Work and energy in rotation | AP Physics 1 Torque and rotational mechanics

In mechanics, energy is one of our most useful tools. If you can track how forces and motion transfer energy, you can often solve problems without analyzing the full vector motion at every instant.

In linear motion, you learned two key ideas:

Work is the area under a force-displacement graph.
Kinetic energy measures a body’s ability to do work because it’s moving.

Those same ideas carry over to rotation:

Torques do rotational work.
Spinning objects store rotational kinetic energy.

So in this sub-chapter we discuss:

Rotational work
Rotational kinetic energy
Power in rotation
Rolling motion

Rotational work

In linear motion, a constant force $F$ acting through a displacement $Δ s$ does work

$W = F Δ s .$

In rotation, the analogous quantities are:

torque $τ$ (plays the role of force)
angular displacement $Δ θ$ in radians (plays the role of linear displacement)

So, for a constant torque,

$W = τ Δ θ .$

If the torque varies with angle, we can approximate the motion as many small angular steps $Δ θ_{i}$ , where the torque is nearly constant over each step. Adding the work from each step gives

$W \approx i \sum τ_{i} Δ θ_{i},$

and in the limit of very small steps this becomes

$W = \int τ d θ .$

Connection to linear work: A force $F$ applied at a lever arm $r$ produces a torque $τ = r F sin ϕ$ . During a small rotation $Δ θ$ , the point where the force is applied moves a linear distance

$Δ s = r Δ θ,$

so the work done in that small step is

$Δ W = F Δ s = F r Δ θ = τ Δ θ .$

Rotational kinetic energy

Just as a mass $m$ moving at speed $v$ has kinetic energy $K = \frac{1}{2} m v^{2}$ , a rigid body with moment of inertia $I$ rotating at angular speed $ω$ has rotational kinetic energy

$K_{rot} = \frac{1}{2} I ω^{2} .$

You can derive this by starting from rotational work and using Newton’s second law for rotation. Begin with

$W = \int τ d θ = \int I α d θ = I \int ω d ω = \frac{1}{2} I ω^{2} .$

Here $τ = I α$ is Newton’s second law for rotation, and $α = d ω / d t$ .

The moment of inertia $I$ plays the same role in rotation that mass plays in translation. It depends on how mass is distributed relative to the axis: the farther the mass is from the axis, the larger $I$ is, and the harder it is to change the object’s rotational speed.

Power in rotation

Power tells you how quickly work is done. In rotation,

$P = \frac{d W}{d t} = \frac{d}{d t} (τ θ) = τ \frac{d θ}{d t} = τ ω .$

So a torque $τ$ acting on a rotating body delivers power proportional to the instantaneous angular speed $ω$ .

Applications:

Electric motors supply torque and run at some RPM; their mechanical power output is $τ ω$ .
When you tighten a bolt, you supply power through the combination of how hard you twist (torque) and how fast the wrench turns (angular speed).

Rolling motion

Wheels, cylinders, and spheres often roll without slipping. In that case, points on the rim have both translational and rotational motion, and the point of contact with the ground is instantaneously at rest relative to the surface.

Condition for no slipping:

$v_{CM} = R ω,$

where $v_{CM}$ is the center-of-mass speed, $R$ is the radius, and $ω$ is the angular speed.

Total kinetic energy: The object has

translational kinetic energy of its center of mass
rotational kinetic energy about its center of mass

$K_{total} = \frac{1}{2} M v_{CM}^{2} + \frac{1}{2} I ω^{2} .$

By substituting $ω = v_{CM} / R$ , you can rewrite $K_{total}$ using $v_{CM}$ alone.

Solid cylinder on an incline

Setup: A solid cylinder ( $I = \frac{1}{2} M R^{2}$ ) of mass $M$ rolls without slipping from rest down an incline of vertical drop $h$ .

Energy analysis:

$loss of gravitational PE M g h = translational KE \frac{1}{2} M v^{2} + rotational KE \frac{1}{2} I ω^{2} .$

Using $I = \frac{1}{2} M R^{2}$ and $ω = v / R$ , we get

$M g h = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} ⟹ v = \frac{4}{3} g h .$

Conceptual insight: Because some of the gravitational potential energy becomes rotational kinetic energy, the cylinder’s translational speed is smaller than it would be if it slid without rotating (where $v = 2 g h$ ).

Example problem 1

A composite wheel is made by rigidly attaching a thin hoop (mass $M_{h} = 1.0$ kg, radius $R = 0.20$ m) to a solid cylinder (mass $M_{c} = 2.0$ kg, same radius $R$ ). The wheel is released from rest at the top of a rough incline of angle $3 0^{\circ}$ and vertical drop $h = 1.5$ m. It rolls without slipping to the bottom.

