7.2 Torque and rotational inertia

Achievable AP Physics 1

7. Torque and rotational mechanics

Our AP Physics 1 course is currently in development and is a work-in-progress.

Torque and rotational inertia

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In linear dynamics, a force $F$ acting on a mass $m$ produces an acceleration $a = F / m$ . In rotational dynamics, torque $τ$ plays the role of force, and the moment of inertia $I$ plays the role of mass. The rotational analogue of Newton’s second law is

$\sum τ = I α,$

where $α$ is the angular acceleration.

In this subchapter:

We define torque and moment of inertia
Develop the key theorems for computing $I$ , and work through representative examples.

Definition of torque

A force $F$ applied at a point $P$ produces a torque about an origin (pivot) $O$ given by the cross product

$τ = r \times F, τ = r F sin ϕ,$

where:

$r$ is the position vector from $O$ to $P$
$ϕ$ is the angle between $r$ and $F$

By the right-hand rule, a positive $τ$ corresponds to counterclockwise rotation.

Lever arm form. You can also write torque in terms of the lever arm (the perpendicular distance from the pivot to the force’s line of action):

$τ = F ℓ, ℓ = r sin ϕ,$

where $ℓ$ is the perpendicular distance (lever arm) from $O$ to the line of action of $F$ .

Sign conventions

Choose a right-handed coordinate system. Positive torque is the torque that, by the right-hand rule (thumb out of the page), tends to rotate the object counterclockwise about the axis.

Moment of inertia

Just as mass $m$ measures an object’s resistance to linear acceleration ( $F = ma$ ), the moment of inertia $I$ measures a body’s resistance to angular acceleration ( $τ = I α$ ). The farther the mass is distributed from the axis, the larger $I$ becomes.

Definitions

Discrete point masses: For $N$ point masses $m_{i}$ each at perpendicular distance $r_{i}$ from the axis,

$I = i = 1 \sum N m_{i} r_{i}^{2} .$

Definitions

Continuous distribution: For a rigid body of total mass $M$ with density $ρ (r)$ , (integral solving not in AP Physics 1 scope)

$I = \int r^{2} d m = \int_{V} r^{2} ρ (r) d V, M = \int_{V} ρ (r) d V .$

Common moments of inertia

$Thin hoop or ring, radius R : Solid disk, radius R : Thin rod about center: Thin rod about end: I = M R^{2}, I = \frac{1}{2} M R^{2}, I = \frac{1}{12} M L^{2}, I = \frac{1}{3} M L^{2} .$

Parallel‐axis and shell theorems Parallel‐axis theorem

If $I_{cm}$ is the moment of inertia about an axis through the center of mass, then the moment about a parallel axis at distance $d$ is

$I = I_{cm} + M d^{2} .$

Thin cylindrical shell theorem. A thin cylindrical shell of mass $M$ and radius $R$ has

$I_{shell} = M R^{2} .$

A full solid cylinder may be built by stacking shells.

Reasoning Practice:

A uniform solid disk and a thin hoop both have mass $M$ and radius $R$ . A tangential force $F$ is applied at the rim of each to produce the same angular acceleration $α$ . Explain which object requires a larger torque to achieve the same $α$ , and why.

Solution: The required torque is $τ = I α$ . For a solid disk, $I_{disk} = \frac{1}{2} M R^{2}$ . For a thin hoop, $I_{hoop} = M R^{2}$ . To get the same $α$ , the hoop requires

$τ_{hoop} = I_{hoop} α = (M R^{2}) α, τ_{disk} = (\frac{1}{2} M R^{2}) α .$

Since $M R^{2} > \frac{1}{2} M R^{2}$ , the thin hoop requires twice the torque of the solid disk to achieve the same $α$ .

A mechanic uses a wrench of length $L = 0.30$ m to loosen a bolt by applying a force $F = 100$ N perpendicular to the handle.
1. Calculate the torque $τ$ produced.
2. If the mechanic instead uses a wrench of length $L /2$ and applies the same force $F$ , how does the torque change? Discuss the reasoning in terms of lever arm and rotational inertia.

