Question

As you finish listening to your favorite compact disc(CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise at 500rpm while the last song is playing, and then spins down to a zero angular speed in 2.60s with constant angular acceleration.

a. If the CD rotates clockwise at 500rpm while the last song is playing, and then spins down to zero angular speed in 2.60s with constant angular acceleration, what is the magnitude of the angular acceleration of the CD, as it spins to a stop?

b.How many revolutions does the CD make as it spins to stop?

Answer 1

a) 500 RPM works out to ω = 2π 500/60 = 50π/3 rad/s. The vector direction is along the axis of rotation and, according to the right-hand-rule, pointing away from the viewer observing clockwise rotation.

α = Δω/Δt = (-50π/3)/(2.60 s) ~~ -20.1 rad/s²

The negative sign means that the angular acceleration vector is directed opposite to the original angular velocity vector, along the axis of CD rotation but toward the viewer.

b)

The acceleration is in the opposite direction to the velocity {the disc is slowing down} so the acceleration should be a = -20.1 rad/s² when you plug it in eq. 1-

θ = 52.36 * 2.6 + ½ * (-20.1) * 2.6² = 68.2 rad {rounding errors}

There are 2π radians in 1 revolution, so 68.2 rad / 2π = 10.83 revolutions

That's 10 complete revolutions plus a bit!