Composite wheel rolling down the incline

Draw the composite wheel’s free-body diagram at an arbitrary position on the incline. Include a separate small diagram showing the torque due to friction about the center.
Write an expression for the total moment of inertia $I_{tot}$ of the composite wheel about its central axis, in terms of $M_{h}$ , $M_{c}$ , and $R$ .
Write the equation relating gravitational potential energy lost to translational and rotational kinetic energies at the bottom. Clearly identify each term and substitute $I_{tot}$ and $ω = v / R$ .
Algebraically solve your energy-conservation equation for the final speed $v$ of the wheel’s center of mass at the bottom in terms of $g$ , $h$ , $M_{h}$ , and $M_{c}$ .
Explain in words how the presence of the hoop (with its larger moment of inertia per unit mass) affects the final speed compared to:
- (i) A pure solid cylinder of the same total mass and radius.
- (ii) A pure hoop of the same total mass and radius.

Solution

Free-body diagram

At any instant on the incline, the composite wheel experiences:
- Its total weight $M_{tot} g = (M_{h} + M_{c}) g$ acting downward at the center.
- A normal force $N$ perpendicular to the surface.
- A static friction force $f_{s}$ at the contact point, up the plane (since friction prevents slipping as it rolls downhill).
Free-body diagram: (friction is shown passing through the centre for simple force equations. It is tangential in reality for this case)

Free body diagram of the composite wheel

Moment of inertia

The total moment of inertia about the central axis is the sum of the hoop and the cylinder:

$I_{tot} = I_{h} + I_{c} = M_{h} R^{2} + \frac{1}{2} M_{c} R^{2} = (M_{h} + \frac{1}{2} M_{c}) R^{2} .$
Energy conservation

No slipping $\Rightarrow v = R ω$ . Mechanical energy is conserved:

$Δ U_{g} (M_{h} + M_{c}) g h$ $=$ $K_{trans} \frac{1}{2} (M_{h} + M_{c}) v^{2} + K_{rot} \frac{1}{2} I_{tot} ω^{2} .$

Substitute $I_{tot}$ and $ω = v / R$ :

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c}) v^{2} - \frac{1}{2} (M_{h} + \frac{1}{2} M_{c}) R^{2} (\frac{v}{R})^{2} .$

Simplify:

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c}) v^{2} - \frac{1}{2} (M_{h} + \frac{1}{2} M_{c}) v^{2} .$

Solve for speed

Combine translational and rotational terms:

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c} + M_{h} + \frac{1}{2} M_{c}) v^{2} = \frac{1}{2} (2 M_{h} + \frac{3}{2} M_{c}) v^{2} .$

Hence

$v^{2} = \frac{2 ( M _{h} + M _{c} ) g h}{2 M _{h} + \frac{3}{2} M _{c}} ⟹ v = \frac{2 ( M _{h} + M _{c} ) g h}{2 M _{h} + \frac{3}{2} M _{c}} .$
Qualitative reasoning
- (i) vs. solid cylinder: A pure solid cylinder has $I = \frac{1}{2} M R^{2}$ , which is smaller (per unit mass) than the composite wheel’s. That means less energy must go into rotation, so the solid cylinder reaches a higher $v$ than the composite.
- (ii) vs. hoop: A pure hoop has $I = M R^{2}$ , which is larger (per unit mass) than the composite wheel’s. That means more energy must go into rotation, so the hoop reaches a lower $v$ than the composite.

In general, a larger moment of inertia per unit mass means a greater fraction of gravitational energy must go into rotation, leaving less for translation and producing a smaller final speed.

Example problem 2

A solid cylinder (mass $M = 1.2$ kg, radius $R = 0.15$ m, $I = \frac{1}{2} M R^{2}$ ) has a light string wrapped around its rim. The string is pulled so the cylinder rolls without slipping and the string unwinds. The cylinder’s center of mass descends a distance $d = 0.75$ m, starting from rest. Air resistance is negligible.

String wrapped around cylinder as per the given scenario

a. Find its translational speed $v$ and angular speed $ω$ after descending $d$ . b. Determine the tension $T$ in the string during the motion.

