Classical-Mechanics · Unit 25 · Video 6 · Interactive Practice

Friction Has No Torque About the Line of Contact: Why a Sliding Bowling Ball Rolls at Five Sevenths of Its Speed

IKey Formulas

Formula	Name	What you need
$\vec{\tau}_S = d\,mg\,\hat{\mathbf{k}} - d\,N\,\hat{\mathbf{k}} + \vec{\mathbf{0}} = \vec{\mathbf{0}}$	Torque about a fixed $S$ on the line of contact	$N = mg$ ; $\vec{\mathbf{f}}_k$ lies along the floor through $S$
$mR\,v_{\text{cm},i} = mR\,v_{\text{cm},f} + I_{\text{cm}}\,\omega_{z,f}$	$L_{S,i} = L_{S,f}$	Released without spin; rolling at the end: $v_{\text{cm},f} = R\,\omega_{z,f}$
$R\,f_k = I_{\text{cm}}\,\alpha_z,\ \ a_{\text{cm},x} = -\dfrac{f_k}{m}$	Equations of motion	Torque about the cm and $\vec{F}^{\,\text{ext}} = m\vec{\mathbf{a}}_{\text{cm}}$ , while skidding
$v_{\text{cm},f} = \dfrac{v_{\text{cm},i}}{1 + I_{\text{cm}}/mR^2} = \tfrac{5}{7}\,v_{\text{cm},i}$	Final rolling speed	Uniform sphere, $I_{\text{cm}} = \tfrac{2}{5}mR^2$ ; then $K_f/K_i = \tfrac{5}{7}$

Key Insight: Choose the point so the unknown force drops out. About a fixed point on the floor, friction's line of action passes through the point and the torques of gravity and the normal force cancel, so one conservation law gives the final speed without ever knowing $f_k$ or $t_f$ .

IIVisualization 1 — Torque About a Point $S$

About a fixed point $S$ at height $h$ above the floor, the total torque on the skidding ball is $f_k h$ .

IIIVisualization 2 — From Skidding to Rolling

Friction slows the center and spins the ball up until $v_{\text{cm}} = R\,\omega_z$ , while $L_S$ never changes.

Time t (s) 0.00 Lane friction μ_k 0.20

IVVisualization 3 — Any Round Body

Released sliding without spin, how much speed a body keeps depends only on the ratio $I_{\text{cm}}/mR^2$ .

Body

💡 Challenge: predict which of the four bodies dissipates exactly as much kinetic energy as it keeps, then check it.

VQuiz Questions

Problem 1 · Final Rolling Speed

Given: A uniform bowling ball ( $I_{\text{cm}} = \tfrac{2}{5}mR^2$ ) is released on a level lane with $v_{\text{cm},i} = 8.4\ \text{m/s}$ and no spin. Find its speed when it starts to roll without slipping.

✅ Correct!

mR\,v_{\text{cm},i} = \tfrac{7}{5}mR\,v_{\text{cm},f}

gives

v_{\text{cm},f} = \tfrac{5}{7}(8.4) = 6.0\ \text{m/s}

❌ Not quite. Halving the speed is the result for

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

(a hoop). A solid sphere's spin part is only

\tfrac{2}{5}mR\,v_{\text{cm},f}

❌ Energy is not conserved. Setting

\tfrac{1}{2}mv_{\text{cm},i}^2 = \tfrac{7}{10}mv_{\text{cm},f}^2

ignores the work kinetic friction does while the bottom of the ball slides. Use angular momentum about a point on the floor.

❌ That is the speed lost.

