Unit 5 · Topic 5.6 Beta

Newton's Second Law for Rotation

Newton’s second law in rotational form says α = τ_net/I: the net torque sets the angular acceleration, and a larger rotational inertia means a smaller α.

Practice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

Drop a bucket into a well with the rope wound on a heavy drum, and the bucket falls far slower than a dropped stone: part of gravity’s pull goes into spinning up the drum. One law, net torque equals rotational inertia times angular acceleration, explains the drum, the bucket and the tension between them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A 2.0 kg cart has a net force of 6.0 N on it. Its acceleration is

  1. 3.0 m/s²
  2. 12 m/s²
  3. 0.33 m/s²
  4. 6.0 m/s²
Show the answer

a = F_net/m = 6.0/2.0.

  • Correct: 3.0 m/s²:
  • 12 m/s²:
  • 0.33 m/s²:
  • 6.0 m/s²:

2. A string unwinds from a wheel of radius R without slipping. The string’s acceleration a and the wheel’s α are linked by

  1. a = Rα
  2. a = α/R
  3. a = R²α
  4. a = α
Show the answer

The string moves with the rim: a = Rα.

  • Correct: a = Rα:
  • a = α/R:
  • a = R²α:
  • a = α:

3. Rotational inertia depends on

  1. mass and its distance from the axis
  2. mass alone
  3. how fast the object turns
  4. the net torque
Show the answer

I = Σmr².

  • Correct: mass and its distance from the axis:
  • mass alone:
  • how fast the object turns:
  • the net torque:

Part 4 · See it

See it first

A pulley of mass M and radius R turns on a fixed axle hung from the ceiling. A light string wrapped around its rim holds a block of mass m, which accelerates downward with acceleration a while the pulley gains angular acceleration α, with a = Rα. Middle: the force diagram of the pulley shows the string pulling down at the rim, Earth’s gravitational force down at the center and the axle’s force up at the center; only the string’s pull has a lever arm, so F_T R = Iα. Right: the free-body diagram of the block shows the string’s tension up, shorter than Earth’s gravitational force down, so mg − F_T = ma.
A pulley with mass and a hanging block: the string’s pull gives the pulley its torque, F_T R = Iα, while the block obeys mg − F_T = ma. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A net torque acts on an object that can turnits angular velocity changes
  2. The net torque is largerthe angular acceleration is larger: α = τ_net/I
  3. The rotational inertia is largerthe same torque gives a smaller angular acceleration
  4. A string pulls a pulley with mass while a block hangs from itthe tension is less than the block’s weight, and the two sides of a string over the pulley carry different tensions
  5. The string does not slipa = Rα links the block’s motion to the pulley’s

Part 6 · Key ideas

Key ideas

  • α = τnet/I, in the direction of the net torque: the rotational partner of a = Fnet/m.
  • For a system with parts that move and parts that turn, write ΣF = ma for each moving part and Στ = Iα for each turning part, then link them with a = Rα.
  • A pulley with mass needs a net torque, so the tensions on its two sides differ.
  • Same torque, larger I (mass farther out): smaller α.
  • α against τ is a line through the origin with slope 1/I.

Part 7 · Misconception

A common mistake

The wrong idea: The tension in a string holding a falling block equals the block’s weight, and is the same on both sides of any pulley.

What actually happens: If the block speeds up downward, the tension is less than its weight. Over a pulley with mass, the two sides differ, because their difference supplies the torque that spins the pulley up.

Part 8 · Check yourself

Check yourself

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

Diagram

A block on a pulley with mass

mPulley: uniform disk,M = 1.50 kg, R = 0.100 m,I = ½MR² (given)Turns on a frictionless axle.Block: m = 0.500 kgLight string, no slipping.Released from rest. Use g = 9.8 m/s².

1. What is the size of the block’s acceleration?

Type a number in m/s^2.

Show the answer

Block: mg − F_T = ma. Pulley: F_T R = Iα = (½MR²)(a/R), so F_T = ½Ma. Adding: mg = (m + ½M)a, so a = (0.500 kg)(9.8 m/s²)/(0.500 kg + 0.750 kg) = 3.92 m/s².

  • Answer: 3.92 m/s^2

2. What is the tension in the string?

Type a number in N.

Show the answer

F_T = ½Ma = (0.750 kg)(3.92 m/s²) = 2.94 N; check with the block: m(g − a) = (0.500 kg)(9.8 − 3.92) m/s² = 2.94 N.

  • Answer: 2.94 N

3. How does the tension compare with the block’s weight while the block falls?

  1. Less than the weight, because the block accelerates downward
  2. Equal to the weight, because the string is light
  3. More than the weight, because the string also turns the pulley
  4. Zero, because the block is falling
Show the answer

The block speeds up downward, so the net force on it is down: mg − F_T = ma > 0. The tension (2.94 N) is less than the weight (4.90 N).