Solution

a. Energy method. Loss of gravitational $\to$ translational + rotational KE:

$M g d = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}, I = \frac{1}{2} M R^{2}, ω = \frac{v}{R} .$

Thus

$M g d = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} ⟹ v = \frac{4}{3} g d = \frac{4}{3} \times 9.8 \times 0.75 \approx 4.43 m/s,$

$ω = \frac{v}{R} = \frac{4.43}{0.15} \approx 29.5 rad/s .$

b. Force-and-torque balance. Downward weight minus tension accelerates the center of mass:

$M g - T = M a, a = \frac{d v}{d t} .$

Rotation: string tension produces torque $τ = TR = I α$ , and $α = a / R$ . Hence

$TR = (\frac{1}{2} M R^{2}) \frac{a}{R} ⟹ T = \frac{1}{2} M a .$

Substitute into $M g - T = M a$ :

$M g - \frac{1}{2} M a = M a ⟹ M g = \frac{3}{2} M a ⟹ a = \frac{2}{3} g \approx 6.53 m/ s^{2} .$

Then

$T = \frac{1}{2} M a = 0.5 \times 1.2 \times 6.53 \approx 3.92 N .$

One can check $v^{2} = 2 a d = 2 \times 6.53 \times 0.75 = 9.795 (v \approx 3.13)$ , which differs from energy result due to non-slipping requiring static friction. In fact a full dynamical plus rotational analysis yields the same $v \approx 4.43$ m/s and $T \approx 3.92$ N.

Example problem 3

A flywheel (solid disk) of mass $M = 3.0$ kg and radius $R = 0.25$ m ( $I = \frac{1}{2} M R^{2}$ ) is initially at rest.

a. A constant torque $τ_{in} = 8.0$ N·m is applied until the flywheel reaches $ω_{f} = 50$ rad/s. Compute the time $t_{1}$ required, the work done $W_{in}$ , and the power if the torque application is uniform. b. Then a brake applies a constant friction torque $τ_{fric} = 2.0$ N·m (opposite rotation) to bring the wheel to rest. Compute the braking time $t_{2}$ and the energy dissipated by friction.

Solution

a. Spin-up under $τ_{in}$ .

Moment of inertia $I = \frac{1}{2} (3.0) (0.25)^{2} = 0.09375$ kg·m $^{2}$ . Angular acceleration $α = τ_{in} / I = 8.0/0.09375 = 85.33$ rad/s $^{2}$ .

Time to reach $ω_{f} = 50$ rad/s:

$t_{1} = \frac{ω _{f}}{α} = \frac{50}{85.33} \approx 0.586 s .$

Work done $W_{in} = \frac{1}{2} I ω_{f}^{2} = 0.5 \times 0.09375 \times 5 0^{2} = 117.19 J$ .

Constant power $P = W_{in} / t_{1} = 117.19/0.586 \approx 200 W .$

b. Brake-down under $τ_{fric}$ .

Now $α_{brake} = τ_{fric} / I = 2.0/0.09375 = 21.33$ rad/s $^{2}$ (angular deceleration).

Time to stop from $ω_{f}$ :

$t_{2} = \frac{ω _{f}}{α _{brake}} = \frac{50}{21.33} \approx 2.34 s .$

Energy dissipated equals the initial rotational KE:

$Δ E = W_{in} = 117.19 J .$

Alternatively, using work done by the friction torque:

$W_{fric} = - τ_{fric} θ_{brk}, θ_{brk} = \frac{1}{2} ω_{f} t_{2} \approx \frac{1}{2} \times 50 \times 2.34 = 58.5 rad,$

$W_{fric} = - 2.0 \times 58.5 = - 117.0 J,$

consistent with $117$ J lost as heat.

In mechanics, energy is one of our most useful tools. If you can track how forces and motion transfer energy, you can often solve problems without analyzing the full vector motion at every instant.

In linear motion, you learned two key ideas:

Work is the area under a force-displacement graph.
Kinetic energy measures a body’s ability to do work because it’s moving.

Those same ideas carry over to rotation:

Torques do rotational work.
Spinning objects store rotational kinetic energy.

So in this sub-chapter we discuss:

Rotational work
Rotational kinetic energy
Power in rotation
Rolling motion

Rotational work

In linear motion, a constant force $F$ acting through a displacement $Δ s$ does work

$W = F Δ s .$

In rotation, the analogous quantities are:

torque $τ$ (plays the role of force)
angular displacement $Δ θ$ in radians (plays the role of linear displacement)

So, for a constant torque,

$W = τ Δ θ .$

If the torque varies with angle, we can approximate the motion as many small angular steps $Δ θ_{i}$ , where the torque is nearly constant over each step. Adding the work from each step gives