Solution: (a) When the force is perpendicular, torque is $τ = r F$ . With $r = L = 0.30$ m and $F = 100$ N,

$τ = (0.30 m) (100 N) = 30 N \cdot m .$

(b) If $r = L /2 = 0.15$ m and the same $F = 100$ N, then

$τ^{'} = (0.15 m) (100 N) = 15 N \cdot m .$

Halving the lever arm halves the torque. The bolt’s rotational inertia is unchanged, but the applied torque is smaller, so you’d need either a larger force or a longer wrench to get the same twisting effect.

A rectangular door of width $w$ and mass $m$ swings on hinges at one edge. Explain why applying a force near the hinges requires a larger force to generate the same torque compared to applying the force near the free edge. Relate your answer to lever arm and moment of inertia.

Solution: Torque from a force $F$ applied a distance $r$ from the hinge is $τ = r F$ . Near the hinges, $r$ is small, so to produce the same torque you must use a larger $F$ . Near the free edge (where $r = w$ ), the same force produces a larger torque because the lever arm is larger. The door’s moment of inertia about the hinge is fixed by its geometry, so changing where you push changes the lever arm (and therefore the torque), not the door’s $I$ .

A uniform thin rod of length $L$ and mass $M$ rotates about its center with angular acceleration $α_{1}$ under a constant torque $τ$ . If the same rod instead rotates about one end under the same torque $τ$ , what will be its angular acceleration $α_{2}$ ? Explain the relationship between the rod’s moment of inertia in each case and the resulting angular acceleration.

Solution: When rotating about the center, $I_{center} = \frac{1}{12} M L^{2}$ . Thus

$α_{1} = \frac{τ}{I _{center}} = \frac{τ}{\frac{1}{12} M L ^{2}} = \frac{12 τ}{M L ^{2}} .$

When rotating about one end, $I_{end} = \frac{1}{3} M L^{2}$ . Then

$α_{2} = \frac{τ}{I _{end}} = \frac{τ}{\frac{1}{3} M L ^{2}} = \frac{3 τ}{M L ^{2}} .$

Comparing,

$α_{2} = \frac{3 τ}{M L ^{2}}, α_{1} = \frac{12 τ}{M L ^{2}}, ⟹ α_{2} = \frac{α _{1}}{4} .$

The rod’s moment of inertia about the end is four times larger than about its center, so the same torque produces one-quarter the angular acceleration.

Example problem 1

A uniform rod of length $L = 2.0$ m and mass $M = 3.0$ kg.

Moment of inertia across two different axes

We wish to compute:

$I_{cm} (about midpoint axis), I_{end} (about axis at one end).$

Solution:

Moment about the center. For a uniform rod of length $L$ , the standard result is

$I_{cm} = \frac{1}{12} M L^{2} .$

Substituting $M = 3.0$ kg and $L = 2.0$ m:

$I_{cm} = \frac{1}{12} (3.0) (2.0)^{2} = \frac{1}{12} \times 3.0 \times 4.0 = 1.0 kg \cdot m^{2} .$
Moment about one end (Parallel‐Axis Theorem). The parallel-axis theorem states

$I_{end} = I_{cm} + M d^{2},$

where $d$ is the distance from the center-of-mass axis to the new axis. Here $d = \frac{L}{2} = 1.0$ m, so

$I_{end} = 1.0 + (3.0) (1.0)^{2} = 1.0 + 3.0 = 4.0 kg \cdot m^{2} .$

Final Answers:

$I_{cm} = 1.0 kg \cdot m^{2}, I_{end} = 4.0 kg \cdot m^{2} .$

Example problem 2

A uniform rod of mass $m_{1} = 2.0$ kg and length $L = 1.2$ m is pivoted frictionlessly about one end. Two small masses, $m_{2} = 1.0$ kg and $m_{3} = 0.5$ kg, are rigidly attached to the rod at distances $r_{2} = \frac{L}{3}$ and $r_{3} = L$ , respectively, from the pivot. A force $F = 10$ N is applied perpendicular to the rod at a point $r_{F} = 0.8 L$ from the pivot.

Uniform rod pivoted at one end with two small masses attached

Compute the total moment of inertia $I_{tot}$ of the system about the pivot.
Find the initial angular acceleration $α$ of the system.