  • Correct: Less than the weight, because the block accelerates downward: Right: the net force on the block is down, so mg > F_T.
  • Equal to the weight, because the string is light: A light string transmits tension, but the block is not in equilibrium.
  • More than the weight, because the string also turns the pulley: Turning the pulley takes tension, but the block still speeds up downward, so F_T < mg.
  • Zero, because the block is falling: If the tension were zero, the pulley would not speed up.

4. What is the pulley’s angular acceleration?

Type a number in rad/s^2.

Show the answer

The string does not slip on the rim, so α = a/R = (3.92 m/s²)/(0.100 m) = 39.2 rad/s².

  • Answer: 39.2 rad/s^2

5. The disk pulley is replaced by a hoop pulley with the same mass and radius (I = MR², given). The block’s acceleration is

  1. smaller, because more of the pulley’s mass is at its rim
  2. larger, because a hoop has less material near the axle
  3. the same, because the pulley’s mass is the same
  4. zero, because the hoop’s inertia balances the block
Show the answer

With I = MR², a = mg/(m + M) = (4.90 N)/(2.00 kg) = 2.45 m/s², less than 3.92 m/s². A larger I needs more torque for the same α, so less of the weight is left to accelerate the block.

  • Correct: smaller, because more of the pulley’s mass is at its rim: Right: a = mg/(m + I/R²) falls when I rises.
  • larger, because a hoop has less material near the axle: Less mass near the axle means more at the rim, so a larger I.
  • the same, because the pulley’s mass is the same: Same mass, larger I; the pulley is harder to spin up.
  • zero, because the hoop’s inertia balances the block: Inertia resists changes in motion; it does not balance a weight.

6. According to Newton’s second law in rotational form, an object’s angular acceleration is

  1. the net torque divided by its rotational inertia
  2. the net torque multiplied by its rotational inertia
  3. the net force on it divided by its total mass
  4. in the direction the object is turning, whatever the torque
Show the answer

α = τ_net/I. A larger net torque gives a larger α; a larger I gives a smaller α; α points the way the net torque does (clockwise or counterclockwise).

  • Correct: the net torque divided by its rotational inertia: Right: α = τ_net/I, in the direction of the net torque.
  • the net torque multiplied by its rotational inertia: τ_net = Iα, so α = τ_net/I.
  • the net force on it divided by its total mass: That is the translational second law.
  • in the direction the object is turning, whatever the torque: α follows the net torque; it can oppose the turning and slow it.

7. A solid disk and a hoop have the same mass and radius. Equal net torques act on each for the same time, starting from rest. Which ends up turning faster?

  1. The disk, because its rotational inertia is smaller
  2. The hoop, because its mass is farther out from the axle
  3. They match, because the torques and the masses are equal
  4. The hoop, because a hoop has less material near the axle
Show the answer

The same torque gives α = τ/I. The disk’s I = ½MR² is half the hoop’s MR², so its α and its final ω are twice as large.

  • Correct: The disk, because its rotational inertia is smaller: Right: α = τ/I, and ½MR² < MR².
  • The hoop, because its mass is farther out from the axle: Mass farther out raises I, which lowers α.
  • They match, because the torques and the masses are equal: Equal mass does not mean equal I.
  • The hoop, because a hoop has less material near the axle: Less mass near the axle means more at the rim and a larger I.

8. Two blocks of different mass hang on either side of a pulley that has mass, connected by one string that does not slip. The tensions in the two parts of the string are

  1. different, because the pulley needs a net torque to speed up
  2. equal, because a single string runs over the pulley
  3. equal, because the pulley turns at a steady angular velocity
  4. both equal to the weight of the heavier of the two blocks
Show the answer

The pulley’s angular acceleration needs a net torque: (F_T,heavy side − F_T,light side)R = Iα. If the pulley had no mass (I = 0), the tensions would be equal; with mass they differ.

  • Correct: different, because the pulley needs a net torque to speed up: Right: (F_T1 − F_T2)R = Iα ≠ 0.
  • equal, because a single string runs over the pulley: One string over a pulley with mass carries different tensions on its two sides.
  • equal, because the pulley turns at a steady angular velocity: The blocks speed up, so the pulley speeds up too.
  • both equal to the weight of the heavier of the two blocks: The heavier block accelerates downward, so the tension on its side is less than its weight.

Part 9 · Summary

Summary

Newton’s second law in rotational form says α = τ_net/I: the net torque sets the angular acceleration, and a larger rotational inertia means a smaller α. Systems with a pulley that has mass need both the translational law for each block and the rotational law for the pulley, linked by a = Rα; the string’s tension is then less than a falling block’s weight.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections