Unit 5.4 | Rotational Inertia and Newton’s Second Law
- 18 minutes
- 3 formulas
- 25 questions
- 2 videos
When torques do not cancel, rotation changes. This lesson adds rotational inertia and Στ = Iα, then solves pulleys with mass and rods that swing down.
Unit 5 Breakdown
In this lesson:
- What rotational inertia is, and why mass far from the axis counts more: [katex] I = \sum mr^2 [/katex]
- Rotational inertia of common shapes, and the parallel axis theorem
- Newton’s second law for rotation: [katex] \alpha = \frac{\sum\tau}{I} [/katex]
- The 4 problem types: finding [katex] I [/katex], torque to angular acceleration, pulleys with mass, and rods that swing down
Introduction
This lesson covers rotational inertia and Newton’s second law in rotational form. It is the last lesson of Unit 5, and it ties the unit together.
Unit 5.1 described rotation, 5.2 measured torque, and 5.3 balanced it. Now the torques do not cancel. How fast does the rotation change? To answer that you need one new idea: rotational inertia.
Rotational Inertia
Mass measures how hard it is to change an object’s linear motion. Rotational inertia [katex] I [/katex] measures how hard it is to change its rotation. It depends on the mass and on where that mass sits relative to the axis.
[katex] I = \sum m r^2 [/katex]
Each bit of mass counts by the square of its distance from the axis. Mass twice as far out counts four times as much. The units are [katex] \text{kg·m}^2 [/katex].
Common shapes
For an object with its mass spread out, the sum becomes a formula. The AP exam gives you these when you need them, but you should know the pattern: the more mass near the rim, the larger [katex] I [/katex].
| Object (axis through the center unless noted) | Rotational inertia |
| Point mass at distance [katex] r [/katex] | [katex] mr^2 [/katex] |
| Thin hoop or ring | [katex] MR^2 [/katex] |
| Solid disk or cylinder | [katex] \tfrac{1}{2}MR^2 [/katex] |
| Solid sphere | [katex] \tfrac{2}{5}MR^2 [/katex] |
| Thin rod about its center | [katex] \tfrac{1}{12}ML^2 [/katex] |
| Thin rod about one end | [katex] \tfrac{1}{3}ML^2 [/katex] |
Moving the axis: the parallel axis theorem
The same object is harder to turn about an axis away from its center of mass. If you know [katex] I_{cm} [/katex] about the center of mass, then about a parallel axis a distance [katex] d [/katex] away:
[katex] I = I_{cm} + Md^2 [/katex]
Check it with a rod: about its end, [katex] d = \tfrac{L}{2} [/katex], so [katex] I = \tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{3}ML^2 [/katex], matching the table.
Rotational inertia depends on the axis. Always ask “about which axis?”
Newton’s Second Law in Rotational Form
Start with one small mass [katex] m [/katex] on a light rod of length [katex] r [/katex], pushed by a force [katex] F [/katex] perpendicular to the rod.
- Newton’s second law along the push: [katex] F = ma_t [/katex].
- Use the link from Unit 5.1, [katex] a_t = r\alpha [/katex]: [katex] F = mr\alpha [/katex].
- Multiply both sides by [katex] r [/katex]: [katex] rF = mr^2\alpha [/katex]. The left side is the torque, and [katex] mr^2 [/katex] is the rotational inertia.
[katex] \alpha = \frac{\sum \tau}{I} [/katex]
It has the same shape as [katex] a = \frac{\sum F}{m} [/katex]. A net torque causes angular acceleration, and rotational inertia resists it. Zero net torque gives [katex] \alpha = 0 [/katex], which is Unit 5.3.
| Linear | Rotational |
| Force [katex] F [/katex] | Torque [katex] \tau [/katex] |
| Mass [katex] m [/katex] | Rotational inertia [katex] I [/katex] |
| [katex] \sum F = ma [/katex] | [katex] \sum \tau = I\alpha [/katex] |
Every rotational dynamics question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Finding and comparing [katex] I [/katex] | “rotational inertia”, ranking shapes, moving masses or the axis | [katex] I = \sum mr^2 [/katex], shape formulas, [katex] I_{cm} + Md^2 [/katex] |
| 2. Torque to angular acceleration | a torque or rim force spins up or slows down a wheel | [katex] \alpha = \frac{\sum\tau}{I} [/katex], then Unit 5.1 kinematics |
| 3. Pulleys with mass | a block on a string around a pulley “of mass M” | Block: [katex] \sum F = ma [/katex]. Pulley: [katex] \sum\tau = I\alpha [/katex]. Link: [katex] a = R\alpha [/katex] |
| 4. Rods that swing down | a rod pivoted at one end and released | Gravity acts at the middle; [katex] \alpha = \frac{\tau}{I} [/katex] at that instant |
Problem Type 1: Finding and comparing rotational inertia
How to spot it: The question asks for [katex] I [/katex], ranks shapes of the same mass and radius, moves masses in or out, or moves the axis.
Setup: Add [katex] mr^2 [/katex] for point masses, use the given formula for each shape, and add [katex] Md^2 [/katex] when the axis is not through the center of mass. For a ratio, write the formula and see what scales.
Two [katex] 0.50\,\text{kg} [/katex] balls are fixed to the ends of a light [katex] 0.80\,\text{m} [/katex] rod. Find the rotational inertia about the center of the rod, and about one end.
About the center, each ball is [katex] 0.40\,\text{m} [/katex] out: [katex] I = 2(0.50)(0.40)^2 = 0.16 \text{ kg·m}^2 [/katex]. About one end, one ball sits on the axis ([katex] r = 0 [/katex]) and the other is [katex] 0.80\,\text{m} [/katex] out: [katex] I = (0.50)(0.80)^2 = 0.32 \text{ kg·m}^2 [/katex]. Same object, twice the rotational inertia.
