Classical-Mechanics · Unit 22 · Video 1 · Interactive Practice

Torque Through an Angle: Rotational Work and Power

IKey Formulas

Formula	Name	What it says
$dW_{\text{rot}} = \tau_{S,z}\,d\theta,\qquad W_{\text{rot}} = \displaystyle\int_{\theta_i}^{\theta_f} \tau_{S,z}\,d\theta$	Rotational work	Torque about the axis times the angle turned; $\text{N}\cdot\text{m} \times \text{rad} = \text{J}$ , because the radian is a pure number
$W_{\text{rot}} = \tfrac{1}{2} I_{\text{cm}}\,\omega_f^2 - \tfrac{1}{2} I_{\text{cm}}\,\omega_i^2 = \Delta K_{\text{rot}}$	Rotational work–kinetic energy theorem	For an axis through the centre of mass; $\omega$ enters squared, so only its magnitude matters
$P_{\text{rot}} = \tau_{S,z}\,\omega_z,\qquad P_{\text{ave}} = \Delta W_{\text{rot}}/\Delta t$	Rotational power	In watts; positive when the torque shares the sense of the rotation, negative when it opposes it
$F_x \to \tau_{S,z},\quad m \to I_S,\quad x \to \theta,\quad v_x \to \omega_z$	Translation → rotation dictionary	Each result above is the twin of a translational one under this substitution

Key Insight: Only the tangential component of a force does work on a rigid body turning about a fixed axis — and $r_i F_{\theta,i}$ is precisely that force's torque about the axis. Work through a distance becomes torque through an angle.

IIWhich Part of the Force Does Work

The element moves along its circle, so only $F_\theta$ has any projection on the displacement $r\,\Delta\theta\,\hat{\theta}$ .

\Delta\theta

(rad) 0.20

IIITorque Through an Angle Is the Change in $K_{\text{rot}}$

A wheel with $I_0 = 0.020\ \text{kg}\cdot\text{m}^2$ spins at $\omega_0 = 30\ \text{rad/s}$ when the axle torque takes over.

t

(s) 0.00 Axle torque

IVOne Equation, Two Unknowns

A second washer, $I_w = 0.01125\ \text{kg}\cdot\text{m}^2$ , drops onto the wheel at $\omega_a = 20\ \text{rad/s}$ and they reach a common $\omega_b$ in $0.50\ \text{s}$ .

\omega_b

(rad/s) 12.0

💡 The relation that does fix $\omega_b$ comes from angular momentum, a quantity introduced in a later chapter; the energy equation then measures $W_{\text{slip}}$ , the heat made between the two washers.

VQuiz Questions

Problem 1 · Work Done by a Constant Torque

Given: A motor drives a wheel with a constant torque $\tau_{S,z} = +0.050\ \text{N}\cdot\text{m}$ while the wheel turns through $\Delta\theta = 140\ \text{rad}$ in the same sense — find the rotational work $W_{\text{rot}}$ .

✅ Correct!

W_{\text{rot}} = \tau_{S,z}\,\Delta\theta = (0.050\ \text{N}\cdot\text{m})(140\ \text{rad}) = 7.0\ \text{J}

, and

\text{N}\cdot\text{m} \times \text{rad} = \text{J}

because the radian is a pure number.

❌ Not quite. That is the answer for

140

revolutions' worth of angle divided out:

140\ \text{rad} = 22.3\ \text{rev}

. The angle in

W_{\text{rot}} = \tau_{S,z}\,\Delta\theta

is already in radians — use it as it stands.

❌ You converted the wrong way. Multiplying by

2\pi

turns revolutions into radians, but

140\ \text{rad}

is already in radians. No conversion is needed.

❌ Check the sign. The work is negative only when the torque opposes the rotation. Here the torque acts in the same sense as the turning, so it adds energy:

W_{\text{rot}} > 0

Show solution

The torque is constant, so it comes straight out of the integral:

W_{\text{rot}} = \int_{\theta_i}^{\theta_f} \tau_{S,z}\,d\theta = \tau_{S,z}\,\Delta\theta = (0.050\ \text{N}\cdot\text{m})(140\ \text{rad}) = 7.0\ \text{J}

The sign is positive because $\tau_{S,z}$ and $\Delta\theta$ have the same sense, so the rotational kinetic energy grows by $7.0\ \text{J}$ . For a wheel with $I_0 = 0.020\ \text{kg}\cdot\text{m}^2$ starting at $30\ \text{rad/s}$ :

\tfrac{1}{2}(0.020)(30)^2 + 7.0 = 9.0 + 7.0 = 16.0\ \text{J} = \tfrac{1}{2}(0.020)\,\omega_f^2 \ \Rightarrow\ \omega_f = 40\ \text{rad/s}

Problem 2 · Work Done by Bearing Friction

Given: The motor is switched off and the same wheel, $I_0 = 0.020\ \text{kg}\cdot\text{m}^2$ , coasts from $\omega_0 = 30\ \text{rad/s}$ down to $\omega_a = 20\ \text{rad/s}$ in $\Delta t_1 = 4.0\ \text{s}$ — find the work done by the bearing friction over that coast.

