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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 5 — The Banked Rotor Platform

Spinning up a system with torque, then using the banked-surface geometry to see how the normal force alone supplies the centripetal force.

A flat, circular turntable of mass \(M = 15\text{ kg}\) and radius \(R = 1.2\text{ m}\) sits on a frictionless vertical axle through its center.

Around the outer rim, the surface is not flat but banked — tilted at \(\theta = 37^\circ\) from the horizontal, like the banking on a racetrack curve. This banked surface is frictionless. A small block of mass \(m = 5\text{ kg}\) rests on this banked rim at a distance \(r = 1.2\text{ m}\) from the central axis.

(For a solid disk, \(I_\text{disk} = \tfrac{1}{2}MR^2\). Treat the block as a point mass, \(I_\text{point} = mr^2\). Use \(g = 10\text{ m/s}^2\), \(\sin 37^\circ = 0.6\), and \(\cos 37^\circ = 0.8\).)

Spinning Up (Torque & Kinematics). The platform starts at rest. A small retractable peg holds the block in place so it can't slide down the frictionless ramp. A constant torque \(\tau = 9\text{ N·m}\) is applied to the axle to spin the system up.

  1. Calculate the total moment of inertia \(I_\text{total}\) of the system (platform + block).

  2. Calculate the angular acceleration \(\alpha\) of the system.

  3. How long does it take to reach the target angular velocity \(\omega = 2.5\text{ rad/s}\)?

  4. Spinning Freely. Once the system reaches \(\omega = 2.5\text{ rad/s}\), the torque is removed and the peg is retracted; the block stays exactly at \(r = 1.2\text{ m}\). Draw a Free-Body Diagram of the block.

  5. Resolve the forces on the block into horizontal (radial) and vertical components. Explain why tilting your axes parallel to the ramp is a bad choice here.

  6. Calculate the magnitude of the Normal Force \(F_N\) on the block.

  7. Show mathematically that at \(\omega = 2.5\text{ rad/s}\), the net horizontal force equals exactly the centripetal force needed to keep the block moving in a circle of radius \(r = 1.2\text{ m}\).

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

- Topic: Moments of inertia, rotational motion, and centripetal forces.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Vector projections, and centripetal forces generated by non-constant normal or tension forces.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part 1 — Total moment of inertia

\[I_\text{disk} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(15)(1.2)^2 = 10.8\ \text{kg·m}^2\]
\[I_\text{block} = mr^2 = (5)(1.2)^2 = 7.2\ \text{kg·m}^2\]
\[I_\text{total} = 10.8 + 7.2 = 18\ \text{kg·m}^2\]

Part 2 — Angular acceleration

\[\tau = I_\text{total}\,\alpha \quad\Rightarrow\quad \alpha = \frac{\tau}{I_\text{total}} = \frac{9}{18} = 0.5\ \text{rad/s}^2\]

Part 3 — Time to reach target speed

Starting from rest, \(\omega = \alpha t\):

\[t = \frac{\omega}{\alpha} = \frac{2.5}{0.5} = 5\text{ s}\]

Part 4 — Free-Body Diagram

Only two forces act on the block (the ramp is frictionless):

  • Weight \(mg\), straight down.
  • Normal force \(F_N\), perpendicular to the banked surface. Because the surface tilts at \(37^\circ\), \(F_N\) points up and inward (toward the axis), making angle \(\theta\) with the vertical.

Part 5 — Choosing axes

Use horizontal (\(x\)) and vertical (\(y\)) axes, not axes tilted along the ramp.

Reason: the block moves in a horizontal circle, so its acceleration is purely horizontal (centripetal, pointing toward the axis) and vertical acceleration is zero. Horizontal/vertical axes align with the actual acceleration, keeping the equations clean.

If you tilted the axes along the ramp instead, the centripetal acceleration would split into components along both tilted axes, creating extra terms and a messier system — the opposite of what you want.

Resolve the two forces:

  • Weight: \((0,\ -mg)\)
  • Normal force: horizontal component \(F_N\sin\theta\) (inward), vertical component \(F_N\cos\theta\) (up).

Part 6 — Normal force

Vertically, the block has no acceleration, so vertical forces balance:

\[F_N\cos\theta = mg\]
\[F_N = \frac{mg}{\cos\theta} = \frac{(5)(10)}{0.8} = 62.5\text{ N}\]

Part 7 — Horizontal force = centripetal requirement

The net horizontal force is the inward component of the normal force:

\[F_\text{horizontal} = F_N\sin\theta = (62.5)(0.6) = 37.5\text{ N}\]

The centripetal force required to hold the block in its circle:

\[F_c = m\omega^2 r = (5)(2.5)^2(1.2) = (5)(6.25)(1.2) = 37.5\text{ N}\]
\[F_\text{horizontal} = 37.5\text{ N} = F_c\]

The inward push from the normal force is exactly what's needed — which is why the block sits at \(r = 1.2\text{ m}\) without sliding, even on a frictionless ramp. This works only at this specific \(\omega\); spin faster and it slides outward/up, slower and it slides inward/down.

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!