Classical-Mechanics · Unit 20 · Video 5 · Interactive Practice

Conservation of Energy in Fixed-Axis Rotation: The Pulley and Incline Problem

IKey Formulas

Formula	Name	What it gives
$U_f + \tfrac{1}{2} I_S\,\omega_f^2 = U_i + \tfrac{1}{2} I_S\,\omega_i^2$	Energy conservation, fixed axis	The balance for a closed system with conservative internal forces; translating bodies add their own $\tfrac{1}{2}mv^2$
$v = R\omega \;\Rightarrow\; \tfrac{1}{2} I_P\,\omega^2 = \tfrac{1}{2}\dfrac{I_P}{R^2}v^2 = \tfrac{1}{4} m_P v^2$	No-slip condition	The disc's kinetic energy in terms of the cord speed: an extra $\tfrac{1}{2}m_P$ moving with the cord
$d = x_2 - x_{2,0} = y_{1,0} - y_1$	Cord constraint	One distance for both blocks — block 2 slides down $d$ , block 1 rises $d$
$v = \sqrt{\dfrac{2gd\,(-m_1 + m_2\sin\theta)}{m_1 + m_2 + \tfrac{1}{2}m_P}}$	Speed after sliding $d$	Numerator: potential energy released per unit distance. Denominator: everything the cord moves

Key Insight: The disc enters the energy balance only through $I_P/R^2 = \tfrac{1}{2}m_P$ , so its radius cancels: a bigger disc has a larger moment of inertia but turns proportionally slower, and $\omega^2$ removes the $R^2$ exactly.

IIVisualization 1 — Where the Released Energy Goes

Released from rest, block 2 slides down and the potential energy it gives up is split three ways.

Slide block 2 0.00 m

💡 The cord's tension never appears in the balance: it does negative work on block 2 and exactly equal positive work on block 1 and the disc, so only gravity moves energy into the system.

IIIVisualization 2 — The Disc as an Extra Half-Mass

The disc enters only as $I_P/R^2 = \tfrac{1}{2}m_P$ — an extra mass carried along by the cord.

Pulley mass mP 4.0 kg Pulley radius R 0.25 m

IVVisualization 3 — Which Way Does It Go?

The sign of $-m_1 + m_2\sin\theta$ decides the motion; at zero the blocks never move at all.

Hanging mass m1 2.0 kg

💡 Challenge: with $m_1 = 2.0\ \mathrm{kg}$ , find the angle that holds the system at rest. The balance condition $m_2\sin\theta = m_1$ contains no $m_P$ : the disc sets how fast the system moves, never which way.

VQuiz Questions

Problem 1 · The Video's Result in Numbers

Given: $m_1 = 2.00\ \mathrm{kg}$ hangs from the cord, $m_2 = 6.00\ \mathrm{kg}$ rests on a frictionless incline at $\theta = 30\degree$ , and the pulley is a uniform disc of mass $m_P = 4.00\ \mathrm{kg}$ and radius $R = 0.25\ \mathrm{m}$ . The cord does not slip. Released from rest, find the speed of block 2 after it has slid $d = 1.00\ \mathrm{m}$ down the slope. Take $g = 9.8\ \mathrm{m/s^2}$ .

✅ Correct!

\;v = \sqrt{2(9.8)(1.00)(1.00)/10.0} = \sqrt{1.96} = 1.40\ \mathrm{m/s}

, with

1.00\ \mathrm{kg}

of driving mass against

10.0\ \mathrm{kg}

of moving inertia.

❌ Not quite. That is the whole pulley mass in the denominator. The no-slip condition gives

\tfrac{1}{2}I_P\omega^2 = \tfrac{1}{4}m_P v^2

, so the pulley contributes

I_P/R^2 = \tfrac{1}{2}m_P = 2.00\ \mathrm{kg}

, not

4.00\ \mathrm{kg}

❌ That is the massless-pulley answer. Dropping

\tfrac{1}{2}I_P\omega^2

leaves only

m_1 + m_2 = 8.00\ \mathrm{kg}

in the denominator and predicts a speed that is too high.

