Midterm III Practice

This collection features solved problems covering work and energy, linear momentum and its conservation, and rotational dynamics—including rotational motion, moments of inertia, and angular momentum.

Problem 4 — The Slide-to-Roll Transition

By tracking a cylinder's transition from sliding to pure rolling, this problem demonstrates how to partition total kinetic energy into translational and rotational components while illustrating how kinetic friction does both negative and positive work.

ExercisePromptSolution

A uniform solid cylinder of mass \(M = 2\text{ kg}\) and radius \(R = 0.5\text{ m}\) is released from rest at the top of a frictionless hill of height \(h = 1.8\text{ m}\) (Point A).

At the bottom of the hill (Point B), the cylinder smoothly transitions onto a flat, horizontal track. Unlike the hill, this horizontal track is rough, with a coefficient of kinetic friction \(\mu_k = 0.2\).

As the cylinder slides along the horizontal track, friction slows its translational motion while simultaneously exerting a torque that causes it to spin. Eventually, at Point C, the cylinder achieves perfect rolling without slipping (\(v = \omega R\)) and continues rolling at a constant velocity.

(For a solid cylinder, \(I_{\text{com}} = \frac{1}{2}MR^2\). Use \(g = 10\text{ m/s}^2\).)

Part A: Down the Frictionless Hill

Find the translational speed (\(v_0\)) and angular velocity (\(\omega_0\)) of the cylinder at the bottom of the hill (Point B). Explain why the gravitational potential energy did not produce any rotational kinetic energy during the descent.

Part B: The Slide-to-Roll Transition

As the cylinder slides on the rough horizontal track between Points B and C:

Calculate the magnitude of the kinetic friction force (\(f_k\)) acting on the cylinder, the translational deceleration (\(a\)) of its COM, and the angular acceleration (\(\alpha\)) about its COM.
Calculate the time (\(t\)) it takes for the cylinder to transition from sliding to pure rolling without slipping.
Calculate the final translational velocity (\(v_f\)) and angular velocity (\(\omega_f\)) at Point C.

Part C: Energy Decomposition & The Work of Friction

Calculate:
- The total mechanical energy of the cylinder at Point B.
- The total mechanical energy (translational \(KE\) + rotational \(KE\)) of the cylinder at Point C.
- The total mechanical energy lost (\(\Delta E\)) due to non-conservative forces.
Calculate the distance (\(d\)) the COM of the cylinder traveled from B to C.
- Calculate the total angle (\(\theta\) in radians) the cylinder rotated through from B to C.
- Show that the total energy lost (\(\Delta E\)) is equal to the net work done by friction: \(W_{\text{fric}} = W_{\text{trans}} + W_{\text{rot}} = -f_k d + \tau_k \theta\). Explain why simply calculating \(-f_k d\) gives the wrong answer.

I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

- Topic: Rotational kinetic energy vs. center-of-mass (COM) kinetic energy vs. potential energy.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Decomposition of motion into COM and rotational components, total energy, and the loss of mechanical energy by non-conservative forces.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part A: Down the Frictionless Hill

Velocity at Point B

Because the hill is frictionless, there is no torque to make the cylinder spin (\(\omega_0 = 0\text{ rad/s}\)).

All gravitational potential energy (\(PE\)) converts only into translational kinetic energy (\(KE_{\text{trans}}\)):

\[Mgh = \frac{1}{2}Mv_0^2\]

\[v_0 = \sqrt{2gh} = \sqrt{2(10)(1.8)} = \sqrt{36} = 6\text{ m/s}\]

Part B: The Slide-to-Roll Transition

Forces and accelerations
- Normal force: \(F_N = Mg = (2)(10) = 20\text{ N}\)
- Kinetic friction: \(f_k = \mu_k F_N = 0.2 \times 20 = 4\text{ N}\) (pointing opposite to motion)
- Translational acceleration (\(a\)):
  
  \[a = \frac{-f_k}{M} = \frac{-4\text{ N}}{2\text{ kg}} = -2\text{ m/s}^2\]
- Moment of inertia (\(I_{\text{com}}\)):
  
  \[I_{\text{com}} = \frac{1}{2}MR^2 = \frac{1}{2}(2\text{ kg})(0.5\text{ m})^2 = 0.25\text{ kg}\cdot\text{m}^2\]
- Rotational acceleration (\(\alpha\)):
  
