A point moves in the plane xy according to the law x = a sin ωt, y = a(1 − cos ωt), where a and ω are positive constants. Find: (a) the distance s traversed by the point during the time τ; (b) the angle between the point’s velocity and acceleration vectors.
(a) The point covers the distance s = aωT in the time T. (b) The angle between its velocity and its acceleration is π/2 (90°) at every moment. With the numbers on screen: the speed is 3 m/s, so s = m at T = s.
Problem statement
Problem 1.26. A point moves in the plane xy according to the law x = a sin ωt, y = a(1 − cos ωt), where a and ω are positive constants. Find: (a) the distance s traversed by the point during the time τ; (b) the angle between the point’s velocity and acceleration vectors.
Interpretation. The point moves in the xy plane. Part (a) asks for the length of the path it travels in the time Part (a) asks for the length of the path it travels in the time T,tau;, which this page calls T, which is the distance along the path, not the straight-line distance from the start. Part (b) asks for the angle between the velocity vector and the acceleration vector, and the page shows it at every moment.
We assume a = 1.5 m and ω = 2 rad/s, because the problem gives no numbers. a has the unit of length so that a sin ωt is a length, and ω is a rate, so that ωt has no unit. Both are sliders. The graphs run from T = 0 to 6 s.
Symbols: a is the radius of the circle, ω the angular rate, T the time, s the distance along the path, v the velocity, w the acceleration.
The big idea in plain words
The two formulas look like waves, but together they describe a circle. The point has x = a sin ωt and y = a − a cos ωt, so x² + (y − a)² = a². It is a circle of radius a with its centre at (0, a), and the point goes round it at a steady rate. Steady speed on a circle means the speed does not change, only its direction does. So the distance is just speed times time. And an acceleration that only turns the velocity, without changing its size, must be at right angles to it. That is why the angle is 90°.
The solution in each frame
Only the ground frame is needed. The problem gives the motion in one frame and asks for the distance and the angle in that frame, so there is no second frame.
The solution step by step
Step 1, the velocity. Differentiate x = a sin ωt and y = a(1 − cos ωt): vx = aω cos ωt and vy = aω sin ωt. So v = aω cos ωt i + aω sin ωt j.
Step 2, the speed. |v| = aω√(cos²ωt + sin²ωt) = aω = 3 m/s. It is the same at every moment.
Step 3, the distance (part a). The distance along the path is the sum of the speed over the time: s = ∫0T |v| dt = ∫0T aω dt = aωT. At the time on the canvas, s = m. Because s counts the path length, it keeps growing after a full lap: one lap is 2πa = 9.425 m and takes 2π/ω = 3.142 s.
Step 4, the acceleration. Differentiate v: wx = −aω² sin ωt and wy = aω² cos ωt. So |w| = aω² = 6 m/s².
Step 5, the angle (part b). Use the dot product: v·w = (aω cos ωt)(−aω² sin ωt) + (aω sin ωt)(aω² cos ωt) = 0. A zero dot product means cos θ = 0, so the angle is θ = π/2, that is 90°. This holds for every t, a and ω.
Step 6, the cases. The problem has one situation, so no cases are needed. One thing to notice: after a time longer than one lap, the distance s is larger than the circumference. The straight-line distance from the start would only go up to 2a, the diameter.
A separate calculation does not use the formulas for v, w or s. It takes the positions x(t) and y(t), finds the speed from small differences, and adds up the small steps of path in 200000 pieces at T = 1, 3 and 6 s. It also gets the angle from the dot product of the numerical v and w at four moments. The largest relative difference from the formulas is 6.7 × 10⁻⁹.
Five methods
1. Force method
No mass or force is given, so this method does not apply. The problem is about motion only (kinematics). In a real circular motion the force on the point would also point to the centre.
2. Energy method
Energy does not help here, since no mass is given. The constant speed shows that the acceleration does no work, which fits with it being at 90° to the velocity.
3. Graph method
The first graph shows s growing in a straight line with slope aω. The second shows vx and vy as a cosine and a sine, while the speed stays flat. The third graph shows the angle fixed at 90°.
4. Calculus method
Differentiate for v, take its size, integrate it for s, differentiate again for w and use the dot product. These are steps 1 to 5.
5. Check method
Units: aω is in m/s and aωT is in metres. Limits: for ω = 0 the point stays at the origin and s = 0. For T = 2π/ω the distance is 2πa, the circumference, as it should be. Geometry: the radius to the point is at 90° to the velocity, and w points along the radius, so w is at 90° to v.
Fastest: the geometry check, since uniform circular motion has w at right angles to v. Most reliable: the calculus method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Radius a | Yes | The size of the circle, the speed aω, the acceleration aω², the distance s and the graphs. The angle stays 90° |
| Rate ω | Yes | The speed, the acceleration, the lap time, the distance s and the graphs. The angle stays 90° |
| Time T shown | No | The point, the arrows, the traced path, s = aωT, the dots on the graphs and the live readings |
| Given-answer mode | Sets T | The page finds the time needed for a chosen distance, and warns if it is beyond 6 s |
| Play and time bar | No | Play sweeps T from 0 to 6 s, holds 2 s at one full lap (if it fits in 6 s) and 1.5 s at the end, then repeats. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Radius of the circle | a | 1.5 m |
| Rate | ω | 2 rad/s |
| Speed | aω | 3 m/s |
| Acceleration size | aω² | 6 m/s² |
| Time for one lap | 2π/ω | 3.142 s |
| Distance in one lap | 2πa | 9.425 m |
Exam traps
- Giving the straight-line distance from the start. The problem asks for the path length, which is aωT, even beyond one lap.
- Integrating the velocity vector instead of its size. The integral of v dt is the displacement, not the distance.
- Forgetting that aω is the speed, not aω². The speed is aω. The acceleration is aω².
- Thinking the angle changes with time. The velocity turns, but the acceleration turns with it, so the angle between them stays 90°.
- Mixing up the sign of sin and cos in w. A wrong sign gives a dot product that is not zero.