# Homework Help: Speed/energy problem, could someone please check my work

1. Sep 7, 2007

### munkachunka

1. The problem statement, all variables and given/known data

when a car has travelled 100m up a 1 in 10 (sine slope) the driver stops and has a drink, he places an empty can of cola on the ground so that it can roll, the can weighs 50g and the radius of rotation k of the can is r

a, calculate the energy of the can at the top and bottom of the slope and the form it takes

b. calculate the speed of the can at the bottom of the slope

2. Relevant equations

E=1/2mv^2+1/2jw^2
Pe = MGH

3. The attempt at a solution

a, because the car has travelled 100m up the slope i calculated the height that the car was at. 100*1/10 = 10m

50 grams = 0.050gk

potential energy = mgh = 0.050*9.8*10 = 4.905Joules,

answer to a = potential energy at the top of 4.905Joules, this energy is changed to kinetic energy at the bottom and is also 4.905Joules

b, Kinetic energy gained ...E=1/2mv^2+1/2Jw^2

= E=1/2mv^2+1/2mk^2*w^2... where m = mass and K= radius of gyration

motion without slip w=v/r
= E=1/2mv^2+1/2mk^2*(v^2/r^2)

because the question states that the radius of rotaion of the can =r, I set these terms to 1

therefore E=1/2mv^2+1/2m*v^2

E=0.025*v^2+0.025*v^2
4.905=0.05v^2
98.1=v^2
v=9.9m/s

thankyou

2. Sep 7, 2007

### learningphysics

Everything looks good except I think should be J = 1/2 mk^2... I think you made a mistake when substituting this into 1/2 Jw^2. that'll change you numbers afterwards.

3. Sep 7, 2007

4. Sep 7, 2007

### learningphysics

Can you describe how you got from the first line to the second...

5. Sep 8, 2007

### munkachunka

yeah, point taken, yet another silly mistake that I am prone to making

6. Sep 8, 2007

### munkachunka

having a look thrugh a text book it turns out that I am not sure what is correct.

the book says E=1/2mv^2+1/2Jw^2 = E=1/2mv^2+1/2mk^2*w^2

7. Sep 8, 2007

### learningphysics

I apologize munkachunka. I was wrong. The textbook is right. The moment of inertia of a thin hollow cylinder is I = mr^2 (not I = 1/2 mr^2 as I thought).

Last edited: Sep 8, 2007
8. Sep 8, 2007

### munkachunka

no problem, thanks again for your help

9. Sep 8, 2007

### learningphysics

you're welcome.