✅ Correct!

W_f = \Delta K_{\text{rot}} = 4.00 - 9.00 = -5.0\ \text{J}

, and the same number follows from

-\tau_f\,\Delta\theta_1 = -(0.050)(100\ \text{rad})

❌ Check the sign. A frictional torque always opposes the rotation, so its work is negative: it removes

5.0\ \text{J}

of rotational kinetic energy and deposits it as thermal energy at the bearing.

❌ The difference is not squared.

\tfrac{1}{2}I_0(\omega_0 - \omega_a)^2

is not the change in kinetic energy. Square each angular velocity first, then subtract:

\omega_f^2 - \omega_i^2 = 400 - 900

, not

(30-20)^2

❌ Not quite.

\tfrac{1}{2}I_0(\omega_0^2 + \omega_a^2)

adds the two kinetic energies. The theorem asks for the change: kinetic energy after minus kinetic energy before.

Show solution

Bearing friction is the only torque acting, so the rotational work–kinetic energy theorem gives its work directly:

W_f = \tfrac{1}{2} I_0\,\omega_a^2 - \tfrac{1}{2} I_0\,\omega_0^2 = \tfrac{1}{2}(0.020)(20)^2 - \tfrac{1}{2}(0.020)(30)^2 = 4.00 - 9.00 = -5.0\ \text{J}

Cross-check with torque times angle. The angular acceleration is constant, so

\tau_f = I_0\,\frac{\omega_0 - \omega_a}{\Delta t_1} = (0.020)\frac{10}{4.0} = 0.050\ \text{N}\cdot\text{m}, \qquad \Delta\theta_1 = \tfrac{1}{2}(\omega_0 + \omega_a)\,\Delta t_1 = 100\ \text{rad}

W_f = -\tau_f\,\Delta\theta_1 = -(0.050)(100) = -5.0\ \text{J}\ \checkmark

Problem 3 · The Angle Turned During the Collision

Given: At $\omega_a = 20\ \text{rad/s}$ a second washer is dropped onto the coasting wheel. Over $\Delta t_{\text{int}} = 0.50\ \text{s}$ the two reach a common angular speed $\omega_b = 12\ \text{rad/s}$ , while the bearing torque stays at $\tau_f = 0.050\ \text{N}\cdot\text{m}$ . Model the collision as constant angular acceleration — find the angle the rotor turns through and the work the bearing friction does.

Angle turned during the collision

Work done by the bearing friction

✅ Correct! The rotor turns

\Delta\theta_2 = \tfrac{1}{2}(20 + 12)(0.50) = 8.0\ \text{rad}

, and the bearing torque opposes that turning, so

W_{f,b} = -(0.050)(8.0) = -0.40\ \text{J}

❌ Check the angle. With constant angular acceleration the angle is the average angular speed times the time,

\tfrac{1}{2}(\omega_a + \omega_b)\,\Delta t_{\text{int}}

— neither endpoint speed on its own.

❌ The factor $\tfrac{1}{2}$ is missing.

(\omega_a + \omega_b)\,\Delta t_{\text{int}} = 16\ \text{rad}

uses the sum of the speeds; the angle uses their average,

\tfrac{1}{2}(\omega_a + \omega_b)\,\Delta t_{\text{int}}

❌ Check the work.

W_{f,b} = (\tau_f)_z\,\Delta\theta_2

with

(\tau_f)_z = -0.050\ \text{N}\cdot\text{m}

and

\Delta\theta_2 = 8.0\ \text{rad}

❌ Check the sign. Counterclockwise is positive here, so the bearing torque has

z

-component

-\tau_f

while

\Delta\theta_2 > 0

. Their product, and hence the work, is negative.

Show solution

Part (a) — the angle. The average angular acceleration over the collision is $\alpha_2 = (\omega_b - \omega_a)/\Delta t_{\text{int}} < 0$ , and the constant-acceleration formula gives

\Delta\theta_2 = \omega_a\,\Delta t_{\text{int}} + \tfrac{1}{2}\alpha_2\,\Delta t_{\text{int}}^2 = \omega_a\,\Delta t_{\text{int}} + \tfrac{1}{2}(\omega_b - \omega_a)\,\Delta t_{\text{int}} = \tfrac{1}{2}(\omega_a + \omega_b)\,\Delta t_{\text{int}}

\Delta\theta_2 = \tfrac{1}{2}(20 + 12)(0.50) = 8.0\ \text{rad}

Part (b) — the work. With counterclockwise positive the bearing torque has $z$ -component $-\tau_f$ :

W_{f,b} = -\tau_f\,\Delta\theta_2 = -(0.050\ \text{N}\cdot\text{m})(8.0\ \text{rad}) = -0.40\ \text{J}

Negative, as it must be for a torque that opposes the rotation.

