Problem Statement
Solve the kinematics problem: A point moves on a circle of radius $R$ with constant angular acceleration $\beta = 0.10\,\text{rad/s}^2$ starting from $\omega_0 = 0.5\,\text{rad/s}$. Find the angle $\alpha$ between total acceleration and velocity at $t = 2.0\,\text{s}$. Angular velocity: $\omega = \omega_0+\beta t = 0.5+0.10(2.0)
Given Information
- $\beta = 0.10\,\text{rad/s}$
- $\omega_0 = 0.5\,\text{rad/s}$
- $t = 2.0\,\text{s}$
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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