Solution:

Moment of inertia.

$I_{rod} = \frac{1}{3} m_{1} L^{2} = \frac{1}{3} (2.0) (1.2)^{2} = 0.96 kg \cdot m^{2},$

$I_{2} = m_{2} r_{2}^{2} = (1.0) (\frac{1.2}{3})^{2} = 0.16 kg \cdot m^{2},$

$I_{3} = m_{3} r_{3}^{2} = (0.5) (1.2)^{2} = 0.72 kg \cdot m^{2} .$

Total:

$I_{tot} = 0.96 + 0.16 + 0.72 = 1.84 kg \cdot m^{2} .$
Angular acceleration. Torque due to $F$ :

$τ = F r_{F} = 10 (0.8 \times 1.2) = 9.6 N \cdot m .$

Thus

$α = \frac{τ}{I _{tot}} = \frac{9.6}{1.84} \approx 5.22 rad/ s^{2} .$

Example problem 3

A composite wheel consists of a solid disk of mass $M = 3.0$ kg and radius $R = 0.20$ m rigidly fastened to a thin hoop of mass $m = 1.0$ kg and radius $2 R = 0.40$ m. The assembly rotates about its central axis. A string wrapped around the outer hoop exerts a constant tangential force $F = 15$ N on the hoop.

The composite wheel rotating about its central axis

Determine the moment of inertia $I_{comp}$ of the composite wheel about its axis.
Find the angular acceleration $α$ produced by the applied force.

Solution:

Composite disk + hoop.

$I_{disk} = \frac{1}{2} M R^{2} = \frac{1}{2} (3.0) (0.20)^{2} = 0.06 kg \cdot m^{2},$

$I_{hoop} = m (2 R)^{2} = 1.0 (0.40)^{2} = 0.16 kg \cdot m^{2},$

$I_{comp} = 0.06 + 0.16 = 0.22 kg \cdot m^{2} .$
Angular acceleration.

$τ = F (2 R) = 15 (0.40) = 6.0 N \cdot m,$

$α = \frac{τ}{I _{comp}} = \frac{6.0}{0.22} \approx 27.3 rad/ s^{2} .$

Torque\

Rotational analogue of force; $τ = r \times F$
Magnitude: $τ = r F sin ϕ$ or $τ = F ℓ$ (lever arm form)
Direction by right-hand rule; positive = counterclockwise

Sign Conventions\

Use right-handed coordinate system
Positive torque: counterclockwise about axis

Moment of Inertia ( $I$ )\

Rotational analogue of mass; measures resistance to angular acceleration
For $N$ point masses: $I = \sum m_{i} r_{i}^{2}$
For continuous bodies: $I = \int r^{2} d m$
$I$ increases as mass is distributed farther from axis

Common Moments of Inertia\

Thin hoop/ring (radius $R$ ): $I = M R^{2}$
Solid disk (radius $R$ ): $I = \frac{1}{2} M R^{2}$
Thin rod (center): $I = \frac{1}{12} M L^{2}$
Thin rod (end): $I = \frac{1}{3} M L^{2}$

Parallel-Axis and Shell Theorems\

Parallel-Axis: $I = I_{cm} + M d^{2}$ ( $d$ = distance from center of mass axis)
Thin cylindrical shell: $I_{shell} = M R^{2}$
Solid cylinder: sum of shells

Reasoning Practice Takeaways\

Larger $I$ → more torque needed for same angular acceleration ( $τ = I α$ )
Torque increases with longer lever arm; halving lever arm halves torque
Applying force farther from pivot increases torque (lever arm effect)
Rotating about axis farther from center of mass increases $I$ , reduces angular acceleration for same torque

Example Problem Key Results\

For rods: $I_{cm} = \frac{1}{12} M L^{2}$ , $I_{end} = \frac{1}{3} M L^{2}$
For composite systems: sum moments of inertia of all parts about same axis
Angular acceleration: $α = τ / I_{tot}$

Summary Concepts\

Torque: depends on force, distance from pivot, and angle
Moment of inertia: depends on mass distribution relative to axis
Use correct $I$ formulas and sign conventions for rotational motion problems

Next | 7.3 Work and energy in rotation

Torque and rotational inertia

$\sum τ = I α,$

where $α$ is the angular acceleration.

