Classical-Mechanics · Unit 21 · Video 4 · Interactive Practice

Gravity Acts at One Point, and a Heavy Pulley Makes the String Tensions Differ

IKey Formulas

Formula	Name	What you need
$\vec{\tau}_O = \displaystyle\sum_i \vec{r}_i \times m_i\vec{g} = \vec{R}_{cm} \times M_T\vec{g}$	Centre of gravity	$\vec{g}$ the same at every particle; the whole weight $M_T\vec{g}$ then acts at the centre of mass
$\tau_{S,z} = I_S\,\alpha_z, \qquad \|\tau_z^{\text{fric}}\| = I_S\,\dfrac{\omega_i}{\Delta t}$	Coast-down torque	The moment of inertia and a timed stop at constant angular acceleration
$R(T_1 - T_2) = I_z\,\alpha_z, \qquad a = R\,\alpha_z$	Pulley with inertia	The torques of the two tensions about the axle, clockwise positive ( $\hat{k}$ into the page) as block 1 falls; a string that does not slip
$a = \dfrac{m_1 g - \mu_k m_2 g}{m_1 + m_2 + I_z/R^2},$ $\qquad T_1 - T_2 = \dfrac{I_z}{R^2}\,a$	Hanging and sliding blocks	Newton's second law on each block joined to the pulley row; valid when $m_1 > \mu_k m_2$

Key Insight: equal tensions on the two sides of a pulley are the special case $I_z = 0$ . A pulley with inertia must itself be accelerated: it adds the effective mass $I_z/R^2$ to the denominator, and the tension difference $(I_z/R^2)\,a$ supplies exactly the torque that spins it.

IIVisualization 1 — Many Weights, One Torque

About any origin, the particle weights' torques add up to the torque of $M_T\vec{g}$ applied at the centre of mass.

Origin O: x (m) −0.2 Origin O: y (m) −0.1

💡 The derivation used only that $\vec{g}$ is the same vector at every particle; for a body large enough that $\vec{g}$ varies across it, the centre of gravity and the centre of mass can be different points.

IIIVisualization 2 — Timing the Coast-Down

Friction is the only torque about the axle, so $I_S$ times the measured deceleration $\omega_i/\Delta t$ is the frictional torque.

Time t (s) 0.0 Stopping time Δt (s) 8.0

IVVisualization 3 — Why the Two Tensions Differ

Spinning up the pulley takes a net torque $R(T_1 - T_2) = I_z\alpha_z$ , so the tensions match only when $I_z = 0$ .

Pulley moment of inertia I_z (kg·m²) 0.030

💡 The same $I_z/R^2$ appeared in the energy method of the previous lecture: the pulley's $\tfrac{1}{2}I_z\omega^2$ with $\omega = v/R$ is $\tfrac{1}{2}(I_z/R^2)v^2$ , the kinetic energy of an extra mass moving with the string.

VQuiz Questions

Problem 1 · Coast-Down Friction Torque

Given: a grinding wheel modelled as a uniform disc of mass $2.0\ \text{kg}$ and radius $0.20\ \text{m}$ spins at $60$ rev/min; with the motor off, friction at the axle brings it to rest in $10\ \text{s}$ with constant angular acceleration — find the magnitude of the frictional torque.

✅ Correct!

I_S = \tfrac{1}{2}(2.0)(0.20)^2 = 0.040\ \text{kg}\cdot\text{m}^2

and

|\alpha_z| = 2\pi/10 \approx 0.63\ \text{rad/s}^2

, so

|\tau_z^{\text{fric}}| \approx 2.5 \times 10^{-2}\ \text{N}\cdot\text{m}

❌ The revolutions were never converted to radians.

60

rev/min is

1

rev/s, and one revolution is

2\pi

rad, so

\omega_i = 2\pi\ \text{rad/s}

, not

1\ \text{rad/s}

❌ That is a hoop's moment of inertia. A uniform disc has

I_S = \tfrac{1}{2}MR^2 = 0.040\ \text{kg}\cdot\text{m}^2

, half of

MR^2

❌ That is $I_S\,\omega_i$ , an angular momentum, not a torque. The torque needs the angular acceleration

\omega_i/\Delta t

, so divide by the

10\ \text{s}

❌ Not quite. Find

I_S = \tfrac{1}{2}MR^2

, convert

60

rev/min to rad/s, divide by

\Delta t

for

|\alpha_z|

, then multiply.

Show solution

Step 1: the moment of inertia of a uniform disc.

I_S = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0\ \text{kg})(0.20\ \text{m})^2 = 0.040\ \text{kg}\cdot\text{m}^2

Step 2: the angular acceleration. One revolution is $2\pi$ rad, so $60$ rev/min is $2\pi$ rad/s:

\alpha_z = \frac{0 - \omega_i}{\Delta t} = -\frac{2\pi\ \text{rad/s}}{10\ \text{s}} \approx -0.63\ \text{rad/s}^2

Step 3: the torque. Gravity acts at the centre of gravity, on the axis, and the axle pushes at the axis, so neither has a torque about it; friction is the only torque:

|\tau_z^{\text{fric}}| = I_S|\alpha_z| = (0.040\ \text{kg}\cdot\text{m}^2)(0.63\ \text{rad/s}^2) \approx 2.5 \times 10^{-2}\ \text{N}\cdot\text{m}

$\alpha_z$ is negative because the wheel is slowing, and the frictional torque points against the rotation.

