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Inertia, frictionless pulley, and a weight

  • Thread starter jaredmt
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1. Homework Statement
A frictionless pulley has the shape of a uniform sold disk of mass 2.5 kg and radius .2m. A 1.5kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.5J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?


2. Homework Equations
I for solid disk: I = (1/2)(MR^2)



3. The Attempt at a Solution

first i plugged in this equation:
K = (1/2)IW^2 + (1/2)MV^2
i replaced W with V/r
K = (1/2)I(V/r)^2 + (1/2)MV^2
first i had to find I:
using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
then i plugged in everything to find final velocity :
4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
then i solved for V and got V = 1.8 m/s

ok now im not entirely sure what to do... i need the distance that the rock drops. but to do that i need to find the acceleration.
radial acceleration is V^2/r = 16.2 rad/s^2

then i tried V^2 = Vo^2 + 2(a)(X-Xo)
Vo=0 and Xo=0 so it becomes:
V^2 = 2(a)(x)

i have value for V. i know im supposed to be able to figure out A so i can solve for x. but im not sure if im converting from radial acceleration to regular acceleration correctly because i keep getting the wrong answer.

i thought this was the formula for that:
a^2 = Arad^2 + Atan^2
and Atan = Arad(r) right? because im doin something wrong... idk what it is
the answer is supposed to be 67.3 cm
 
Last edited:

Answers and Replies

Doc Al
Mentor
44,827
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first i plugged in this equation:
K = (1/2)IW^2 + (1/2)MV^2
i replaced W with V/r
K = (1/2)I(V/r)^2 + (1/2)MV^2
first i had to find I:
using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
then i plugged in everything to find final velocity :
4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
then i solved for V and got V = 1.8 m/s
4.5J is the KE of the pulley, not the total KE.

Figure out the total KE. Then ask where that energy came from. (Hint: Conservation of mechanical energy.)
 
122
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thanks for the help. i got part (b) which is Kpulley/Ktotal = 4.5/9.89 x 100 = 45.5% :)

but im still stuck on part (a). since i got the second part right i know that my velocity is right now. the W = 13.4 rad/s and V = Wr = 13.4(.2) = 2.68 m/s.
this time i tried solving the distance by finding the change in the angle:
Arad = (W^2)(r) = 36 rad/s^2
W^2 = 2(Arad)(feta)
13.4^2 = 2(36)(feta)
feta = 2.494 rads
this doesnt come out to 67.3m :(

edit: SWEET! i figured it out cus of ur mechanical energy hint!
i did:
Ktotal = mgh
9.89 = 1.5(9.8)h
h = .673m = 67.3 cm

thanks a lot!
 
Last edited:
Doc Al
Mentor
44,827
1,083
I can't quite follow what you're doing. In any case, instead of messing around with speeds, use conservation of energy. Compare the mechancial energy of the system before and after the stone falls.
 

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