Unit 6 · Topic 6.1 Beta

Rotational Kinetic Energy

An object turning about an axis has rotational kinetic energy K = ½Iω², the sum of ½mv² over all its pieces.

Practice 2: Mathematical Routines

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Question set for this topic

Part 1 · Hook

Why this matters

Some city buses and racing cars store energy in a heavy spinning wheel: brake, and the wheel spins up; accelerate, and it hands the energy back. The wheel never goes anywhere, yet it holds a lot of energy. That is kinetic energy of turning, and it follows the same rule as a moving cart, with new letters.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A 4.0 kg cart moves at 3.0 m/s. Its kinetic energy is

  1. 18 J
  2. 6.0 J
  3. 36 J
  4. 12 J
Show the answer

K = ½mv² = ½(4.0)(3.0)² = 18 J.

  • Correct: 18 J:
  • 6.0 J:
  • 36 J:
  • 12 J:

2. A point 0.50 m from an axis turns at 4.0 rad/s. Its speed is

  1. 2.0 m/s
  2. 8.0 m/s
  3. 0.13 m/s
  4. 4.0 m/s
Show the answer

v = rω = (0.50)(4.0).

  • Correct: 2.0 m/s:
  • 8.0 m/s:
  • 0.13 m/s:
  • 4.0 m/s:

3. A hoop and a disk have the same mass and radius. Which has the larger rotational inertia about its center?

  1. The hoop
  2. The disk
  3. Neither
  4. It depends on how fast they spin
Show the answer

All of the hoop’s mass is at the rim: MR² > ½MR².

  • Correct: The hoop:
  • The disk:
  • Neither:
  • It depends on how fast they spin:

Part 4 · See it

See it first

A flywheel spins counterclockwise on a fixed stand. Its center of mass does not move (v_cm = 0), yet every piece of the rim moves: velocity arrows at the top, bottom, left and right of the rim point in different, opposite directions, tangent to the rim. The wheel therefore has kinetic energy of rotation. A panel derives K_rot = ½Iω² from adding ½m(rω)² for every piece, notes it is a scalar measured in joules, that doubling ω quadruples it, and that moving mass outward at the same ω increases it.
Every piece of a spinning wheel moves, even though the center stays put, so the wheel has kinetic energy K = ½Iω². LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A wheel turns at ω about a fixed axleeach piece of mass m at distance r moves at v = rω
  2. Each piece has kinetic energy ½mv² = ½m r²ω²adding the pieces gives ½(Σmr²)ω²
  3. Σmr² is the rotational inertia Ithe wheel’s kinetic energy is K = ½Iω²
  4. ω doublesK becomes 4 times as large
  5. The same mass sits farther from the axleI and so K are larger at the same ω

Part 6 · Key ideas

Key ideas

  • A turning object has rotational kinetic energy K = ½Iω², even if its center of mass is at rest.
  • It is the sum of ½mv² over every piece, with v = rω.
  • A scalar in joules; the direction of turning does not matter, because ω is squared.
  • K ∝ ω² at fixed I (doubling ω gives 4K) and K ∝ I at fixed ω.
  • K against ω² is a straight line with slope ½I.

Part 7 · Misconception

A common mistake

The wrong idea: A wheel spinning in place has no kinetic energy, because it is not going anywhere.

What actually happens: Every piece of the wheel is moving, so the wheel has kinetic energy ½Iω². Its momentum adds to zero because opposite sides move opposite ways, but kinetic energy is a scalar and does not cancel.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A turntable with a lump of clay

0.15 mSolid disk: M = 3.0 kg,R = 0.20 m, I = ½MR² (given)Clay lump: 0.10 kg,0.15 m from the axisTurning at 6.0 rad/son a fixed vertical axle

1. What is the rotational inertia of the turntable and clay together about the axle?

Type a number in kg·m^2.

Show the answer

I = ½MR² + mr² = ½(3.0 kg)(0.20 m)² + (0.10 kg)(0.15 m)² = 0.0600 + 0.00225 = 0.06225 kg·m², about 0.062 kg·m² (the data have two significant figures).

  • Answer: 0.062 kg·m^2

2. What is the rotational kinetic energy of the turntable and clay? Give the value and its unit.

Type a number and its unit.

Show the answer

K = ½Iω² = ½(0.06225 kg·m²)(6.0 rad/s)² = 1.12 J, about 1.1 J.

  • Answer: 1.1 J

3. The turntable’s center does not move. Which statement about its kinetic energy is correct?

  1. It has kinetic energy, because every piece of it is moving
  2. It has no kinetic energy, because its center of mass is at rest
  3. It has no kinetic energy, because opposite sides cancel out
  4. Its kinetic energy is at the rim, where the speed is largest
Show the answer

Each piece of mass moves at v = rω and has ½mv². Kinetic energy is a scalar, so these add up (to ½Iω²) even though the velocities point in different directions and the center does not move.

  • Correct: It has kinetic energy, because every piece of it is moving: Right: K = ½Iω² counts the motion of every piece around the axle.
  • It has no kinetic energy, because its center of mass is at rest: A body can have rotational kinetic energy while its center of mass stays put.
  • It has no kinetic energy, because opposite sides cancel out: Velocities cancel in the total momentum, but kinetic energy is a scalar; it never cancels.
  • Its kinetic energy is at the rim, where the speed is largest: Every piece moves (v = rω) and contributes ½mv²; the rim pieces just contribute most.

4. The turntable is sped up to 12 rad/s. Its rotational kinetic energy becomes

  1. 4 times as large
  2. 2 times as large
  3. the same
  4. 8 times as large
Show the answer

K = ½Iω² with I fixed, so doubling ω multiplies K by 4 (to about 4.5 J).

  • Correct: 4 times as large: Right: K ∝ ω², and (2)² = 4.
  • 2 times as large: K depends on ω², not ω.
  • the same: A faster spin has more kinetic energy.
  • 8 times as large: There is no cube: K = ½Iω².

Data table

Energy stored in a flywheel

An engineer measures the energy stored in a flywheel (from the work needed to spin it up from rest on a frictionless bearing) at four angular velocities.

Stored energy against angular velocity
Angular velocity (rad/s)Energy stored (J)
1040
20160
30360
40640

5. Which graph of these data gives a straight line through the origin?

  1. K against ω²
  2. K against ω
  3. K against 1/ω
  4. √K against ω²
Show the answer

K = ½Iω², so K against ω² is a straight line through the origin with slope ½I.

  • Correct: K against ω²: Right: K = (½I)ω² is proportional to ω².
  • K against ω: K rises faster than ω; the points curve upward.
  • K against 1/ω: K increases with ω, so it falls along a curve against 1/ω.
  • √K against ω²: √K is proportional to ω, not ω².

6. Use the data to find the flywheel’s rotational inertia.

Type a number in kg·m^2.

Show the answer

From any row, I = 2K/ω²: for example 2(640 J)/(40 rad/s)² = 0.80 kg·m². The slope of K against ω² is ½I = 0.40 kg·m².

  • Answer: 0.80 kg·m^2

7. A wheel with rotational inertia 0.40 kg·m² has 25 J of rotational kinetic energy. How fast is it turning?

Type a number in rad/s.

Show the answer

ω = √(2K/I) = √(2 × 25 J / 0.40 kg·m²) = √125 = 11.2 rad/s, about 11 rad/s.

  • Answer: 11 rad/s

Part 9 · Summary

Summary

An object turning about an axis has rotational kinetic energy K = ½Iω², the sum of ½mv² over all its pieces. It is a scalar in joules, is present even when the center of mass is at rest, grows with the square of ω and in proportion to the rotational inertia.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections