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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 3 — The Sliding Rod on Ice

Understand the physical independence of translational and rotational dynamics by showing how off-center forces can alter an object's rotation and kinetic energy while leaving the motion of its center of mass completely unaffected.

A uniform thin wooden rod of mass \(M = 3\text{ kg}\) and length \(L = 2\text{ m}\) lies flat on a frictionless, horizontal sheet of ice. We define a coordinate system on the ice where the rod initially lies along the North-South axis (the \(y\)-axis), with its center of mass (\(COM\)) exactly at the origin \((0,0)\). Assume that the forces have the same angle with the rod throughout the push.

(For a uniform rod of length \(L\) rotating about its center of mass, \(I_{\text{com}} = \frac{1}{12}ML^2\).)

Part A: The Off-Center Push

A constant horizontal force \(F_1 = 6\text{ N}\) pointing directly East (\(+x\)) is applied to the rod at a point \(0.5\text{ m}\) North of the center of mass. This force is maintained pointing East for a duration of \(t = 2\text{ s}\).

  1. Calculate the translational acceleration of the center of mass (\(a_{\text{com}}\)) during the push.

  2. If the same \(6\text{ N}\) force were instead applied directly to the center of mass (\(y = 0\)), would the center of mass accelerate faster, slower, or the same? Explain the physical law that supports your answer.

  3. Calculate the angular acceleration (\(\alpha\)) of the rod about its center of mass during the push.

  4. Calculate both the translational velocity (\(v_{\text{com}}\)) of the center of mass and the angular velocity (\(\omega\)) of the rod at the instant \(t = 2\text{ s}\) when the force is removed.

Part B: The Force Couple & COM Conservation

The rod is brought back to rest at the origin. This time, we apply two forces simultaneously for a duration of \(t = 2\text{ s}\): Force \(F_1 = 6\text{ N}\) pointing East (\(+x\)) is applied at \(0.5\text{ m}\) North of the center of mass. Force \(F_2 = 6\text{ N}\) pointing West (\(-x\)) is applied at \(0.5\text{ m}\) South of the center of mass.

  1. Describe the motion of the center of mass during and after this double-push. What is its final velocity \(v_{\text{com}}\) at \(t = 2\text{ s}\)?

  2. Calculate the net torque (\(\Sigma \tau_{\text{com}}\)) acting on the rod about its center of mass, and determine the final angular velocity (\(\omega_f\)) of the rod at \(t = 2\text{ s}\).

  3. Explain how the rod gained a significant amount of kinetic energy during Part B, even though its center of mass did not move a single millimeter.

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

- Topic: Rotational and translational dynamics (non-equilibrium with net forces and net torques).
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Center of mass (COM) motion vs. rotational motion, and the conservation of COM motion.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part A: The Off-Center Push

  1. Translational acceleration of the COM

    According to Newton's Second Law for translational motion, the acceleration of the center of mass depends only on the net external force and the total mass:

    \[\Sigma F_x = Ma_{\text{com}}\]
    \[6\text{ N} = (3\text{ kg})a_{\text{com}} \implies a_{\text{com}} = 2\text{ m/s}^2\]

  2. The point-of-application question

    The center of mass accelerates at exactly the same rate (\(2\text{ m/s}^2\)) regardless of where the force is applied.

    Many people intuitively believe that because some of the energy goes into making the rod rotate, there is "less force left over" to make it translate. This is incorrect. Forces do not get divided between translation and rotation. Newton's Second Law (\(\Sigma F = Ma_{\text{com}}\)) cares only about the magnitude of the force, not where it hits.

  3. Rotational acceleration (\(\alpha\))

    First, calculate the moment of inertia about the center of mass (\(I_{\text{com}}\)):

    \[I_{\text{com}} = \frac{1}{12}ML^2 = \frac{1}{12}(3\text{ kg})(2\text{ m})^2 = \frac{1}{12}(12) = 1\text{ kg}\cdot\text{m}^2\]

    Now, find the torque about the COM (\(\tau = F \cdot d\), where \(d = 0.5\text{ m}\) is the lever arm):

    \[\tau_{\text{com}} = F_1 \cdot d = (6\text{ N})(0.5\text{ m}) = 3\text{ N}\cdot\text{m} \quad \text{(Clockwise)}\]

    Using Newton's Second Law for Rotation (\(\tau = I\alpha\)):

    \[\alpha = \frac{\tau_{\text{com}}}{I_{\text{com}}} = \frac{3\text{ N}\cdot\text{m}}{1\text{ kg}\cdot\text{m}^2} = 3\text{ rad/s}^2\]

  4. Kinematics at \(t = 2\text{ s}\)

    Center of mass velocity:

    \[v_{\text{com}} = a_{\text{com}}t = (2\text{ m/s}^2)(2\text{ s}) = 4\text{ m/s}\]

    Angular velocity:

    \[\omega = \alpha t = (3\text{ rad/s}^2)(2\text{ s}) = 6\text{ rad/s}\]

Part B: The Force Couple & COM Conservation

  1. Motion of the COM

    Let's find the net external force acting on the system:

    \[\Sigma F_x = F_1 + F_2 = (+6\text{ N}) + (-6\text{ N}) = 0\text{ N}\]

    Since the net external force is zero:

    \[a_{\text{com}} = 0\text{ m/s}^2 \implies v_{\text{com}} = 0\text{ m/s}\]

    The center of mass remains perfectly stationary at the origin \((0,0)\) at all times.

  2. Net torque & final angular velocity

    Even though the forces cancel out translationally, they cooperate rotationally (this is called a force couple). Both forces tend to rotate the rod clockwise:

    \[\Sigma \tau_{\text{com}} = (F_1 \cdot d_1) + (F_2 \cdot d_2)\]
    \[\Sigma \tau_{\text{com}} = (6\text{ N} \times 0.5\text{ m}) + (6\text{ N} \times 0.5\text{ m}) = 3 + 3 = 6\text{ N}\cdot\text{m}\]

    Now solve for angular acceleration and final velocity:

    \[\alpha = \frac{\Sigma \tau_{\text{com}}}{I_{\text{com}}} = \frac{6\text{ N}\cdot\text{m}}{1\text{ kg}\cdot\text{m}^2} = 6\text{ rad/s}^2\]
    \[\omega_f = \alpha t = (6\text{ rad/s}^2)(2\text{ s}) = 12\text{ rad/s}\]

  3. The energy vs. momentum explanation

    How did the rod gain kinetic energy if \(\Sigma F = 0\)?

    Linear momentum (\(\vec{p} = M\vec{v}_{\text{com}}\)) is a vector quantity, and because the opposing force vectors cancel out, the net linear momentum remains zero.

    However, kinetic energy is a scalar quantity (\(KE = \frac{1}{2}I\omega^2\)). The work done by \(F_1\) and \(F_2\) is positive because the points of application of both forces move in the direction of the forces as the rod begins to rotate. Therefore, net positive work was done on the rod, converting chemical/mechanical energy into rotational kinetic energy:

    \[KE_{\text{rot}} = \frac{1}{2}I_{\text{com}}\omega_f^2 = \frac{1}{2}(1)(12)^2 = 72\text{ J}\]

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!