Answer:
- correct (y, y, CCW, n)
- needs correction (y, y, CW, y)
- correct (n, n, CCW, y)
Explanation:
You want to know the starting position, rotation direction, speed, and acceleration for three points specified by different angle relations.
Setup
To save "ink" and typing effort, we will write the vector ...
r(t) = a·cos(θ(t))i +b·sin(θ(t))j
in the more compact notation
r(t) = √(a²+b²)∠α(t) . . . . . where tan(α(t)) = (b/a)tan(θ(t))
Here, √(a²+b²) = 1 in all cases. Then the initial position and direction of motion are ...
r(0) = 1∠α(0) . . . . . . . . . . . . . . 1∠0 corresponds to 1i+0j = (1, 0)
direction = sign(α'(0⁺)) . . . . . . where positive is counterclockwise
The acceleration is the sum of the radial and tangential accelerations. When the speed is not constant, the tangential acceleration is non-zero, and the acceleration is not orthogonal to the velocity.
A. r(t) = 1∠(t -π/6)
The speed is d(t -π/6)/dt = 1, a constant.
Because the speed is constant, the acceleration is orthogonal to the velocity.
The sign of the speed is positive, so the direction is CCW.
The particle begins at the point 1∠(-π/6) ≠ 1∠0.
B. r(t) = 1∠-t
The speed is d(-t)/dt = -1, a constant.
Because the speed is constant, the acceleration is orthogonal to the velocity.
The sign of the speed is negative, so the direction is CW.
The particle begins at the point 1∠0.
C. r(t) = 1∠t^7
The angular speed is d(t^7)/dt = 7·t^6, not constant.
Because the speed is not constant, the acceleration is not orthogonal to the velocity.
The sign of the angular speed is positive for t>0, so the direction is CCW.
The particle begins at the point 1∠0.