Problem Statement
Solve the rotational mechanics problem: Solve the kinematics problem: Find the angular velocity of the Earth’s rotation about its own axis. See problem statement for all given quantities. Rotational kinematics mirrors linear kinematics with $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$. The angular vel
Given Information
- Mass $m$, geometry (radius $R$, length $L$, etc.)
- Angular velocity $\omega$ or torque $\tau$
- Axis of rotation
Physical Concepts & Formulas
Rotational mechanics is the angular analogue of linear mechanics. The moment of inertia $I = \sum m_i r_i^2$ plays the role of mass; torque $\tau = r\times F$ plays the role of force; angular momentum $L = I\omega$ plays the role of linear momentum. Newton’s 2nd law becomes $\tau_{\text{net}} = I\alpha$. The parallel axis theorem $I = I_{\text{cm}} + Md^2$ allows computing $I$ about any axis from the centre-of-mass value. Conservation of angular momentum ($L = \text{const}$ when $\tau_{\text{ext}} = 0$) explains why spinning skaters speed up when they pull their arms in.
- $I = \int r^2\,dm$ — moment of inertia
- $\tau = I\alpha$ — rotational Newton’s 2nd law
- $L = I\omega$ — angular momentum
- $KE_{\text{rot}} = \frac{1}{2}I\omega^2$
- Solid sphere: $I = \frac{2}{5}MR^2$; hollow sphere: $\frac{2}{3}MR^2$
- Solid cylinder: $I = \frac{1}{2}MR^2$; thin rod (centre): $\frac{1}{12}ML^2$
Step-by-Step Solution
Step 1 — Identify body and axis: Choose the appropriate moment of inertia formula or use the parallel axis theorem.
Step 2 — Torque equation: $\tau_{\text{net}} = I\alpha$
Step 3 — Kinematics: Use rotational analogues: $\omega = \omega_0 + \alpha t$, $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$, $\omega^2 = \omega_0^2 + 2\alpha\theta$.
Step 4 — Energy: $KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$ for rolling bodies.
Worked Calculation
Substituting all values with units:
Solid cylinder ($I = MR^2/2$) rolling down incline $h = 1\,\text{m}$:
$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{MR^2}{2}\cdot\frac{v^2}{R^2} = \frac{3}{4}Mv^2$$
$$v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4\times9.8\times1}{3}} = \sqrt{13.07} \approx 3.61\,\text{m/s}$$
Answer
$$\boxed{v = \sqrt{\dfrac{4gh}{3}}\text{ (solid cylinder rolling)}}$$
Physical Interpretation
A rolling cylinder reaches 3.61 m/s at the bottom of a 1 m incline — compared to $\sqrt{2gh} = 4.43\,\text{m/s}$ for a sliding point mass. Rolling is slower because energy is split between translational and rotational KE. A hollow cylinder ($I = MR^2$) would be even slower ($v = \sqrt{gh}$), while a solid sphere lands between the two. This is why denser, more centrally concentrated objects win in race-down-a-ramp experiments.
Leave a Reply