

IKey Formulas

Formula	Name	What it gives
$Mg\sin\beta - f_s = Ma_x,\ \ f_s R = I_{\text{cm}}\alpha_z,\ \ \alpha_z = a_x/R$	Approach 1: forces + torque about the cm	$a_x = \tfrac{2}{3}g\sin\beta$ and $f_s = \tfrac{1}{3}Mg\sin\beta$ , up the slope
$\tfrac{1}{3}Mg\sin\beta \le \mu_s Mg\cos\beta \;\Rightarrow\; \mu_s \ge \tfrac{1}{3}\tan\beta$	Rolling condition	Static friction can supply at most $\mu_s N$ , with $N = Mg\cos\beta$
$Mgh = \tfrac{1}{2}Mv_{x,f}^{\ 2} + \tfrac{1}{2}I_{\text{cm}}\omega_{z,f}^{\ 2} = \tfrac{3}{4}Mv_{x,f}^{\ 2}$	Approach 2: energy	$v_{x,f} = \sqrt{\tfrac{4}{3}gh}$ ; $N$ and $f_s$ do no work
$RMg\sin\beta = I_{\text{cm}}\alpha_z + RMa_x$	Approach 3: $\vec{\boldsymbol{\tau}}_P = d\vec{\mathbf{L}}_P/dt$ about a fixed $P$ on the contact line	$a_x = \tfrac{2}{3}g\sin\beta$ with no $f_s$ and no $N$

Key Insight: All three routes give $a_x = \tfrac{2}{3}g\sin\beta$ and $v_{x,f} = \sqrt{\tfrac{4}{3}gh}$ . Choose by what is asked: the forces (approach 1), a speed (energy), or an acceleration with the fewest unknowns (torque about $P$ ). A frictionless block reaches $\sqrt{2gh}$ ; the cylinder is slower because one third of its kinetic energy is spin.

IIVisualization 1 — How Much Friction Rolling Needs

Rolling needs $f_s = \tfrac{1}{3}Mg\sin\beta$ up the slope, and static friction supplies at most $\mu_s Mg\cos\beta$ .

Incline angle β 30° Coefficient μ_s 0.25

IIIVisualization 2 — Torque About a Point on the Contact Line

About a fixed $P$ on the contact line, friction has no torque and the $d_P$ terms of gravity and $N$ cancel.

Distance rolled (contact point from the top) 1.00 m

IVVisualization 3 — Energy Race: Cylinder vs. Frictionless Block

Released from the same height $h$ , the cylinder reaches $\sqrt{\tfrac{4}{3}gh}$ and the frictionless block $\sqrt{2gh}$ , at any angle.

Incline angle β 30° Time 0.00 s

VQuiz Questions

Problem 1 · Acceleration Down the Incline

Given: A uniform solid cylinder ( $I_{\text{cm}} = \tfrac{1}{2}MR^2$ ) rolls without slipping down an incline at $\beta = 30\degree$ . Take $g = 9.8\ \text{m/s}^2$ . Find the acceleration $a_x$ of its center of mass down the incline.

✅ Correct!

a_x = \tfrac{2}{3}g\sin\beta = \tfrac{2}{3}(9.8)(0.5) = 3.27\ \text{m/s}^2

, independent of

M

and

R

❌ That is $g\sin\beta$ . It is the frictionless block's acceleration. Rolling needs friction up the slope, which takes away

\tfrac{1}{3}

Mg\sin\beta

❌ That is $\tfrac{1}{2}g\sin\beta$ . It belongs to a thin hoop with

I_{\text{cm}} = MR^2

. With

I_{\text{cm}} = \tfrac{1}{2}MR^2

friction is

\tfrac{1}{2}Ma_x

, not

Ma_x

❌ The factor is upside down.

