Classical-Mechanics · Unit 25 · Video 3 · Interactive Practice

Why Your Outer Foot Pushes Harder: Zero Torque About the Center of Mass on a Turning Train

IKey Formulas

Formula	Equation	What it fixes
$\hat{\mathbf{r}}:\ f_1 + f_2 = m\dfrac{v^2}{R} \qquad \hat{\mathbf{k}}:\ N_1 + N_2 = mg$	$\vec{\mathbf{F}}^{\,\text{ext}} = m\,\vec{\mathbf{a}}_{\text{cm}}$ for the center of mass	The friction sum (centripetal force) and the normal-force sum (weight)
$\dfrac{d}{2}\,(N_1 - N_2) + L\,(f_1 + f_2) = 0$	$\vec{\tau}_{\text{cm}}^{\,\text{total}} = 0$ (no tipping, even though the cm accelerates)	The normal-force difference; the frictions enter only as their sum
$N_{1,2} = \dfrac{1}{2}\left(mg \mp \dfrac{2Lmv^2}{Rd}\right)$	Inner foot $N_1$ , outer foot $N_2$	$\dfrac{Lmv^2}{Rd}$ moves from the inner foot to the outer foot
$v_{\max} = \sqrt{\dfrac{gRd}{2L}}$	$N_1 = 0$ , since the floor can push but never pull	The top speed before the inner foot lifts; independent of $m$

Key Insight: The inward friction acts a height $L$ below the center of mass, so it exerts a torque that would tip him outward. The only force that can cancel it is the extra push $N_2 - N_1 = \dfrac{2Lmv^2}{Rd}$ on the outer foot.

IIVisualization 1 — Speeding Up Shifts the Load

As $v$ grows, the same amount $Lmv^2/(Rd)$ leaves the inner foot and lands on the outer foot.

speed v (m/s) 6.0

IIIVisualization 2 — Balancing the Torque About the Center of Mass

At $v = 8$ m/s, only one split of the weight between the feet makes the torque about the cm zero.

outer foot N₂ (N) 343 inner share f₁/(f₁+f₂) 0.50

💡 The share slider never changes the net torque: the frictions appear only as $f_1 + f_2$ , so the three equations fix both normal forces but neither friction on its own.

IVVisualization 3 — What Sets the Top Speed

The inner foot unloads at $v_{\max} = \sqrt{gRd/(2L)}$ ; stance width, cm height and curve radius each move it.

R (m) 50 d (m) 0.40 L (m) 1.00 v (m/s) 7.0

VQuiz Questions

Problem 1 · Load on the Outer Foot

Given: A $60$ kg passenger stands facing forward in a train rounding an unbanked curve of radius $R = 40$ m at $v = 6$ m/s. His feet are $d = 0.40$ m apart and his center of mass is $L = 1.0$ m above the floor, midway between his feet. Take $g = 9.8$ m/s². Find the normal force $N_2$ on his outer foot.

✅ Correct! Half the weight,

294

N, plus the transferred

\dfrac{Lmv^2}{Rd} = 135

N gives

N_2 = 429

N; the inner foot keeps

N_1 = 159

❌ That is the inner foot.

159

= 294 - 135

N is

N_1

. The outer foot gains the

135

N the inner foot loses.

❌ Check the transfer. You moved

\dfrac{Lmv^2}{2Rd} = 67.5

N. The torque equation gives

N_2 - N_1 = \dfrac{2Lmv^2}{Rd} = 270

N, so each foot changes by half of that,

135

❌ Missing the factor $\tfrac{1}{2}$ .

N_2 = \tfrac{1}{2}\left(mg + \dfrac{2Lmv^2}{Rd}\right)

: the difference

270

N is split between the two feet, so

N_2 = 294 + 135

N, not

294 + 270

Show solution

The weight is $mg = 60 \times 9.8 = 588$ N, and the radial equation gives the friction sum $f_1 + f_2 = \dfrac{mv^2}{R} = \dfrac{60 \times 36}{40} = 54$ N.