Problem 2 · The Table-Side Tension

Given: block 1, of mass $3.0\ \text{kg}$ , hangs from a string that runs over a pulley of radius $R = 0.10\ \text{m}$ and moment of inertia $I_z = 0.020\ \text{kg}\cdot\text{m}^2$ to block 2 on the table; the string does not slip, and the blocks accelerate at $a = 2.8\ \text{m/s}^2$ ( $g = 9.8\ \text{m/s}^2$ ) — find the tension $T_2$ in the table-side segment.

✅ Correct!

T_1 = m_1(g - a) = 21.0\ \text{N}

, and the pulley needs

T_1 - T_2 = (I_z/R^2)\,a = 5.6\ \text{N}

, so

T_2 = 15.4\ \text{N}

❌ That is $T_1$ — the massless-pulley shortcut. Here

I_z \neq 0

and

\alpha_z = a/R \neq 0

, so the pulley needs a net torque

R(T_1 - T_2)

and the two tensions cannot be equal.

❌ The difference has the wrong sign. The pulley turns the way block 1 pulls it, so the hanging side is the tighter one:

T_2 = T_1 - (I_z/R^2)\,a

❌ Check the powers of $R$ .

R(T_1 - T_2) = I_z\alpha_z

with

\alpha_z = a/R

gives

T_1 - T_2 = I_z a/R^2 = 5.6\ \text{N}

;

I_z a/R

has the units of torque,

\text{N}\cdot\text{m}

, not force.

❌ Not quite. Get

T_1

from block 1, then subtract the tension difference the pulley's torque equation demands.

Show solution

Step 1: block 1 (downward positive):

m_1 g - T_1 = m_1 a \;\Longrightarrow\; T_1 = m_1(g - a) = (3.0\ \text{kg})(9.8 - 2.8)\ \text{m/s}^2 = 21.0\ \text{N}

Step 2: the pulley (clockwise positive as block 1 falls). No slip gives $\alpha_z = a/R = 2.8/0.10 = 28\ \text{rad/s}^2$ , and the torque equation $R(T_1 - T_2) = I_z\alpha_z$ becomes

T_1 - T_2 = \frac{I_z}{R^2}\,a = \frac{0.020\ \text{kg}\cdot\text{m}^2}{(0.10\ \text{m})^2}\,(2.8\ \text{m/s}^2) = (2.0\ \text{kg})(2.8\ \text{m/s}^2) = 5.6\ \text{N}

Step 3: the table-side tension.

T_2 = T_1 - 5.6\ \text{N} = 21.0\ \text{N} - 5.6\ \text{N} = 15.4\ \text{N}

Check the torque: $R(T_1 - T_2) = (0.10\ \text{m})(5.6\ \text{N}) = 0.56\ \text{N}\cdot\text{m}$ , and $I_z\alpha_z = (0.020\ \text{kg}\cdot\text{m}^2)(28\ \text{rad/s}^2) = 0.56\ \text{N}\cdot\text{m}$ .

Problem 3 · Heavy Pulley, Sliding Block, Falling Block

Given: block 1 ( $m_1 = 4.0\ \text{kg}$ ) hangs over the table edge; block 2 ( $m_2 = 3.0\ \text{kg}$ ) slides on the table with $\mu_k = 0.40$ ; the pulley is a uniform disc of mass $m_p = 2.0\ \text{kg}$ , so $I_{cm} = \tfrac{1}{2}m_p R^2$ ; the string does not slip, the blocks start from rest, and block 1 hits the ground at $t_1 = 1.5\ \text{s}$ ( $g = 9.8\ \text{m/s}^2$ ) — find the acceleration of the blocks and how far block 1 falls.

Acceleration of the blocks?

Distance block 1 falls?

✅ Correct!

a = 27.44\ \text{N} / 8.0\ \text{kg} = 3.43\ \text{m/s}^2

and

y_1 = \tfrac{1}{2}(3.43)(1.5)^2 \approx 3.86\ \text{m}

❌ That is the massless-pulley answer.

(m_1 - \mu_k m_2)g/(m_1 + m_2)

leaves out the pulley, whose inertia adds

I_z/R^2 = \tfrac{1}{2}m_p = 1.0\ \text{kg}

to the denominator.

❌ That treats the pulley as a hoop.

