Question

1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal

Answer 1

Solve for the linear/tangential speed:

a = v²/r

where a = centripetal acceleration, v = speed, and r = radius.

4.7 m/s² = v²/(0.3 m)

v² = (0.3 m) (4.7 m/s²)

v ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2π (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.