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Midterm II 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 Spinning Satellite and the Probe

Angular momentum conservation through a rotational collision, then the energy consequences of pulling mass toward the axis.

A flat, uniform, solid cylindrical satellite platform of mass \(M = 2\text{ kg}\) and radius \(R = 4\text{ m}\) is floating in deep space. It spins about its central vertical axis (the \(z\)-axis) with angular velocity \(\omega_0 = 10\text{ rad/s}\) in the counter-clockwise (CCW) direction when viewed from above.

A small \(m = 1\text{ kg}\) space-probe (treated as a point mass) flies parallel to the disk's surface. Just as it reaches the outer edge (\(r = 4\text{ m}\)), it is moving tangentially at \(v_0 = 32\text{ m/s}\) in the clockwise (CW) direction relative to the platform's center.

The probe's sticky landing gear engages, and it lands on and sticks to the outer edge of the platform.

(For a solid disk, \(I_\text{disk} = \tfrac{1}{2}MR^2\). For a point mass, \(I_\text{point} = mr^2\).)

  1. The Angular Collision. Define the initial angular momentum vector of the platform (\(\vec{L}_\text{disk}\)) and of the probe (\(\vec{L}_\text{probe}\)) relative to the rotation axis. Express them using \(+z\) (upward) and \(-z\) (downward) notation.

  2. Calculate the final angular velocity \(\omega_1\) (magnitude and direction) of the system immediately after the probe lands.

  3. Calculate the change in rotational kinetic energy (\(\Delta KE_\text{rot}\)) during this collision.

  4. Moving Inward. Once stuck, an internal motorized winch pulls the \(1\text{ kg}\) probe radially inward from the outer edge (\(r_1 = 4\text{ m}\)) all the way to the center (\(r_2 = 0\text{ m}\)).

    • Calculate the new angular velocity \(\omega_2\) of the system.
    • Determine whether the rotational kinetic energy increased, decreased, or stayed the same during this pull, and explain the physical cause.
  I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

  - Topic: Moments of inertia, angular momentum conservation, and rotational motion.
  - Difficulty Level: 4/5 (challenging with common pitfalls).
  - Numbers: Optimized for no-calculator use.
  - Previously Challenging Parts I Want to Study: Directions of angular velocity vectors, and the non-intuitive changes in rotation due to angular momentum conservation.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Convention. CCW (viewed from above) is \(+z\); CW is \(-z\).

Part 1 — Initial angular momentum vectors

Platform (solid disk):

\[I_\text{disk} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2)(4)^2 = 16\ \text{kg·m}^2\]
\[L_\text{disk} = I_\text{disk}\,\omega_0 = (16)(10) = 160\ \text{kg·m}^2/\text{s}\]

Since the platform spins CCW:

\[\vec{L}_\text{disk} = +160\ \hat{z}\ \text{kg·m}^2/\text{s}\]

Probe (point mass moving tangentially CW at the edge):

\[L_\text{probe} = m v_0 r = (1)(32)(4) = 128\ \text{kg·m}^2/\text{s}\]

Since the probe moves CW:

\[\vec{L}_\text{probe} = -128\ \hat{z}\ \text{kg·m}^2/\text{s}\]

Part 2 — Final angular velocity after landing

No external torques, so total angular momentum is conserved:

\[L_\text{total} = +160 - 128 = +32\ \text{kg·m}^2/\text{s}\]

After landing, the probe sticks to the edge, so the total moment of inertia is:

\[I_1 = I_\text{disk} + I_\text{probe} = 16 + (1)(4)^2 = 16 + 16 = 32\ \text{kg·m}^2\]
\[\omega_1 = \frac{L_\text{total}}{I_1} = \frac{32}{32} = +1\text{ rad/s}\]

The positive sign means the system rotates CCW at \(1\text{ rad/s}\) — the platform's original spin wins, but is greatly slowed.

Part 3 — Change in rotational kinetic energy

Initial (platform + probe):

\[KE_i = \tfrac{1}{2}I_\text{disk}\omega_0^2 + \tfrac{1}{2}mv_0^2 = \tfrac{1}{2}(16)(10)^2 + \tfrac{1}{2}(1)(32)^2\]
\[KE_i = 800 + 512 = 1312\text{ J}\]

Final:

\[KE_f = \tfrac{1}{2}I_1\omega_1^2 = \tfrac{1}{2}(32)(1)^2 = 16\text{ J}\]
\[\Delta KE_\text{rot} = 16 - 1312 = \boxed{-1296\text{ J}}\]

A large energy loss, as expected for a sticking (inelastic) collision — and amplified here because the two objects were spinning in nearly opposite senses, so much of the motion cancels.

Part 4 — Winching the probe to the center

New angular velocity. The winch is an internal force, so angular momentum is still conserved. Moving the probe to \(r_2 = 0\) removes its contribution to \(I\):

\[I_2 = I_\text{disk} + m r_2^2 = 16 + (1)(0)^2 = 16\ \text{kg·m}^2\]
\[\omega_2 = \frac{L_\text{total}}{I_2} = \frac{32}{16} = +2\text{ rad/s}\]

The system speeds up to \(2\text{ rad/s}\) CCW (the "ice-skater" effect).

Energy change:

\[KE_2 = \tfrac{1}{2}I_2\omega_2^2 = \tfrac{1}{2}(16)(2)^2 = 32\text{ J}\]

Compared to \(KE_1 = 16\text{ J}\), the rotational kinetic energy increased (from 16 J to 32 J).

Physical cause: the winch motor must do positive work to pull the probe inward against the outward-directed centripetal requirement (in the rotating frame, it pulls against the "centrifugal" tendency). That work is converted into the extra rotational kinetic energy. Angular momentum is conserved, but energy is added by the internal motor.

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!