Zero torque about the center of mass, $\dfrac{d}{2}(N_1 - N_2) + L(f_1 + f_2) = 0$ , gives

N_2 - N_1 = \frac{2L(f_1 + f_2)}{d} = \frac{2 \times 1.0 \times 54}{0.40} = 270\ \text{N}.

With $N_2 + N_1 = mg = 588$ N, adding the two equations and dividing by two:

N_2 = \tfrac{1}{2}(588 + 270) = 429\ \text{N}, \qquad N_1 = \tfrac{1}{2}(588 - 270) = 159\ \text{N}.

Equivalently, $\dfrac{Lmv^2}{Rd} = \dfrac{1.0 \times 60 \times 36}{40 \times 0.40} = 135$ N moves from the inner foot to the outer foot.

Problem 2 · Where Does the Tipping Torque Come From?

Given: A passenger says his outer foot pushes harder "because an outward centrifugal force at my center of mass pushes me toward the outside." Working in the ground frame and taking torques about his center of mass, which torque does the extra push on the outer foot actually cancel?

✅ Correct! The frictions point inward and act at the floor,

L

below the cm, so

L(f_1 + f_2)

turns him in

+\hat{\boldsymbol{\theta}}

, clockwise seen from behind; only

N_2

turns him the other way.

❌ Not quite. Gravity acts at the center of mass itself, so its moment arm about the cm is zero and it exerts no torque there.

❌ There is no such force in the ground frame. The only horizontal forces on him are the two frictions, and their sum points inward because his center of mass accelerates toward the center of the curve.

❌ The torque equation still holds.

\vec{\tau}_{\text{cm}}^{\,\text{total}} = d\vec{\mathbf{L}}_{\text{spin}}/dt

is valid about the center of mass even when it accelerates. He does not tip, so his spin angular momentum is constant and the total torque about the cm is zero.

Show solution

About the center of mass, with $\hat{\mathbf{r}}$ outward, $\hat{\mathbf{k}}$ up and $\hat{\boldsymbol{\theta}}$ along the motion:

\vec{\tau}_{\text{cm},1} = \left(\tfrac{d}{2}N_1 + L f_1\right)\hat{\boldsymbol{\theta}}, \qquad \vec{\tau}_{\text{cm},2} = \left(-\tfrac{d}{2}N_2 + L f_2\right)\hat{\boldsymbol{\theta}}.

Gravity contributes nothing (zero moment arm). Three of the four foot forces, $N_1$ , $f_1$ and $f_2$ , turn him in $+\hat{\boldsymbol{\theta}}$ , which tips him outward; only $N_2$ turns him back. Setting the sum to zero,

\tfrac{d}{2}(N_2 - N_1) = L(f_1 + f_2) = L\,\frac{mv^2}{R}.

The friction torque on the right is what forces $N_2 \gt N_1$ . No outward force is needed, and none exists in the ground frame.

Problem 3 · Top Speed and the Load There

Given: A $60$ kg passenger stands with feet $d = 0.30$ m apart and his center of mass $L = 0.90$ m above the floor, in a train on an unbanked curve of radius $R = 60$ m. Take $g = 9.8$ m/s² and assume friction never slips. Find the top speed before his inner foot lifts, and the normal force on his outer foot at that speed.

What is $v_{\max}$ ?

What is $N_2$ at $v_{\max}$ ?

✅ Correct!

v_{\max} = \sqrt{98} \approx 9.90

m/s, and with

N_1 = 0

the outer foot carries his whole weight,

588

❌ Missing the factor 2.

N_1 = 0

means

mg = \dfrac{2Lmv^2}{Rd}

, so

v_{\max}^2 = \dfrac{gRd}{2L} = 98

m²/s², not

\dfrac{gRd}{L} = 196

m²/s².

❌ $d$ and $L$ are swapped. A wider stance raises

v_{\max}

and a higher center of mass lowers it, so

d

belongs in the numerator:

v_{\max} = \sqrt{gRd/(2L)}

❌ Check $v_{\max}$ . Set

N_1 = \tfrac{1}{2}\left(mg - \dfrac{2Lmv^2}{Rd}\right) = 0

and solve for

v

; the mass cancels.

