A mass of 47.51 kg is attached to a light string which is wrapped around a cylin

Question

A mass of 47.51 kg is attached to a light string which is wrapped around a cylindrical spool of radius 10.0 cm and moment of inertia 4.00 kg*m2. the spool is suspended from the ceiling and the mass is then released from rest a distance 2.40 meters above the floor. How long does it take to reach the floor?
Here is one way to solve it, but it comes up with the wrong answer:

Find final velocity of spool:
initial PE = final KE
mgh = 1/2 mv^2 + 1/2 Iw^2
mgh = 1/2 mv^2 + 1/2 (1/2mr^2)(v/r)^2
cancel m's and r^2s
gh = 3/4 v^2
final velocity = ((4/3) gh) = 5.6 m/s

find avg velocity = 1/2 final v = 2.8 m/s

time to reach floor = distance/avg velocity = 2.4 /2.8 = .857 seconds.

The correct answer is 2.15 seconds, but I don't know why. What's wrong with my solution method?

Explanation / Answer

Find final velocity of spool: initial PE = final KE mgh = 1/2 mv^2 + 1/2 Iw^2 in this problem I is not equal to 0.5mr^2 but I=4. so we have that mgh=0.5*mv^2+0.5*4*v^2/r^2. so gh=0.5*v^2+2*v^2/47.5*0.1^2 so that gh=4.71v^2. from this you can have the final velocity. v=2.23(m/s) so T=2.4*2/2.23=2.15(s)

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