Problem Statement
Solve the kinematics problem: A point moves on a circle of radius $R = 20\,\text{cm}$ with constant speed, making one revolution in $T = 0.40\,\text{s}$. Find: (a) the modulus of mean velocity over $t = T/4$; (b) the modulus of mean acceleration over $t = T/4$. Linear speed: $v = 2\pi R/T = 2\pi(0.20)/0.40 = \pi\,\text{m/s} \app
Given Information
- $R = 20\,\text{cm}$
- $T = 0.40\,\text{s}$
Physical Concepts & Formulas
Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.
- $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
- $F_c = mv^2/R$ — net centripetal force needed
- Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
- Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)
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
Full substitution shown in the steps above.
Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Physical Interpretation
The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.
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