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## Homework Statement

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius of .2 m. A 1.50 kg mass 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 4.50 J of kinetic energy. b) What percent of the total kinetic energy does the pulley have?

## Homework Equations

KE_Rotational = (1/2)Iw^2

I_Disk = (1/2)Mr^2

v=rw

Vf^2=Vi^2 + 2gΔy

## The Attempt at a Solution

Plugging in I_Disk into KE Rotational, I had a single unknown variable, w, which I solved for and got

KE = .5 [ .5(2.5)(.2)^2 ] w^2 w = 13.41 rad/s

V= rw so V=.2(13.41) = 2.683 m/s which is the linear velocity of the falling mass connected in tandem with the pulley.

My solutions manual uses some sort of conservation of energy approach at this point. I tried using

kinematics and the formula Vf^2=Vi^2 + 2gΔy as such.

2.68^2 = 0 + 2(9.8)(y) , and Delta Y = .367 m. I'm posting because my book says this answer is wrong; it says .673m. I understand that their was a energy approach to this problem, but I'm confused why my answer is not the same despite the differing approach. I didn't see any road blocks in my way, and it's making me wonder what I didn't account for trying this with kinematics.

Any help would be appreciated.