\tfrac{2}{7}(8.4) = 2.4\ \text{m/s}

is the drop in speed; the ball keeps

\tfrac{5}{7}

Show solution

About a fixed point $S$ on the floor the total torque is zero, so $L_{S,i} = L_{S,f}$ . Initially there is no spin, only the orbital part; finally both parts, with $\omega_{z,f} = v_{\text{cm},f}/R$ :

mR\,v_{\text{cm},i} = mR\,v_{\text{cm},f} + \tfrac{2}{5}mR^2\cdot\frac{v_{\text{cm},f}}{R} = \tfrac{7}{5}mR\,v_{\text{cm},f}

v_{\text{cm},f} = \tfrac{5}{7}\,v_{\text{cm},i} = \tfrac{5}{7}(8.4\ \text{m/s}) = 6.0\ \text{m/s}

Problem 2 · Choosing the Point

Given: A ball skids on a level floor, acted on by its weight $m\vec{g}$ , the normal force $\vec{N}$ and kinetic friction $\vec{f}_k$ . Find the choice of point about which its angular momentum stays constant throughout the skid.

✅ Correct! Friction lies along the floor through

S

, so its torque vanishes, and the torques

d\,mg

and

-d\,N

cancel because

N = mg

❌ Not quite. Friction acts a distance

R

below the center, perpendicular to the radius, so

\tau_{\text{cm}} = R\,f_k \neq 0

: that torque is what spins the ball up.

❌ Not quite. Friction's line of action is the floor, a perpendicular distance

R

below that point, so the torque about it is

f_k R \neq 0

❌ Not quite. The vertical forces cancel, but friction is unbalanced: about a fixed point at height

h

its torque is

f_k h

, zero only when

h = 0

Show solution

Let $S$ be a fixed point a horizontal distance $d$ behind the contact point and a height $h$ above the floor. Taking clockwise ( $\hat{\mathbf{k}}$ , into the page) as positive:

\tau_{S,mg} = d\,mg,\qquad \tau_{S,N} = -d\,N,\qquad \tau_{S,f_k} = h\,f_k

With $N = mg$ the total is $\tau_S = f_k h$ . It vanishes for every $d$ exactly when $h = 0$ , that is, for any fixed point on the line of contact. The center ( $h = R$ ) gives $\tau = f_k R$ , so $L_{\text{cm}}$ changes.

Problem 3 · How Long, and How Much Energy

Given: A uniform $6.0\ \text{kg}$ bowling ball ( $I_{\text{cm}} = \tfrac{2}{5}mR^2$ ) is released at $v_{\text{cm},i} = 7.0\ \text{m/s}$ with no spin on a lane with $\mu_k = 0.20$ ; take $g = 9.8\ \text{m/s}^2$ . Find the time $t_f$ at which it starts to roll and the kinetic energy dissipated by friction.

How long does it skid?

How much kinetic energy is dissipated?

✅ Correct! The ball skids for

t_f = \dfrac{2v_{\text{cm},i}}{7\mu_k g} = 1.02\ \text{s}

and loses

\tfrac{2}{7}

K_i = 147\ \text{J}

, which is

42\ \text{J}

❌ That is the time to stop.

v_{\text{cm},i}/(\mu_k g) = 3.57\ \text{s}

is when the center would halt with no spin-up; skidding ends much earlier, when

v_{\text{cm},x} = R\,\omega_z

❌ Check the spin-up rate.

R\,\omega_z

grows at

R^2 f_k/I_{\text{cm}} = \tfrac{5}{2}\mu_k g

, not

\tfrac{2}{5}\mu_k g

: divide by

I_{\text{cm}}/mR^2

, don't multiply by it.

❌ Not quite. Set

v_{\text{cm},i} - \mu_k g\,t_f = \tfrac{5}{2}\mu_k g\,t_f

; using

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

instead gives

1.79\ \text{s}

❌ Include the rotation.

147 - 75 = 72\ \text{J}

counts only

\tfrac{1}{2}mv_{\text{cm},f}^2

; the spin carries

\tfrac{1}{5}mv_{\text{cm},f}^2 = 30\ \text{J}

more.

❌ Mechanical energy is not constant. Kinetic friction does work on the skidding ball while its bottom slides over the floor.

❌ That is the energy kept.

K_f = 105\ \text{J}

; the dissipated energy is

K_i - K_f

❌ Not quite. Compute

K_i

and

K_f = \tfrac{7}{10}m v_{\text{cm},f}^2

and subtract.