❌ Check the incline. Only the component of

m_2\vec{g}

along the slope releases energy: the numerator is

-m_1 + m_2\sin\theta = -2.00 + 3.00

, not

-m_1 + m_2

Show solution

Step 1 — the moving inertia. No slip means $\omega = v/R$ , so the pulley's term is

\tfrac{1}{2}I_P\omega^2 = \tfrac{1}{2}\frac{I_P}{R^2}v^2 = \tfrac{1}{2}\big(\tfrac{1}{2}m_P\big)v^2, \qquad \frac{I_P}{R^2} = \tfrac{1}{2}m_P = 2.00\ \mathrm{kg}

$m_1 + m_2 + \tfrac{1}{2}m_P = 2.00 + 6.00 + 2.00 = 10.0\ \mathrm{kg}$

Step 2 — the energy released. Block 2 drops $d\sin\theta$ while block 1 is lifted $d$ :

-\Delta U = gd\,(-m_1 + m_2\sin\theta) = (9.8)(1.00)\big(-2.00 + 6.00(0.500)\big) = 9.80\ \mathrm{J}

Step 3 — solve. All of it becomes kinetic energy:

\tfrac{1}{2}(10.0)v^2 = 9.80\ \mathrm{J} \quad\Longrightarrow\quad v = \sqrt{\frac{2(9.80)}{10.0}} = \sqrt{1.96} = 1.40\ \mathrm{m/s}

Equivalently, straight from the boxed formula:

v = \sqrt{\frac{2gd\,(-m_1 + m_2\sin\theta)}{m_1 + m_2 + \tfrac{1}{2}m_P}} = \sqrt{\frac{2(9.8)(1.00)(1.00)}{10.0}} = 1.40\ \mathrm{m/s}

Problem 2 · A Bigger Disc, Same Mass

Given: the same system, except that the pulley is replaced by a uniform disc of the same mass $m_P = 4.00\ \mathrm{kg}$ and twice the radius, $R = 0.50\ \mathrm{m}$ . Find the speed of block 2 after $d = 1.00\ \mathrm{m}$ .

✅ Correct! The radius never appears in the result:

I_P

grows as

R^2

while

\omega^2 = v^2/R^2

shrinks by the same factor, so

I_P/R^2 = \tfrac{1}{2}m_P

is unchanged.

❌ Not quite. The moment of inertia really does become four times larger — but at the same cord speed the bigger disc turns half as fast, and

\omega^2

removes exactly that factor of four.

❌ Check the units.

I_P = \tfrac{1}{2}m_PR^2 = 0.500\ \mathrm{kg\,m^2}

is a moment of inertia, not a mass; what the energy balance needs is

I_P/R^2 = 2.00\ \mathrm{kg}

❌ Not quite. Write the pulley's kinetic energy in terms of the cord speed before comparing the two discs.

Show solution

With $R = 0.50\ \mathrm{m}$ the moment of inertia is four times what it was:

I_P = \tfrac{1}{2}m_PR^2 = \tfrac{1}{2}(4.00)(0.500)^2 = 0.500\ \mathrm{kg\,m^2} \quad (\text{was } 0.125\ \mathrm{kg\,m^2})

But the cord still runs at $v$ , so the disc turns at $\omega = v/R$ , half as fast as before:

\tfrac{1}{2}I_P\omega^2 = \tfrac{1}{2}\frac{I_P}{R^2}v^2 = \tfrac{1}{2}\Big(\frac{0.500}{0.250}\Big)v^2 = \tfrac{1}{2}(2.00)v^2

$I_P/R^2 = \tfrac{1}{2}m_P = 2.00\ \mathrm{kg}$ exactly as before, so the denominator is still $10.0\ \mathrm{kg}$ and

v = \sqrt{\frac{2(9.8)(1.00)(1.00)}{10.0}} = 1.40\ \mathrm{m/s}

Only the pulley's mass matters, never its radius.

Problem 3 · Splitting the Kinetic Energy

Given: the original system again ( $m_1 = 2.00\ \mathrm{kg}$ , $m_2 = 6.00\ \mathrm{kg}$ , $m_P = 4.00\ \mathrm{kg}$ , $R = 0.25\ \mathrm{m}$ , $\theta = 30\degree$ ), after block 2 has slid $d = 1.00\ \mathrm{m}$ from rest.

What share of the kinetic energy is stored in the spinning disc?

How much kinetic energy is that, in joules?