  The friction force acts at the outer rim, creating a torque \(\tau_k = f_k R = (4\text{ N})(0.5\text{ m}) = 2\text{ N}\cdot\text{m}\).
  
  \[\alpha = \frac{\tau_k}{I_{\text{com}}} = \frac{2\text{ N}\cdot\text{m}}{0.25\text{ kg}\cdot\text{m}^2} = 8\text{ rad/s}^2\]
Time to reach pure rolling

We set up our kinematic equations for both translation and rotation:

\[v(t) = v_0 + at = 6 - 2t\]

\[\omega(t) = \omega_0 + \alpha t = 0 + 8t\]

Pure rolling occurs when \(v(t) = \omega(t)R\):

\[6 - 2t = (8t)(0.5)\]

\[6 - 2t = 4t \implies 6t = 6 \implies t = 1\text{ second}\]
Final velocities at Point C

\[v_f = 6 - 2(1) = 4\text{ m/s}\]

\[\omega_f = 8(1) = 8\text{ rad/s}\]

(Check: \(\omega_f R = 8 \times 0.5 = 4\text{ m/s} = v_f\). It works!)

Part C: Energy Decomposition & The Work of Friction

Energy tracking
- Energy at Point B (\(E_B\)):
  
  \[E_B = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2}Mv_0^2 + 0 = \frac{1}{2}(2)(6)^2 = 36\text{ J}\]
- Energy at Point C (\(E_C\)):
  
  \[KE_{\text{trans}, C} = \frac{1}{2}Mv_f^2 = \frac{1}{2}(2)(4)^2 = 16\text{ J}\]
  
  \[KE_{\text{rot}, C} = \frac{1}{2}I_{\text{com}}\omega_f^2 = \frac{1}{2}(0.25)(8)^2 = 8\text{ J}\]
  
  \[E_C = 16\text{ J} + 8\text{ J} = 24\text{ J}\]
- Energy lost (\(\Delta E\)):
  
  \[\Delta E = E_C - E_B = 24\text{ J} - 36\text{ J} = -12\text{ J}\]
The work-friction decomposition

First, let's find the kinematics:
- Linear distance (\(d\)):
  
  \[d = v_0 t + \frac{1}{2}at^2 = 6(1) + \frac{1}{2}(-2)(1)^2 = 6 - 1 = 5\text{ m}\]
- Angular displacement (\(\theta\)):
  
  \[\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(8)(1)^2 = 4\text{ rad}\]
Friction does two types of work here:
- Translational work (\(W_{\text{trans}}\)): Friction opposes the forward linear slide of the center of mass.
  
  \[W_{\text{trans}} = -f_k \cdot d = -4\text{ N} \times 5\text{ m} = -20\text{ J}\]
- Rotational work (\(W_{\text{rot}}\)): Friction's torque points in the same direction as the rotation, speeding it up.
  
  \[W_{\text{rot}} = \tau_k \cdot \theta = (2\text{ N}\cdot\text{m}) \times (4\text{ rad}) = +8\text{ J}\]
Net work:

\[W_{\text{net}} = W_{\text{trans}} + W_{\text{rot}} = -20\text{ J} + 8\text{ J} = -12\text{ J}\]

\[\Delta E = W_{\text{net}} = -12\text{ J}\]

A common mistake is to calculate the energy loss as \(W = -f_k d = -20\text{ J}\). This fails to recognize that kinetic friction is not just a loss mechanism here. While friction removes \(20\text{ J}\) of energy from the translational motion, it acts like a clutch, transferring \(8\text{ J}\) of that energy back into spinning up the rotation. The net thermal energy dissipated to the environment is only the difference: \(12\text{ J}\).

Feedback & Contributions

Have feedback, want to share your experience, or found a fun variation of this exercise? Email jens.f.mahlmann@dartmouth.edu — we'd love to hear from you!