$W \approx i \sum τ_{i} Δ θ_{i},$

and in the limit of very small steps this becomes

$W = \int τ d θ .$

$Δ s = r Δ θ,$

so the work done in that small step is

$Δ W = F Δ s = F r Δ θ = τ Δ θ .$

Rotational kinetic energy

Just as a mass $m$ moving at speed $v$ has kinetic energy $K = \frac{1}{2} m v^{2}$ , a rigid body with moment of inertia $I$ rotating at angular speed $ω$ has rotational kinetic energy

$K_{rot} = \frac{1}{2} I ω^{2} .$

You can derive this by starting from rotational work and using Newton’s second law for rotation. Begin with

$W = \int τ d θ = \int I α d θ = I \int ω d ω = \frac{1}{2} I ω^{2} .$

Here $τ = I α$ is Newton’s second law for rotation, and $α = d ω / d t$ .

Power in rotation

Power tells you how quickly work is done. In rotation,

$P = \frac{d W}{d t} = \frac{d}{d t} (τ θ) = τ \frac{d θ}{d t} = τ ω .$

So a torque $τ$ acting on a rotating body delivers power proportional to the instantaneous angular speed $ω$ .

Applications:

Electric motors supply torque and run at some RPM; their mechanical power output is $τ ω$ .
When you tighten a bolt, you supply power through the combination of how hard you twist (torque) and how fast the wrench turns (angular speed).

Rolling motion

Condition for no slipping:

$v_{CM} = R ω,$

where $v_{CM}$ is the center-of-mass speed, $R$ is the radius, and $ω$ is the angular speed.

Total kinetic energy: The object has

translational kinetic energy of its center of mass
rotational kinetic energy about its center of mass

$K_{total} = \frac{1}{2} M v_{CM}^{2} + \frac{1}{2} I ω^{2} .$

By substituting $ω = v_{CM} / R$ , you can rewrite $K_{total}$ using $v_{CM}$ alone.

Solid cylinder on an incline

Setup: A solid cylinder ( $I = \frac{1}{2} M R^{2}$ ) of mass $M$ rolls without slipping from rest down an incline of vertical drop $h$ .

Energy analysis:

$loss of gravitational PE M g h = translational KE \frac{1}{2} M v^{2} + rotational KE \frac{1}{2} I ω^{2} .$

Using $I = \frac{1}{2} M R^{2}$ and $ω = v / R$ , we get

$M g h = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} ⟹ v = \frac{4}{3} g h .$

Example problem 1

Draw the composite wheel’s free-body diagram at an arbitrary position on the incline. Include a separate small diagram showing the torque due to friction about the center.
Write an expression for the total moment of inertia $I_{tot}$ of the composite wheel about its central axis, in terms of $M_{h}$ , $M_{c}$ , and $R$ .
Write the equation relating gravitational potential energy lost to translational and rotational kinetic energies at the bottom. Clearly identify each term and substitute $I_{tot}$ and $ω = v / R$ .
Algebraically solve your energy-conservation equation for the final speed $v$ of the wheel’s center of mass at the bottom in terms of $g$ , $h$ , $M_{h}$ , and $M_{c}$ .
Explain in words how the presence of the hoop (with its larger moment of inertia per unit mass) affects the final speed compared to:
- (i) A pure solid cylinder of the same total mass and radius.
- (ii) A pure hoop of the same total mass and radius.

Solution

Free-body diagram

At any instant on the incline, the composite wheel experiences:
- Its total weight $M_{tot} g = (M_{h} + M_{c}) g$ acting downward at the center.
- A normal force $N$ perpendicular to the surface.
- A static friction force $f_{s}$ at the contact point, up the plane (since friction prevents slipping as it rolls downhill).
Free-body diagram: (friction is shown passing through the centre for simple force equations. It is tangential in reality for this case)

Moment of inertia

The total moment of inertia about the central axis is the sum of the hoop and the cylinder:

$I_{tot} = I_{h} + I_{c} = M_{h} R^{2} + \frac{1}{2} M_{c} R^{2} = (M_{h} + \frac{1}{2} M_{c}) R^{2} .$
Energy conservation

No slipping $\Rightarrow v = R ω$ . Mechanical energy is conserved:

$Δ U_{g} (M_{h} + M_{c}) g h$ $=$ $K_{trans} \frac{1}{2} (M_{h} + M_{c}) v^{2} + K_{rot} \frac{1}{2} I_{tot} ω^{2} .$

Substitute $I_{tot}$ and $ω = v / R$ :

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c}) v^{2} - \frac{1}{2} (M_{h} + \frac{1}{2} M_{c}) R^{2} (\frac{v}{R})^{2} .$