✅ Correct! The three kinetic energies stand in the ratio

m_2 : m_1 : \tfrac{1}{2}m_P = 6 : 2 : 2

, so the disc holds

20\%

9.80\ \mathrm{J}

, that is

1.96\ \mathrm{J}

❌ Check the share. Every term carries the same

\tfrac{1}{2}v^2

, so the shares are just

m_1

m_2

and

\tfrac{1}{2}m_P

divided by their sum

10.0\ \mathrm{kg}

— and the pulley's entry is

\tfrac{1}{2}m_P

, not

m_P

❌ Check the joules. The total kinetic energy equals the potential energy released,

9.80\ \mathrm{J}

; multiply it by the disc's share, or evaluate

\tfrac{1}{4}m_Pv^2

directly with

v = 1.40\ \mathrm{m/s}

Show solution

Shares. Every kinetic term has the same factor $\tfrac{1}{2}v^2$ :

K = \tfrac{1}{2}m_1v^2 + \tfrac{1}{2}m_2v^2 + \tfrac{1}{2}\big(\tfrac{1}{2}m_P\big)v^2, \qquad \frac{K_P}{K} = \frac{\tfrac{1}{2}m_P}{m_1 + m_2 + \tfrac{1}{2}m_P} = \frac{2.00}{10.0} = 0.200

Joules. The total is the energy released, $K = -\Delta U = gd(-m_1 + m_2\sin\theta) = 9.80\ \mathrm{J}$ , so

K_P = 0.200 \times 9.80 = 1.96\ \mathrm{J}

Check directly with $v = 1.40\ \mathrm{m/s}$ and $\omega = v/R = 5.60\ \mathrm{rad/s}$ :

K_P = \tfrac{1}{2}I_P\omega^2 = \tfrac{1}{2}(0.125)(5.60)^2 = 1.96\ \mathrm{J} = \tfrac{1}{4}m_Pv^2\ \checkmark

The full ledger: $K_2 = 5.88\ \mathrm{J}$ , $K_1 = 1.96\ \mathrm{J}$ , $K_P = 1.96\ \mathrm{J}$ , summing to $9.80\ \mathrm{J}$ .

Problem 4 · When Does It Move at All?

Given: the same masses and pulley ( $m_1 = 2.00\ \mathrm{kg}$ , $m_2 = 6.00\ \mathrm{kg}$ , $m_P = 4.00\ \mathrm{kg}$ ), but the incline angle can now be set to any value. Released from rest, above which angle does block 2 slide down the slope?

✅ Correct!

\sin\theta > m_1/m_2 = 1/3

gives

\theta > 19.47\degree

— and

m_P

is absent, because the pulley changes how fast the system moves, never which way.

❌ Wrong masses. That is

\sin\theta = m_1/(m_1 + m_2 + \tfrac{1}{2}m_P)

. The denominator of

v

is a sum of masses and is always positive; the sign is set by the numerator

-m_1 + m_2\sin\theta

alone.

❌ Close, but that is $\tan\theta = 1/3$ . The slope component of block 2's weight is

m_2g\sin\theta

, not

m_2g\tan\theta

❌ Wrong component.

m_2g\cos\theta

presses into the surface and is cancelled by the normal force, which does no work; the energy is released by

m_2g\sin\theta

along the slope.

Show solution

Under the square root, $v^2$ cannot be negative:

v^2 = \frac{2gd\,(-m_1 + m_2\sin\theta)}{m_1 + m_2 + \tfrac{1}{2}m_P} \;\ge\; 0

The denominator is a sum of masses, so the sign lives entirely in $d\,(-m_1 + m_2\sin\theta)$ . Block 2 slides down — that is, $d > 0$ — exactly when the bracket is positive:

m_2\sin\theta > m_1 \quad\Longleftrightarrow\quad \sin\theta > \frac{m_1}{m_2} = \frac{2.00}{6.00} = \frac{1}{3}

\theta > \arcsin\tfrac{1}{3} = 19.47\degree

Reading the three cases:

$\theta > 19.47\degree$ : the slope component $m_2g\sin\theta$ beats block 1's weight $m_1g$ , block 2 runs down and block 1 is lifted.
$\theta < 19.47\degree$ : block 1 descends and drags block 2 up the slope; $d$ is then negative, both factors flip sign, and the same formula still gives the speed.
$\theta = 19.47\degree$ : the two weights balance along the cord and $v = 0$ for every $d$ — the blocks never leave their starting positions.

At $\theta = 30\degree$ we are above the threshold, which is why Problem 1 gave a real speed.

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