Problem 4 · Power, and What Energy Cannot Settle

Given: During that same collision the rotational kinetic energy falls from $\tfrac{1}{2}I_0\,\omega_a^2 = 4.00\ \text{J}$ to $\tfrac{1}{2}(I_0 + I_w)\,\omega_b^2 = 2.25\ \text{J}$ , and the bearing friction does $W_{f,b} = -0.40\ \text{J}$ in $\Delta t_{\text{int}} = 0.50\ \text{s}$ — find the average power of the bearing friction and the work $W_{\text{slip}}$ done by the friction between the two washers.

Average power of the bearing friction

Work done by the sliding friction between the washers

✅ Correct!

P_f = W_{f,b}/\Delta t_{\text{int}} = -0.80\ \text{W}

, matching

(\tau_f)_z\,\omega_{\text{ave}} = -(0.050)(16) = -0.80\ \text{W}

; and

W_{\text{slip}} = \Delta K_{\text{rot}} - W_{f,b} = -1.75 + 0.40 = -1.35\ \text{J}

❌ Check the power. Average power is work divided by the time interval:

P_f = W_{f,b}/\Delta t_{\text{int}}

, in watts.

❌ You multiplied by the time.

W_{f,b}\,\Delta t_{\text{int}}

has units of

\text{J}\cdot\text{s}

. Power is a rate: divide the work by the interval,

-0.40/0.50

❌ Check the sign. The magnitude is right, but the bearing torque opposes the rotation, so it does negative work:

W_{f,b} = -0.40\ \text{J}

and therefore

P_f = -0.80\ \text{W}

— energy is leaving the rotor at that rate.

❌ That uses the wrong angular velocity.

(\tau_f)_z\,\omega_a = -1.0\ \text{W}

is the instantaneous power at the start. Averaged over the collision the rotor turns at

\tfrac{1}{2}(\omega_a + \omega_b) = 16\ \text{rad/s}

❌ Check the energy equation.

\Delta K_{\text{rot}} = W_{f,b} + W_{\text{slip}}

, so

W_{\text{slip}} = \Delta K_{\text{rot}} - W_{f,b}

with

\Delta K_{\text{rot}} = 2.25 - 4.00

❌ That is the whole energy change.

-1.75\ \text{J}

\Delta K_{\text{rot}}

, and part of it went to the bearing. Subtract the bearing's share,

W_{f,b} = -0.40\ \text{J}

, to isolate the sliding term.

❌ Check the sign. Sliding friction between the two washers always converts kinetic energy to thermal energy, so

W_{\text{slip}} < 0

— it can never add energy to the pair.

❌ Sign slip in the rearrangement. From

\Delta K_{\text{rot}} = W_{f,b} + W_{\text{slip}}

you subtract

W_{f,b}

, which is itself negative:

-1.75 - (-0.40) = -1.35\ \text{J}

Show solution

Average power. Power is the rate at which the work is done:

P_f = \frac{W_{f,b}}{\Delta t_{\text{int}}} = \frac{-0.40\ \text{J}}{0.50\ \text{s}} = -0.80\ \text{W}

The $\Delta t_{\text{int}}$ cancels against the one hidden in $\Delta\theta_2$ , so the same number comes from $P_{\text{rot}} = \tau_{S,z}\,\omega_z$ evaluated at the average angular velocity:

P_f = -\tau_f\cdot\tfrac{1}{2}(\omega_a + \omega_b) = -(0.050)(16) = -0.80\ \text{W}

The sliding work. Two torques do work on the rotor-plus-washers system during the collision, so the theorem reads

\tfrac{1}{2}(I_0 + I_w)\,\omega_b^2 - \tfrac{1}{2} I_0\,\omega_a^2 = W_{f,b} + W_{\text{slip}}

2.25 - 4.00 = -0.40 + W_{\text{slip}} \ \Rightarrow\ W_{\text{slip}} = -1.35\ \text{J}

Note what this equation cannot do: with $\omega_b$ unknown it holds two unknowns, $\omega_b$ and $W_{\text{slip}}$ , so energy alone never predicts the common angular speed. Dropping $W_{\text{slip}}$ would leave a solvable quadratic, but it would describe a collision with no internal loss — which this one is not. Angular momentum supplies the missing relation; the energy equation then reports the heat, $1.35\ \text{J}$ of it made between the washers.

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Torque Through an Angle: Rotational Work and Power

IKey Formulas

IIWhich Part of the Force Does Work

IIITorque Through an Angle Is the Change in KrotK_{\text{rot}}Krot​

IVOne Equation, Two Unknowns

VQuiz Questions

IIITorque Through an Angle Is the Change in $K_{\text{rot}}$