❌ That is the load at rest. At

v = 0

each foot carries

\tfrac{1}{2}mg = 294

N. At

v_{\max}

the inner foot carries nothing.

❌ Friction is horizontal.

686 = 588 + 98

adds the friction sum to the weight, but the vertical equation

N_1 + N_2 = mg

contains no friction.

❌ Check $N_2$ . At

v_{\max}

N_1 = 0

; use

N_1 + N_2 = mg

Show solution

Step 1: top speed. The inner foot lifts when $N_1 = \tfrac{1}{2}\left(mg - \dfrac{2Lmv^2}{Rd}\right) = 0$ :

v_{\max} = \sqrt{\frac{gRd}{2L}} = \sqrt{\frac{9.8 \times 60 \times 0.30}{2 \times 0.90}} = \sqrt{\frac{176.4}{1.8}} = \sqrt{98} \approx 9.90\ \text{m/s}.

Step 2: outer foot. The vertical equation $N_1 + N_2 = mg$ with $N_1 = 0$ gives $N_2 = mg = 60 \times 9.8 = 588$ N.

Check with the torque equation. The friction sum is $f_1 + f_2 = \dfrac{mv_{\max}^2}{R} = \dfrac{60 \times 98}{60} = 98$ N, and $N_2 - N_1 = \dfrac{2L(f_1 + f_2)}{d} = \dfrac{2 \times 0.90 \times 98}{0.30} = 588$ N, consistent with $N_1 = 0$ , $N_2 = 588$ N.

Problem 4 · Transfer: Does the Car Tip or Skid?

Given: The same analysis applies to a car on an unbanked curve, with the wheels in place of the feet: track width $d = 1.6$ m, center-of-mass height $L = 0.80$ m, radius $R = 50$ m, coefficient of static friction $\mu_s = 0.80$ between tires and road, $g = 9.8$ m/s². The total friction can be at most $\mu_s(N_1 + N_2) = \mu_s mg$ . As the car speeds up, what happens first?

✅ Correct! Skidding starts at

\sqrt{\mu_s gR} \approx 19.8

m/s, below the tipping speed

\sqrt{gRd/(2L)} \approx 22.1

m/s. Tipping comes first only when

\dfrac{d}{2L} \lt \mu_s

; here

\dfrac{d}{2L} = 1.0 \gt 0.80

❌ Compare the two limits.

22.1

m/s is the tipping speed, but the friction needed,

mv^2/R

, already exceeds

\mu_s mg

above

\sqrt{\mu_s gR} \approx 19.8

m/s, so the tires slide first.

❌ Right speed, wrong event.

\sqrt{\mu_s gR} \approx 19.8

m/s is where friction runs out, which is a skid; the tipping condition

N_1 = 0

gives

22.1

m/s.

❌ Check the lever arm.

15.7

m/s comes from using

d/2

in place of

d

. Each wheel sits

d/2

from the cm, and that half is already built into

v_{\max} = \sqrt{gRd/(2L)}

with

d

the full track width.

Show solution

Tipping: zero torque about the center of mass gives the same result as for the passenger, with $N_1 = 0$ at

v_{\text{tip}} = \sqrt{\frac{gRd}{2L}} = \sqrt{\frac{9.8 \times 50 \times 1.6}{2 \times 0.80}} = \sqrt{490} \approx 22.1\ \text{m/s}.

Skidding: the radial equation needs $f_1 + f_2 = \dfrac{mv^2}{R}$ , and static friction can supply at most $\mu_s(N_1 + N_2) = \mu_s mg$ :

v_{\text{skid}} = \sqrt{\mu_s gR} = \sqrt{0.80 \times 9.8 \times 50} = \sqrt{392} \approx 19.8\ \text{m/s}.

Since $v_{\text{skid}} \lt v_{\text{tip}}$ , the car skids first. In general $\dfrac{v_{\text{tip}}^2}{v_{\text{skid}}^2} = \dfrac{d}{2L\mu_s}$ , so a vehicle tips first only when $\dfrac{d}{2L} \lt \mu_s$ : a narrow track, a high center of mass, or very grippy tires.

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