I = m_p R^2

would add

2.0\ \text{kg}

; a uniform disc has

I_{cm} = \tfrac{1}{2}m_p R^2

, which adds

I_z/R^2 = 1.0\ \text{kg}

❌ The friction on block 2 is missing. The net driving force is

m_1 g - \mu_k m_2 g = 39.2\ \text{N} - 11.76\ \text{N} = 27.44\ \text{N}

, not

39.2\ \text{N}

❌ Check the acceleration. Numerator: the weight of block 1 minus the friction on block 2. Denominator:

m_1 + m_2 + I_z/R^2

❌ The factor $\tfrac{1}{2}$ is missing. From rest at constant acceleration,

y_1 = \tfrac{1}{2}a t_1^2

❌ That is $\tfrac{1}{2}a t_1$ , which has the units of speed. The distance needs

t_1^2

y_1 = \tfrac{1}{2}a t_1^2

❌ That distance uses the massless-pulley acceleration, $3.92\ \text{m/s}^2$ . The heavy pulley lowers it to

3.43\ \text{m/s}^2

❌ Check the distance. Block 1 starts from rest with the constant acceleration just found:

y_1 = \tfrac{1}{2}a t_1^2

Show solution

Step 1: the pulley's effective mass. The radius cancels:

\frac{I_z}{R^2} = \frac{\tfrac{1}{2}m_p R^2}{R^2} = \tfrac{1}{2}m_p = 1.0\ \text{kg}

Step 2: the acceleration. The blocks move because $m_1 = 4.0\ \text{kg} > \mu_k m_2 = 1.2\ \text{kg}$ :

a = \frac{m_1 g - \mu_k m_2 g}{m_1 + m_2 + I_z/R^2} = \frac{39.2\ \text{N} - 11.76\ \text{N}}{4.0 + 3.0 + 1.0\ \text{kg}} = \frac{27.44\ \text{N}}{8.0\ \text{kg}} = 3.43\ \text{m/s}^2

Step 3: the distance fallen, from rest in $t_1 = 1.5\ \text{s}$ :

y_1 = \tfrac{1}{2}a t_1^2 = \tfrac{1}{2}(3.43\ \text{m/s}^2)(1.5\ \text{s})^2 = 3.85875\ \text{m} \approx 3.86\ \text{m}

The tensions, as a check:

T_1 = m_1(g - a) = (4.0)(9.8 - 3.43) = 25.48\ \text{N}, \qquad T_2 = m_2(\mu_k g + a) = (3.0)(3.92 + 3.43) = 22.05\ \text{N}

Their difference, $3.43\ \text{N}$ , is $(I_z/R^2)\,a = (1.0\ \text{kg})(3.43\ \text{m/s}^2)$ , as the pulley's torque equation requires.

Problem 4 · A Metre Stick Released from Horizontal

Given: a uniform metre stick ( $L = 1.00\ \text{m}$ , mass $0.20\ \text{kg}$ ) turns freely about a horizontal axle through one end, where its moment of inertia is $I_S = \tfrac{1}{3}ML^2$ ; it is held horizontal and released from rest ( $g = 9.8\ \text{m/s}^2$ ) — find the magnitude of its angular acceleration at the instant of release.

✅ Correct! The weight acts at the centre of gravity,

L/2

from the axle, so

|\alpha_z| = \dfrac{Mg\,(L/2)}{\tfrac{1}{3}ML^2} = \dfrac{3g}{2L} = 14.7\ \text{rad/s}^2

— the mass cancels.

❌ The weight does not act at the far end. With

\vec{g}

uniform, the whole weight acts at the centre of gravity, which is the centre of mass,

L/2

from the axle:

\tau = Mg\,L/2

❌ $\tfrac{1}{12}ML^2$ is the moment of inertia about the centre. The stick turns about its end, so

\tau_{S,z} = I_S\alpha_z

needs

I_S = \tfrac{1}{3}ML^2

❌ The weight may be placed at the centre of mass, but the mass may not.

M(L/2)^2 = \tfrac{1}{4}ML^2

ignores how the mass is spread along the stick; the torque is

Mg\,L/2

, but the moment of inertia stays

\tfrac{1}{3}ML^2

❌ Not quite. Put the whole weight at the centre of mass for the torque, use

\tfrac{1}{3}ML^2

for the inertia, and divide.

Show solution

Step 1: the gravitational torque about the axle. Take the stick to the right of the axle, clockwise positive. With $\vec{g}$ uniform, the weights of all the pieces of the stick act as one force $Mg$ at the centre of mass, a lever arm $L/2$ from the axle:

\tau_{S,z} = Mg\,\frac{L}{2} = (0.20\ \text{kg})(9.8\ \text{m/s}^2)(0.50\ \text{m}) = 0.98\ \text{N}\cdot\text{m}

Step 2: the moment of inertia about the end.

I_S = \tfrac{1}{3}ML^2 = \tfrac{1}{3}(0.20\ \text{kg})(1.00\ \text{m})^2 = \tfrac{1}{15}\ \text{kg}\cdot\text{m}^2

Step 3: the angular acceleration.

\alpha_z = \frac{\tau_{S,z}}{I_S} = \frac{0.98\ \text{N}\cdot\text{m}}{\tfrac{1}{15}\ \text{kg}\cdot\text{m}^2} = 14.7\ \text{rad/s}^2 = \frac{3g}{2L}

The far end starts with tangential acceleration $\alpha_z L = 14.7\ \text{m/s}^2 = 1.5g$ , faster than free fall.

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