Simplify:

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c}) v^{2} - \frac{1}{2} (M_{h} + \frac{1}{2} M_{c}) v^{2} .$

Solve for speed

Combine translational and rotational terms:

$(M_{h} + M_{c}) g h = \frac{1}{2} (M_{h} + M_{c} + M_{h} + \frac{1}{2} M_{c}) v^{2} = \frac{1}{2} (2 M_{h} + \frac{3}{2} M_{c}) v^{2} .$

Hence

$v^{2} = \frac{2 ( M _{h} + M _{c} ) g h}{2 M _{h} + \frac{3}{2} M _{c}} ⟹ v = \frac{2 ( M _{h} + M _{c} ) g h}{2 M _{h} + \frac{3}{2} M _{c}} .$
Qualitative reasoning
- (i) vs. solid cylinder: A pure solid cylinder has $I = \frac{1}{2} M R^{2}$ , which is smaller (per unit mass) than the composite wheel’s. That means less energy must go into rotation, so the solid cylinder reaches a higher $v$ than the composite.
- (ii) vs. hoop: A pure hoop has $I = M R^{2}$ , which is larger (per unit mass) than the composite wheel’s. That means more energy must go into rotation, so the hoop reaches a lower $v$ than the composite.

In general, a larger moment of inertia per unit mass means a greater fraction of gravitational energy must go into rotation, leaving less for translation and producing a smaller final speed.

Example problem 2

a. Find its translational speed $v$ and angular speed $ω$ after descending $d$ . b. Determine the tension $T$ in the string during the motion.

Solution

a. Energy method. Loss of gravitational $\to$ translational + rotational KE:

$M g d = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}, I = \frac{1}{2} M R^{2}, ω = \frac{v}{R} .$

Thus

$M g d = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} ⟹ v = \frac{4}{3} g d = \frac{4}{3} \times 9.8 \times 0.75 \approx 4.43 m/s,$

$ω = \frac{v}{R} = \frac{4.43}{0.15} \approx 29.5 rad/s .$

b. Force-and-torque balance. Downward weight minus tension accelerates the center of mass:

$M g - T = M a, a = \frac{d v}{d t} .$

Rotation: string tension produces torque $τ = TR = I α$ , and $α = a / R$ . Hence

$TR = (\frac{1}{2} M R^{2}) \frac{a}{R} ⟹ T = \frac{1}{2} M a .$

Substitute into $M g - T = M a$ :

$M g - \frac{1}{2} M a = M a ⟹ M g = \frac{3}{2} M a ⟹ a = \frac{2}{3} g \approx 6.53 m/ s^{2} .$

Then

$T = \frac{1}{2} M a = 0.5 \times 1.2 \times 6.53 \approx 3.92 N .$

Example problem 3

A flywheel (solid disk) of mass $M = 3.0$ kg and radius $R = 0.25$ m ( $I = \frac{1}{2} M R^{2}$ ) is initially at rest.

Solution

a. Spin-up under $τ_{in}$ .

Moment of inertia $I = \frac{1}{2} (3.0) (0.25)^{2} = 0.09375$ kg·m $^{2}$ . Angular acceleration $α = τ_{in} / I = 8.0/0.09375 = 85.33$ rad/s $^{2}$ .

Time to reach $ω_{f} = 50$ rad/s:

$t_{1} = \frac{ω _{f}}{α} = \frac{50}{85.33} \approx 0.586 s .$

Work done $W_{in} = \frac{1}{2} I ω_{f}^{2} = 0.5 \times 0.09375 \times 5 0^{2} = 117.19 J$ .

Constant power $P = W_{in} / t_{1} = 117.19/0.586 \approx 200 W .$

b. Brake-down under $τ_{fric}$ .

Now $α_{brake} = τ_{fric} / I = 2.0/0.09375 = 21.33$ rad/s $^{2}$ (angular deceleration).

Time to stop from $ω_{f}$ :

$t_{2} = \frac{ω _{f}}{α _{brake}} = \frac{50}{21.33} \approx 2.34 s .$

Energy dissipated equals the initial rotational KE:

$Δ E = W_{in} = 117.19 J .$

Alternatively, using work done by the friction torque:

$W_{fric} = - τ_{fric} θ_{brk}, θ_{brk} = \frac{1}{2} ω_{f} t_{2} \approx \frac{1}{2} \times 50 \times 2.34 = 58.5 rad,$

$W_{fric} = - 2.0 \times 58.5 = - 117.0 J,$

consistent with $117$ J lost as heat.