Mg\sin\beta = \tfrac{3}{2}Ma_x

gives

a_x = \tfrac{2}{3}g\sin\beta

, not

\tfrac{3}{2}g\sin\beta

. The acceleration cannot exceed the frictionless value

g\sin\beta

❌ Not quite. Combine

Mg\sin\beta - f_s = Ma_x

with

f_s = \tfrac{1}{2}Ma_x

Show solution

Newton's second law along the slope and the torque about the center of mass:

Mg\sin\beta - f_s = Ma_x, \qquad f_s R = I_{\text{cm}}\alpha_z = \tfrac{1}{2}MR^2\,\frac{a_x}{R} \;\Rightarrow\; f_s = \tfrac{1}{2}Ma_x

Substituting:

Mg\sin\beta = \tfrac{3}{2}Ma_x \;\Rightarrow\; a_x = \tfrac{2}{3}g\sin\beta = \tfrac{2}{3}(9.8)(0.5) = 3.27\ \text{m/s}^2

Problem 2 · Size and Direction of Static Friction

Given: A uniform solid cylinder of mass $M = 4.0\ \text{kg}$ rolls without slipping down an incline with $\sin\beta = 0.6$ , $\cos\beta = 0.8$ . Take $g = 9.8\ \text{m/s}^2$ . Find the static friction force on the cylinder.

✅ Correct!

f_s = \tfrac{1}{3}Mg\sin\beta = \tfrac{1}{3}(4.0)(9.8)(0.6) = 7.84\ \text{N}

, pointing up the slope.

❌ Right size, wrong direction. Friction is the only force with a torque about the center. The spin must speed up clockwise as the cylinder rolls down, which needs friction pointing up the slope.

❌ That is $Mg\sin\beta$ . If friction balanced the whole down-slope pull, the center would not accelerate at all.

❌ You used $a_x = g\sin\beta$ in $f_s = \tfrac{1}{2}Ma_x$ . The rolling acceleration is

\tfrac{2}{3}g\sin\beta

, so

f_s = \tfrac{1}{3}Mg\sin\beta

❌ Not quite. Use

f_s = \tfrac{1}{2}Ma_x

with

a_x = \tfrac{2}{3}g\sin\beta

Show solution

From the torque about the center, $f_s R = \tfrac{1}{2}MR^2(a_x/R)$ , so $f_s = \tfrac{1}{2}Ma_x$ . With $a_x = \tfrac{2}{3}g\sin\beta$ :

f_s = \tfrac{1}{2}M\left(\tfrac{2}{3}g\sin\beta\right) = \tfrac{1}{3}Mg\sin\beta = \tfrac{1}{3}(4.0)(9.8)(0.6) = 7.84\ \text{N}

Direction: about the center, the weight and the normal force have no torque. Friction alone must increase the clockwise spin $\alpha_z$ , so it points up the slope.

Problem 3 · Speed at the Bottom and the Friction It Takes

Given: A uniform solid cylinder starts from rest on an incline at $\beta = 30\degree$ and rolls without slipping until its center has dropped $h = 2.4\ \text{m}$ . Take $g = 9.8\ \text{m/s}^2$ . Find the speed of its center at the bottom and the smallest $\mu_s$ that keeps it rolling.

Speed of the center of mass at the bottom?

Smallest coefficient of static friction?

✅ Correct! Energy gives

v_{x,f} = \sqrt{\tfrac{4}{3}gh} = 5.60\ \text{m/s}

, and approach 1 gives

\mu_s \ge \tfrac{1}{3}\tan 30\degree = 0.192

❌ That is $\sqrt{2gh}$ . It is the frictionless block's speed. The rolling cylinder keeps

\tfrac{1}{3}

of its kinetic energy in spin.

❌ That is $\sqrt{gh}$ . It is a thin hoop's speed. For the solid cylinder the spin term is

\tfrac{1}{4}Mv^2

, so

K = \tfrac{3}{4}Mv^2

❌ Check the energy equation.

Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}\left(\tfrac{1}{2}MR^2\right)(v/R)^2 = \tfrac{3}{4}Mv^2

❌ That is $\tan\beta$ . It is the condition for a block to stay at rest. A rolling cylinder needs only

f_s = \tfrac{1}{3}Mg\sin\beta

❌ You used $N = Mg$ . On the incline

N = Mg\cos\beta

, so the condition becomes

\mu_s \ge \tfrac{1}{3}\tan\beta

, not

\tfrac{1}{3}\sin\beta

❌ Check the friction condition. Rolling needs

f_s = \tfrac{1}{3}Mg\sin\beta \le \mu_s Mg\cos\beta

Show solution

Speed (energy): neither $N$ nor static friction does work, so

Mgh = \tfrac{1}{2}Mv_{x,f}^{\ 2} + \tfrac{1}{2}\left(\tfrac{1}{2}MR^2\right)\left(\frac{v_{x,f}}{R}\right)^2 = \tfrac{3}{4}Mv_{x,f}^{\ 2}

v_{x,f} = \sqrt{\tfrac{4}{3}gh} = \sqrt{\tfrac{4}{3}(9.8)(2.4)} = \sqrt{31.36} = 5.60\ \text{m/s}

Friction (approach 1): $f_s = \tfrac{1}{3}Mg\sin\beta$ must not exceed $\mu_s N = \mu_s Mg\cos\beta$ :

\mu_s \ge \tfrac{1}{3}\tan\beta = \tfrac{1}{3}\tan 30\degree = \tfrac{1}{3}(0.5774) = 0.192

The angle does not enter the speed, but it sets the friction needed.

Problem 4 · Torque About P for a Solid Sphere

Given: A uniform solid sphere ( $I_{\text{cm}} = \tfrac{2}{5}MR^2$ ) rolls without slipping down an incline at angle $\beta$ . Take torques about a fixed point $P$ on the line the contact point travels along. Find the acceleration $a_x$ of its center.

✅ Correct!

RMg\sin\beta = \tfrac{2}{5}MRa_x + RMa_x = \tfrac{7}{5}MRa_x

, so

a_x = \tfrac{5}{7}g\sin\beta

. Neither

f_s

nor

N

was needed.

❌ That is the cylinder's result. The spin term

I_{\text{cm}}\alpha_z

depends on the body: for a sphere it is

\tfrac{2}{5}MRa_x

, not

\tfrac{1}{2}MRa_x

❌ You dropped the orbital term.

\vec{\mathbf{L}}_P

also contains

\vec{\mathbf{r}}_{P,\text{cm}} \times M\vec{\mathbf{V}}_{\text{cm}}

, whose rate is

RMa_x

. Without it,

a_x

would exceed

g\sin\beta

, which is impossible.

❌ You dropped the spin term.

d\vec{\mathbf{L}}_P/dt

includes

I_{\text{cm}}\alpha_z

as well as

RMa_x

. Only a frictionless slider reaches

g\sin\beta

❌ Not quite. Use

RMg\sin\beta = I_{\text{cm}}\alpha_z + RMa_x

with

\alpha_z = a_x/R

Show solution

About $P$ , friction has zero moment arm and the $d_P$ terms of gravity and $N$ cancel, so $\tau_P = RMg\sin\beta$ . The angular momentum about $P$ is $L_P = I_{\text{cm}}\omega_z + RMv_x$ , so

RMg\sin\beta = I_{\text{cm}}\alpha_z + RMa_x = \tfrac{2}{5}MR^2\,\frac{a_x}{R} + RMa_x = \tfrac{7}{5}MRa_x

a_x = \tfrac{5}{7}g\sin\beta

The sphere ( $\tfrac{5}{7} \approx 0.714$ ) beats the cylinder ( $\tfrac{2}{3} \approx 0.667$ ): a smaller $I_{\text{cm}}/MR^2$ means less of the energy goes into spin.

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