Problem Statement
A uniform disc of mass m and radius R rolls down an incline of angle $\alpha$ without slipping. Find the acceleration and the friction force.
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems apply the rotational analogue of Newton’s laws about a chosen axis, using moment of inertia and torque.
- $M = I\beta$ — equation of rotational dynamics
- $I = \sum m_i r_i^2$ — moment of inertia
- $E_k = \tfrac{1}{2}I\omega^2$ — rotational kinetic energy
- $L = I\omega$ — angular momentum of a rigid body
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$Newton along incline: \text{mg} \sin \alpha – f = ma$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra.
$$Torque about CM: fR = I \alpha _ang = (mR^{2}/2)(a/R) → f = ma/2$$
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation.
$$\text{mg} \sin \alpha – ma/2 = ma → \text{mg} \sin \alpha = 3ma/2 → a = 2g \sin \alpha /3$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find.
$$f = ma/2 = \text{mg} \sin \alpha /3$$
Worked Calculation
$$a = 2g \sin \alpha /3$$
$$f = \text{mg} \sin \alpha /3$$
Answer
$$\boxed{a = 2g \sin \alpha /3; f = \text{mg} \sin \alpha /3 (up the incline)}$$
This is the quantity the problem asked for, expressed in terms of the given data: $a = 2g \sin \alpha /3; f = \text{mg} \sin \alpha /3 (up the incline)$.
Physical Interpretation
Friction is directed up the slope and provides the torque for rolling — without friction the disc would slide without rotating. Maximum static friction $\mu$ mgcos $\alpha$ ≥ tan $\alpha$ $/3$ required. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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