In this subchapter:

We define torque and moment of inertia
Develop the key theorems for computing $I$ , and work through representative examples.

Definition of torque

A force $F$ applied at a point $P$ produces a torque about an origin (pivot) $O$ given by the cross product

$τ = r \times F, τ = r F sin ϕ,$

where:

$r$ is the position vector from $O$ to $P$
$ϕ$ is the angle between $r$ and $F$

By the right-hand rule, a positive $τ$ corresponds to counterclockwise rotation.

Lever arm form. You can also write torque in terms of the lever arm (the perpendicular distance from the pivot to the force’s line of action):

$τ = F ℓ, ℓ = r sin ϕ,$

where $ℓ$ is the perpendicular distance (lever arm) from $O$ to the line of action of $F$ .

Sign conventions

Choose a right-handed coordinate system. Positive torque is the torque that, by the right-hand rule (thumb out of the page), tends to rotate the object counterclockwise about the axis.

Moment of inertia

Definitions

Discrete point masses: For $N$ point masses $m_{i}$ each at perpendicular distance $r_{i}$ from the axis,

$I = i = 1 \sum N m_{i} r_{i}^{2} .$

Definitions

Continuous distribution: For a rigid body of total mass $M$ with density $ρ (r)$ , (integral solving not in AP Physics 1 scope)

$I = \int r^{2} d m = \int_{V} r^{2} ρ (r) d V, M = \int_{V} ρ (r) d V .$

Common moments of inertia

$Thin hoop or ring, radius R : Solid disk, radius R : Thin rod about center: Thin rod about end: I = M R^{2}, I = \frac{1}{2} M R^{2}, I = \frac{1}{12} M L^{2}, I = \frac{1}{3} M L^{2} .$

Parallel‐axis and shell theorems Parallel‐axis theorem

If $I_{cm}$ is the moment of inertia about an axis through the center of mass, then the moment about a parallel axis at distance $d$ is

$I = I_{cm} + M d^{2} .$

Thin cylindrical shell theorem. A thin cylindrical shell of mass $M$ and radius $R$ has

$I_{shell} = M R^{2} .$

A full solid cylinder may be built by stacking shells.

Reasoning Practice:

A uniform solid disk and a thin hoop both have mass $M$ and radius $R$ . A tangential force $F$ is applied at the rim of each to produce the same angular acceleration $α$ . Explain which object requires a larger torque to achieve the same $α$ , and why.

Solution: The required torque is $τ = I α$ . For a solid disk, $I_{disk} = \frac{1}{2} M R^{2}$ . For a thin hoop, $I_{hoop} = M R^{2}$ . To get the same $α$ , the hoop requires

$τ_{hoop} = I_{hoop} α = (M R^{2}) α, τ_{disk} = (\frac{1}{2} M R^{2}) α .$

Since $M R^{2} > \frac{1}{2} M R^{2}$ , the thin hoop requires twice the torque of the solid disk to achieve the same $α$ .

A mechanic uses a wrench of length $L = 0.30$ m to loosen a bolt by applying a force $F = 100$ N perpendicular to the handle.
1. Calculate the torque $τ$ produced.
2. If the mechanic instead uses a wrench of length $L /2$ and applies the same force $F$ , how does the torque change? Discuss the reasoning in terms of lever arm and rotational inertia.

Solution: (a) When the force is perpendicular, torque is $τ = r F$ . With $r = L = 0.30$ m and $F = 100$ N,

$τ = (0.30 m) (100 N) = 30 N \cdot m .$

(b) If $r = L /2 = 0.15$ m and the same $F = 100$ N, then

$τ^{'} = (0.15 m) (100 N) = 15 N \cdot m .$

A rectangular door of width $w$ and mass $m$ swings on hinges at one edge. Explain why applying a force near the hinges requires a larger force to generate the same torque compared to applying the force near the free edge. Relate your answer to lever arm and moment of inertia.

A uniform thin rod of length $L$ and mass $M$ rotates about its center with angular acceleration $α_{1}$ under a constant torque $τ$ . If the same rod instead rotates about one end under the same torque $τ$ , what will be its angular acceleration $α_{2}$ ? Explain the relationship between the rod’s moment of inertia in each case and the resulting angular acceleration.