AP trap: Scaling with [katex] r [/katex] instead of [katex] r^2 [/katex]. A mass moved twice as far from the axis adds four times as much rotational inertia, not twice.
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Problem Type 2: Torque to angular acceleration
How to spot it: A known torque, or a force at the rim, spins something up or slows it down. Often followed by “how long” or “how many revolutions”.
Setup: Find the net torque first (a friction torque at the axle counts as negative). Then [katex] \alpha = \frac{\sum\tau}{I} [/katex], then a kinematic equation from Unit 5.1.
A [katex] 4.0\,\text{kg} [/katex] solid disk of radius [katex] 0.25\,\text{m} [/katex] ([katex] I = \tfrac{1}{2}MR^2 [/katex]) starts at rest. A [katex] 5.0\,\text{N} [/katex] force pulls on a string wrapped around its rim. How fast is it spinning after [katex] 4.0\,\text{s} [/katex]?
[katex] I = \tfrac{1}{2}(4.0)(0.25)^2 = 0.125 \text{ kg·m}^2 [/katex]. [katex] \tau = rF = (0.25)(5.0) = 1.25 \text{ N·m} [/katex]. [katex] \alpha = \frac{1.25}{0.125} = 10 \text{ rad/s}^2 [/katex]. [katex] \omega = \alpha t = (10)(4.0) = 40 \text{ rad/s} [/katex].
AP trap: Dividing the force by [katex] I [/katex]. Newton’s second law for rotation needs the torque, [katex] rF [/katex], not the force.
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Problem Type 3: Pulleys with mass
How to spot it: A block hangs from a string wrapped around a pulley, drum or cylinder that has mass: “a pulley of mass M and radius R”.
Setup: Two objects, two equations, one link. Block: [katex] mg – T = ma [/katex]. Pulley: [katex] TR = I\alpha [/katex]. No slipping: [katex] a = R\alpha [/katex].
A [katex] 2.0\,\text{kg} [/katex] block hangs from a light string wrapped around a solid disk pulley of mass [katex] 6.0\,\text{kg} [/katex] ([katex] I = \tfrac{1}{2}MR^2 [/katex]). Find the block’s acceleration and the tension.
Pulley: [katex] TR = \left(\tfrac{1}{2}MR^2\right)\frac{a}{R} [/katex], so [katex] T = \tfrac{1}{2}Ma = 3.0a [/katex]. Block: [katex] (2.0)(9.8) – 3.0a = 2.0a [/katex], so [katex] a = \frac{19.6}{5.0} \approx 3.9 \text{ m/s}^2 [/katex]. Then [katex] T = 3.0a \approx 12 \text{ N} [/katex], less than the block’s weight of [katex] 19.6\,\text{N} [/katex]. The radius cancels.
In general [katex] a = \frac{mg}{m + \frac{1}{2}M} [/katex]: the pulley acts like extra mass that also has to be sped up. With a massless pulley you get back free fall for one block, or the Atwood machine results for two.
AP trap: Setting the tension equal to [katex] mg [/katex]. If the block speeds up downward, [katex] T [/katex] must be less than [katex] mg [/katex]. With two blocks, the tensions on the two sides of a massive pulley are not equal: their difference is what turns it.
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Problem Type 4: Rods that swing down
How to spot it: A rod, beam or door is pivoted at one end, held level and released: “just after release”, “initial angular acceleration”, or how [katex] \alpha [/katex] changes as it swings.
Setup: Gravity acts at the center of mass, [katex] \tfrac{L}{2} [/katex] from the pivot. Level: [katex] \tau = Mg\tfrac{L}{2} [/katex]. Tilted: the lever arm shrinks to [katex] \tfrac{L}{2}\cos\theta [/katex]. Then [katex] \alpha = \frac{\tau}{I} [/katex] at that instant.
A uniform [katex] 1.2\,\text{m} [/katex] rod is pivoted at one end, held level and released. Find its angular acceleration just after release. ([katex] I = \tfrac{1}{3}ML^2 [/katex] about the end.)
[katex] \tau = Mg(0.60) [/katex] and [katex] I = \tfrac{1}{3}M(1.2)^2 = 0.48M [/katex], so [katex] \alpha = \frac{(9.8)(0.60)M}{0.48M} \approx 12 \text{ rad/s}^2 [/katex]. The mass cancels. In symbols, [katex] \alpha = \frac{3g}{2L} [/katex], so a longer rod swings down more slowly.
The swing gets gentler
AP trap: Treating [katex] \alpha [/katex] as constant during the swing. The torque changes with the angle, so the constant-acceleration equations do not apply. Use energy for the speed at the bottom.
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Unit 5 Wrap-Up
- Describe rotation with [katex] \theta [/katex], [katex] \omega [/katex] and [katex] \alpha [/katex], and link it to a point with [katex] v = r\omega [/katex] (5.1).
- Torque is what changes rotation: [katex] \tau = rF\sin\theta [/katex] (5.2).
- Zero net torque means no change in rotation, which is how seesaws, shelves and ladders stay put (5.3).
- Net torque causes angular acceleration, resisted by rotational inertia: [katex] \sum\tau = I\alpha [/katex] (5.4).
For the whole unit on one page, see the rotational kinematics and torque speed review.
Unit 6 Preview
Next up is Unit 6: Energy and Momentum of Rotating Systems. Energy and momentum from Units 3 and 4 get their rotational twins: rotational kinetic energy [katex] \tfrac{1}{2}I\omega^2 [/katex], angular momentum [katex] L = I\omega [/katex], and the spinning skater who speeds up by pulling in their arms.
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This article was written by Jason