A CD playing in a CD player is spinning at 50.0 rev/s when a power failure sudde

Question

A CD playing in a CD player is spinning at 50.0 rev/s when a power failure suddenly occurs. The power is off for 5.0 seconds, and during this time the CD slows down uniformly due to friction. During the time the poer if off the CD makes 150.0 complete revolutions.

A) At what rate is the CD spinning when the power comes back on?

B) How long after the beginning of the power failure would it have taken the CD to stop if the power had not come back on?

Thanks! Will rate A+ tonight, need explanation please. Thank you.

Explanation / Answer

A.

= 0t + (1/2)t2

==> = 2( - 0t) / t2 =  2((2 150.0 rad) - (2 50.0 rad/s)(5.0 s)) / (5.0 s)2 = -50 rad/s2

= (f - 0) / t

==> f = t + 0 = (-50 rad/s2)(5.0 s) + (2 50.0 rad/s) = 64 rad/s

B.

= (f - 0)/t

==> t = (f - 0) / = (0 - 2 50.0 rad/s) / (-50 rad/s2) = 6.28 s

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