Rotational Motion Essentials: Torque, Angular Quantities, and Energy in Rotation

1) Angular quantities: position, velocity, acceleration

Rotational motion uses angular versions of the familiar linear ideas. Instead of tracking a point moving along a line, we track how an object turns about an axis.

Angular position (θ)

Angular position θ tells how far an object has rotated from a reference direction. The standard unit is the radian (rad).

• Key idea: radians are based on arc length: θ = s/r, where s is arc length and r is radius.
• Unit check: s and r are both meters, so s/r is dimensionless; we still label it rad to emphasize “angle.”
• Useful facts: 2π rad = 360°, so π rad = 180°.

Angular velocity (ω)

Angular velocity ω describes how fast the angle changes:

ω = dθ/dt (instantaneous), or ω_avg = Δθ/Δt (average).

• Units: rad/s.
• Sign convention: choose a positive rotation direction (commonly counterclockwise when viewed from the axis). Clockwise then becomes negative.

Angular acceleration (α)

Angular acceleration α describes how fast angular velocity changes:

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α = dω/dt (instantaneous), or α_avg = Δω/Δt (average).

• Units: rad/s^2.
• Interpretation: if α and ω have the same sign, the object speeds up its rotation; if opposite signs, it slows down.

2) Connecting linear and angular motion

For a rigid object rotating about a fixed axis, points farther from the axis move faster in a linear sense, even though they share the same ω and α.

Arc length and tangential speed

A point at distance r from the axis travels along a circle. Its arc length is s = rθ. Differentiating with respect to time gives the link between linear speed and angular speed:

v = ωr

• Meaning: doubling r doubles the linear speed for the same ω.
• Units check: (rad/s)·m = m/s (radians act like “no unit” in multiplication).

Tangential vs centripetal acceleration

A point on a rotating object can have two perpendicular acceleration components:

• Tangential acceleration (changes speed along the circle): a_t = αr
• Centripetal acceleration (changes direction, points toward the center): a_c = ω^2 r

These are different ideas that often get mixed up:

• a_t exists only if ω is changing (nonzero α).
• a_c exists whenever the point is moving in a circle (nonzero ω), even if ω is constant.

Quick example: a point on a spinning wheel

A wheel has radius r = 0.30 m and angular speed ω = 10 rad/s. Then

• v = ωr = (10)(0.30) = 3.0 m/s
• a_c = ω^2 r = (10^2)(0.30) = 30 m/s^2

If the wheel is also speeding up with α = 4 rad/s^2, then a_t = αr = (4)(0.30) = 1.2 m/s^2.

3) Torque: the rotational analog of force

For translation, forces cause changes in motion. For rotation, the “turning effect” is described by torque τ (Greek letter tau).

What torque depends on

Torque depends on:

• How big the force is (F)
• Where you apply it (distance from the axis)
• How you apply it (direction relative to the lever arm)

In magnitude form for a force applied at distance r from the axis:

τ = r F sin(φ)

where φ is the angle between the radius vector (from axis to point of application) and the force direction.

A very practical way to think about it is using the lever arm (also called moment arm): the perpendicular distance from the axis to the force’s line of action. Then:

τ = F · (lever arm)

• Units: N·m (newton-meters). This is not the same as a joule, even though J = N·m in energy contexts; torque is a turning tendency, not energy.
• Direction/sign: torque can be positive or negative depending on whether it tends to rotate the object in your chosen positive direction.

Step-by-step example: wrench on a bolt

You push on a wrench of length r = 0.25 m with a force F = 80 N.

• Case A (best): push perpendicular to the wrench handle (φ = 90°).
  τ = rF sin(90°) = (0.25)(80)(1) = 20 N·m
• Case B (worse): push at φ = 30° relative to the handle.
  τ = (0.25)(80)sin(30°) = 20(0.5) = 10 N·m

Same force, same wrench length, but half the torque because the force was not applied as effectively.

Step-by-step example: pushing a door

A door rotates about its hinges. If you push near the hinges, r is small, so torque is small. If you push near the doorknob, r is larger, so torque is larger. This is why doors are easier to open by pushing far from the hinges.

Common misconception: “torque is the same as force”

• Force is a push/pull.
• Torque is a push/pull applied with a lever arm that tends to rotate an object.

You can apply a large force that produces almost no torque if its line of action passes through the axis (lever arm nearly zero). For example, pushing directly toward the hinge line of a door produces little rotation.

4) Moment of inertia: how mass distribution affects rotation

In translation, mass measures resistance to changes in velocity. In rotation, the analogous quantity is the moment of inertia I, which measures resistance to changes in angular velocity about a chosen axis.

Key idea: I depends not just on total mass, but on how that mass is distributed relative to the axis. Mass farther from the axis increases I a lot.