Show solution

Time. While skidding, $a_{\text{cm},x} = -\mu_k g$ and $\alpha_z = R f_k/I_{\text{cm}} = \dfrac{5\mu_k g}{2R}$ , so

v_{\text{cm},x}(t) = v_{\text{cm},i} - \mu_k g\,t,\qquad R\,\omega_z(t) = \tfrac{5}{2}\mu_k g\,t

They meet when $v_{\text{cm},i} = \tfrac{7}{2}\mu_k g\,t_f$ :

t_f = \frac{2v_{\text{cm},i}}{7\mu_k g} = \frac{2(7.0)}{7(0.20)(9.8)} = \frac{14}{13.72} \approx 1.02\ \text{s}

Energy. $K_i = \tfrac{1}{2}(6.0)(7.0)^2 = 147\ \text{J}$ and $v_{\text{cm},f} = \tfrac{5}{7}(7.0) = 5.0\ \text{m/s}$ , so

K_f = \tfrac{7}{10}m\,v_{\text{cm},f}^2 = \tfrac{7}{10}(6.0)(5.0)^2 = 105\ \text{J},\qquad K_i - K_f = 42\ \text{J} = \tfrac{2}{7}K_i

Problem 4 · Released With Backspin

Given: A uniform ball ( $I_{\text{cm}} = \tfrac{2}{5}mR^2$ ) is released on a level floor with $v_{\text{cm},i} = 7.0\ \text{m/s}$ forward and a backspin (counterclockwise, with the ball moving right) of $R\,|\omega_{z,i}| = 1.4\ \text{m/s}$ . Find its speed when it starts to roll without slipping.

✅ Correct! The same point

S

still feels no torque, and the initial spin part

I_{\text{cm}}\omega_{z,i}

is negative:

7.0 - 0.56 = 1.4\,v_{\text{cm},f}

❌ The initial spin counts.

\tfrac{5}{7}v_{\text{cm},i}

assumes

\omega_{z,i} = 0

; here

L_{S,i}

has a spin part too.

❌ Check the sign. With

\hat{\mathbf{k}}

into the page, clockwise is positive, so backspin makes

I_{\text{cm}}\omega_{z,i}

negative and lowers

L_{S,i}

❌ Weight the spin by $I_{\text{cm}}$ . The spin part is

\tfrac{2}{5}mR\,(R\omega_{z,i})

, not

mR\,(R\omega_{z,i})

: subtract

0.56\ \text{m/s}

, not

1.4\ \text{m/s}

Show solution

The torque about a fixed $S$ on the floor is still zero (friction lies along the floor, $N = mg$ ), whatever the spin. Clockwise is positive, so $R\,\omega_{z,i} = -1.4\ \text{m/s}$ :

L_{S,i} = mR\,v_{\text{cm},i} + \tfrac{2}{5}mR^2\,\omega_{z,i} = mR\big(7.0 + \tfrac{2}{5}(-1.4)\big) = mR\,(6.44\ \text{m/s})

L_{S,f} = \tfrac{7}{5}mR\,v_{\text{cm},f} \;\Rightarrow\; v_{\text{cm},f} = \frac{6.44}{1.4} = 4.6\ \text{m/s}

Check with the equations of motion: the bottom slides forward at $7.0 + 1.4 = 8.4\ \text{m/s}$ , so friction points backward; $v_{\text{cm},x} - R\omega_z$ shrinks at $\tfrac{7}{2}\mu_k g$ , giving $t_f = \dfrac{8.4}{3.5\,\mu_k g}$ and $v_{\text{cm},f} = 7.0 - \dfrac{8.4}{3.5} = 4.6\ \text{m/s}$ for any $\mu_k$ .

Solved: 0 / 4

Friction Has No Torque About the Line of Contact: Why a Sliding Bowling Ball Rolls at Five Sevenths of Its Speed

IKey Formulas

IIVisualization 1 — Torque About a Point SSS

IIIVisualization 2 — From Skidding to Rolling

IVVisualization 3 — Any Round Body

VQuiz Questions

IIVisualization 1 — Torque About a Point $S$