Solution: When rotating about the center, $I_{center} = \frac{1}{12} M L^{2}$ . Thus

$α_{1} = \frac{τ}{I _{center}} = \frac{τ}{\frac{1}{12} M L ^{2}} = \frac{12 τ}{M L ^{2}} .$

When rotating about one end, $I_{end} = \frac{1}{3} M L^{2}$ . Then

$α_{2} = \frac{τ}{I _{end}} = \frac{τ}{\frac{1}{3} M L ^{2}} = \frac{3 τ}{M L ^{2}} .$

Comparing,

$α_{2} = \frac{3 τ}{M L ^{2}}, α_{1} = \frac{12 τ}{M L ^{2}}, ⟹ α_{2} = \frac{α _{1}}{4} .$

The rod’s moment of inertia about the end is four times larger than about its center, so the same torque produces one-quarter the angular acceleration.

Example problem 1

A uniform rod of length $L = 2.0$ m and mass $M = 3.0$ kg.

We wish to compute:

$I_{cm} (about midpoint axis), I_{end} (about axis at one end).$

Solution:

Moment about the center. For a uniform rod of length $L$ , the standard result is

$I_{cm} = \frac{1}{12} M L^{2} .$

Substituting $M = 3.0$ kg and $L = 2.0$ m:

$I_{cm} = \frac{1}{12} (3.0) (2.0)^{2} = \frac{1}{12} \times 3.0 \times 4.0 = 1.0 kg \cdot m^{2} .$
Moment about one end (Parallel‐Axis Theorem). The parallel-axis theorem states

$I_{end} = I_{cm} + M d^{2},$

where $d$ is the distance from the center-of-mass axis to the new axis. Here $d = \frac{L}{2} = 1.0$ m, so

$I_{end} = 1.0 + (3.0) (1.0)^{2} = 1.0 + 3.0 = 4.0 kg \cdot m^{2} .$

Final Answers:

$I_{cm} = 1.0 kg \cdot m^{2}, I_{end} = 4.0 kg \cdot m^{2} .$

Example problem 2

Compute the total moment of inertia $I_{tot}$ of the system about the pivot.
Find the initial angular acceleration $α$ of the system.

Solution:

Moment of inertia.

$I_{rod} = \frac{1}{3} m_{1} L^{2} = \frac{1}{3} (2.0) (1.2)^{2} = 0.96 kg \cdot m^{2},$

$I_{2} = m_{2} r_{2}^{2} = (1.0) (\frac{1.2}{3})^{2} = 0.16 kg \cdot m^{2},$

$I_{3} = m_{3} r_{3}^{2} = (0.5) (1.2)^{2} = 0.72 kg \cdot m^{2} .$

Total:

$I_{tot} = 0.96 + 0.16 + 0.72 = 1.84 kg \cdot m^{2} .$
Angular acceleration. Torque due to $F$ :

$τ = F r_{F} = 10 (0.8 \times 1.2) = 9.6 N \cdot m .$

Thus

$α = \frac{τ}{I _{tot}} = \frac{9.6}{1.84} \approx 5.22 rad/ s^{2} .$

Example problem 3

Determine the moment of inertia $I_{comp}$ of the composite wheel about its axis.
Find the angular acceleration $α$ produced by the applied force.

Solution:

Composite disk + hoop.

$I_{disk} = \frac{1}{2} M R^{2} = \frac{1}{2} (3.0) (0.20)^{2} = 0.06 kg \cdot m^{2},$

$I_{hoop} = m (2 R)^{2} = 1.0 (0.40)^{2} = 0.16 kg \cdot m^{2},$

$I_{comp} = 0.06 + 0.16 = 0.22 kg \cdot m^{2} .$
Angular acceleration.

$τ = F (2 R) = 15 (0.40) = 6.0 N \cdot m,$

$α = \frac{τ}{I _{comp}} = \frac{6.0}{0.22} \approx 27.3 rad/ s^{2} .$

Achievable AP Physics 1

7. Torque and rotational mechanics

Our AP Physics 1 course is currently in development and is a work-in-progress.