• Units: kg·m^2.
• Axis matters: the same object has different I values about different axes.

Given moments of inertia for common shapes (about central symmetry axis)

Notice how the hoop has larger I than a solid disk of the same M and R: more mass is located farther from the axis.

5) Rotational kinetic energy and rolling without slipping

Rotational kinetic energy

A rotating rigid body has rotational kinetic energy:

K_rot = (1/2) I ω^2

If the object’s center of mass is also moving, it can have both translational and rotational kinetic energy:

K_total = (1/2) M v_cm^2 + (1/2) I_cm ω^2

Rolling without slipping

When an object rolls without slipping, the point of contact with the ground is instantaneously at rest relative to the ground. The key constraint is:

v_cm = ωR

This connects translation and rotation. It also means that when something rolls, some energy goes into translation and some into rotation. How the energy splits depends on I, so shape matters.

Misconception: “all objects roll down the same way regardless of shape”

If two objects have the same mass and radius but different moments of inertia (disk vs hoop), the one with smaller I needs less energy to spin up, so more of the gravitational energy becomes translational speed. It reaches the bottom faster.

Problem 1: Rolling solid cylinder down a ramp (no slipping)

Goal: find the speed at the bottom after descending a vertical height h, starting from rest, for a solid cylinder of mass M and radius R.

Given: I = (1/2)MR^2, and rolling constraint v = ωR.

Step-by-step solution

• Step 1: Write energy conversion. The loss in gravitational potential energy becomes translational plus rotational kinetic energy:
  Mgh = (1/2)Mv^2 + (1/2)Iω^2
• Step 2: Substitute the cylinder’s I.
  Mgh = (1/2)Mv^2 + (1/2)(1/2 MR^2)ω^2
• Step 3: Use rolling without slipping: ω = v/R.
  Mgh = (1/2)Mv^2 + (1/4)MR^2 (v^2/R^2)
• Step 4: Simplify.
  Mgh = (1/2)Mv^2 + (1/4)Mv^2 = (3/4)Mv^2
• Step 5: Solve for v.
  v^2 = (4/3)gh so v = sqrt((4/3)gh)

Comparison check: if it were sliding without rotating, you’d get v = sqrt(2gh), which is larger. Rolling “stores” some energy in rotation, reducing the translational speed.

Shape comparison (same M, R, h)

Using the same method, you can show:

• Solid cylinder/disk (I = (1/2)MR^2): v = sqrt((4/3)gh)
• Hoop (I = MR^2): v = sqrt(gh) (slower)
• Solid sphere (I = (2/5)MR^2): v = sqrt((10/7)gh) (faster than cylinder)

Problem 2: Spinning disk with an applied torque

Scenario: A solid disk (mass M, radius R) starts from rest. A constant torque τ is applied about its central axis for a time t. Find its angular speed after time t and the rotational kinetic energy.

Given: I = (1/2)MR^2.

Step-by-step solution (using rotational dynamics)

• Step 1: Connect torque and angular acceleration. For rotation about a fixed axis:
  τ = Iα so α = τ/I
• Step 2: Use constant α to get ω. Starting from rest (ω_0 = 0):
  ω = ω_0 + αt = (τ/I)t
• Step 3: Substitute I for a solid disk.
  ω = (τ / ((1/2)MR^2)) t = (2τt)/(MR^2)
• Step 4: Compute rotational kinetic energy.
  K_rot = (1/2)Iω^2
  Substitute I and ω:
  K_rot = (1/2)(1/2 MR^2) ( (2τt)/(MR^2) )^2
• Step 5: Simplify.
  K_rot = (1/4)MR^2 · (4 τ^2 t^2)/(M^2 R^4) = (τ^2 t^2)/(M R^2)

Numerical example

Let M = 2.0 kg, R = 0.20 m, τ = 0.80 N·m, t = 3.0 s.

• ω = (2τt)/(MR^2) = (2·0.80·3.0)/(2.0·0.20^2) = 4.8/0.08 = 60 rad/s
• K_rot = (τ^2 t^2)/(MR^2) = (0.80^2·3.0^2)/(2.0·0.20^2) = (0.64·9)/(2.0·0.04) = 5.76/0.08 = 72 J

Common pitfalls to watch for

• Mixing up degrees and radians: formulas like v = ωr and K_rot = (1/2)Iω^2 assume ω is in rad/s. If you use degrees per second, results will be wrong by a factor of π/180.
• Confusing torque with force: a bigger force does not guarantee a bigger torque; the lever arm and angle matter.
• Assuming rolling speed depends only on mass: for rolling without slipping, I (shape and mass distribution) changes the acceleration and final speed.
• Forgetting there are two accelerations in rotation: tangential (a_t = αr) changes speed; centripetal (a_c = ω^2 r) changes direction.