Unit 6.1 | Rotational Kinetic Energy and Work

Problem Type 1: Finding rotational kinetic energy

How to spot it: Something spins, and the question asks for its kinetic energy, or ranks objects by it.

Setup: Find [katex] I [/katex] (point masses, or the given shape formula) and [katex] \omega [/katex] in rad/s. Then [katex] K = \tfrac{1}{2}I\omega^2 [/katex]. For rankings, compare [katex] I\omega^2 [/katex] case by case.

A [katex] 2.0\,\text{kg} [/katex] solid disk of radius [katex] 0.30\,\text{m} [/katex] ([katex] I = \tfrac{1}{2}MR^2 [/katex]) spins at [katex] 20\,\text{rad/s} [/katex]. Find its rotational kinetic energy. What happens to it if the disk spins twice as fast?

[katex] I = \tfrac{1}{2}(2.0)(0.30)^2 = 0.090 \text{ kg·m}^2 [/katex]. [katex] K = \tfrac{1}{2}(0.090)(20)^2 = 18 \text{ J} [/katex]. At [katex] 40\,\text{rad/s} [/katex], [katex] K = 72 \text{ J} [/katex]: twice the spin, four times the energy.

A hoop of the same mass and radius has twice the rotational inertia ([katex] MR^2 [/katex]), so at the same [katex] \omega [/katex] it stores twice the energy. Mass far from the axis stores more.

AP trap: Forgetting to square [katex] \omega [/katex], or plugging in rpm. Doubling [katex] \omega [/katex] quadruples [katex] K [/katex], and [katex] \omega [/katex] must be in rad/s.

Problem Type 2: Work done by a torque

How to spot it: A motor, a brake or friction at the axle acts on something spinning “through an angle” or “for so many revolutions”, and the question asks for a speed, an angle or an energy.

Setup: Write the work-energy theorem: [katex] \tau\Delta\theta = \tfrac{1}{2}I\omega_f^2 – \tfrac{1}{2}I\omega_i^2 [/katex]. A torque against the rotation does negative work. Angles in radians.

A bicycle wheel with [katex] I = 0.12 \text{ kg·m}^2 [/katex] spins at [katex] 25\,\text{rad/s} [/katex]. A brake pad presses on it with a constant friction torque of [katex] 1.5\,\text{N·m} [/katex]. Through what angle does the wheel turn before it stops?

The wheel starts with [katex] K = \tfrac{1}{2}(0.12)(25)^2 = 37.5 \text{ J} [/katex] and ends with none. The brake does all the (negative) work: [katex] -(1.5)\Delta\theta = 0 – 37.5 [/katex], so [katex] \Delta\theta = 25 \text{ rad} [/katex], about 4.0 revolutions.

You could also find [katex] \alpha = \frac{\tau}{I} [/katex] and use a rotational kinematic equation. The energy route is shorter whenever the question gives or asks for an angle instead of a time.

AP trap: Using degrees or revolutions in [katex] W = \tau\Delta\theta [/katex]. The formula only works with [katex] \Delta\theta [/katex] in radians: one revolution is [katex] 2\pi [/katex] rad.

Problem Type 3: Torque vs. angle graphs

How to spot it: A graph of torque against angular position or angular displacement, often followed by “how fast is it spinning” or “how much work”.

Setup: The area under the [katex] \tau [/katex] vs. [katex] \theta [/katex] graph is the work. Area above the axis adds energy, area below takes it away. Then [katex] W = \Delta K_{rot} [/katex].

A flywheel with [katex] I = 0.50 \text{ kg·m}^2 [/katex] starts from rest. The net torque on it changes with angle as in the graph above. How fast is it spinning after [katex] 6\,\text{rad} [/katex]?

Area: [katex] (6)(2) + \tfrac{1}{2}(4)(6) = 24 \text{ J} [/katex]. All of it becomes rotational kinetic energy: [katex] 24 = \tfrac{1}{2}(0.50)\omega^2 [/katex], so [katex] \omega = \sqrt{96} \approx 9.8 \text{ rad/s} [/katex].

The flywheel is still speeding up at the end, even though the torque is shrinking. It speeds up for as long as the torque is positive, just more and more gently.

AP trap: Reading the slope, or the height of the graph at the end. Work is the area. And the area gives the energy, not [katex] \omega [/katex]: take a square root at the end.

Problem Type 4: Energy with rotation

How to spot it: Something falls or swings while something else spins: a block on a string over a pulley “of mass M”, a spool or yo-yo unwinding, a rod pivoted at one end and released. The question asks for a speed after a drop.

Setup: Use conservation of energy with every kind of kinetic energy: [katex] mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 [/katex]. Link the two speeds with [katex] v = r\omega [/katex] and solve for one of them.

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]). The block is released from rest. How fast is it moving after it falls [katex] 1.5\,\text{m} [/katex]?

Energy lost: [katex] mgh = (2.0)(9.8)(1.5) = 29.4 \text{ J} [/katex]. With [katex] \omega = \frac{v}{R} [/katex], the pulley’s energy is [katex] \tfrac{1}{2}\left(\tfrac{1}{2}MR^2\right)\frac{v^2}{R^2} = \tfrac{1}{4}Mv^2 = 1.5v^2 [/katex], and the block’s is [katex] \tfrac{1}{2}(2.0)v^2 = v^2 [/katex].

So [katex] 29.4 = 2.5v^2 [/katex] and [katex] v \approx 3.4 \text{ m/s} [/katex]. Check it against Unit 5.4, where the same setup gave [katex] a \approx 3.9 \text{ m/s}^2 [/katex]: [katex] v^2 = 2a\Delta y = 2(3.92)(1.5) = 11.8 [/katex], the same answer. The pulley ends up with 60% of the energy and the block with 40%.

▶ Watch: the same kind of pulley with mass, solved with forces and torques (starts at 10:25).

Rods that swing down

In Unit 5.4 the angular acceleration of a falling rod kept changing, so kinematics could not give its speed at the bottom. Energy can. A uniform rod of length [katex] L [/katex], pivoted at one end and released from horizontal, drops its center of mass by [katex] \tfrac{L}{2} [/katex]:

[katex] Mg\frac{L}{2} = \tfrac{1}{2}\left(\tfrac{1}{3}ML^2\right)\omega^2 [/katex], so [katex] \omega = \sqrt{\frac{3g}{L}} [/katex]. For a [katex] 1.2\,\text{m} [/katex] rod, [katex] \omega \approx 4.9 \text{ rad/s} [/katex] at the bottom, and its tip moves at [katex] v = \omega L \approx 5.9 \text{ m/s} [/katex], faster than a ball dropped [katex] 1.2\,\text{m} [/katex] ([katex] 4.8\,\text{m/s} [/katex]).

AP trap: Leaving out the rotational kinetic energy and getting [katex] v = \sqrt{2gh} [/katex]. If something with mass spins, it takes a share of the energy, so the block (or the rod’s center of mass) ends up slower than free fall. For a rod, use the drop of its center of mass, not of its end.

Lesson 6.2 Preview

In the next lesson we give momentum the same treatment: angular momentum, [katex] L = I\omega [/katex], and the angular impulse that changes it. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-6-1-rotational-kinetic-energy-and-work/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-6-1-rotational-kinetic-energy-and-work/