Problem Statement
Solve the rotational mechanics problem: Find the dimensions and SI unit of torque. Torque $\tau = r \times F$ Same dimensions as energy but physically distinct Step 1: $\tau = F \times r$ (force × perpendicular distance). Step 2: $[F] = MLT^{-2}$, $[r] = L$. Step 3: $[\tau] = ML^2T^{-2}$. $$\boxed{[\tau] = ML^2T^{-2},\quad \text{SI unit}
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{[\tau] = ML^2T^{-2},\quad \text{SI unit}
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 $h = 1\,\text{m}$:
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Answer
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Answer
$$\boxed{v = \sqrt{\dfrac{4gh}{3}}\text{ (solid cylinder rolling)}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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