Problem 1.6 — Circular motion from parametric equations

Problem Statement

A point moves in plane $xy$: $x = a\sin\omega t$, $y = a(1-\cos\omega t)$. Find: (a) distance traversed in time $\tau$; (b) angle between velocity and acceleration.

Given Information

• See problem statement for all given quantities.

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 — Identify given quantities and set up the problem: Velocity components:

Step 2 — Apply the relevant physical law or equation: $$\dot x = a\omega\cos\omega t,\quad \dot y = a\omega\sin\omega t$$
$$v = \sqrt{\dot x^2+\dot y^2} = a\omega = \text{const}$$

Step 3 — Solve algebraically for the unknown: (a) Distance: Since speed is constant,

Step 4 — Substitute numerical values with units: $$\boxed{s = a\omega\tau}$$

Step 5 — Compute and check the result: Acceleration components:

Step 6: $$\ddot x = -a\omega^2\sin\omega t,\quad \ddot y = a\omega^2\cos\omega t$$

Worked Calculation

$$\dot x = a\omega\cos\omega t,\quad \dot y = a\omega\sin\omega t$$

$$v = \sqrt{\dot x^2+\dot y^2} = a\omega = \text{const}$$

$$\boxed{s = a\omega\tau}$$

Velocity components:

$$\dot x = a\omega\cos\omega t,\quad \dot y = a\omega\sin\omega t$$
$$v = \sqrt{\dot x^2+\dot y^2} = a\omega = \text{const}$$

(a) Distance: Since speed is constant,

$$\boxed{s = a\omega\tau}$$

Acceleration components:

$$\ddot x = -a\omega^2\sin\omega t,\quad \ddot y = a\omega^2\cos\omega t$$

(b) Dot product $\vec v\cdot\vec w$:

$$\vec v\cdot\vec w = a\omega(-a\omega^2)\cos\omega t\sin\omega t + a\omega(a\omega^2)\sin\omega t\cos\omega t = 0$$
$$\boxed{\text{Angle} = 90°}$$

The trajectory is a circle of radius $a$ centered at $(0,a)$; constant-speed circular motion always has $\vec v \perp \vec w$.

Answer

$$\boxed{s = a\omega\tau}$$

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.