- #1

as2528

- 40

- 9

- Homework Statement
- A metal can containing condensed mushroom soup has a mass of 215 g, a height of 10.8 cm, and a diameter of 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at an angle of 25.0° to the horizontal and is then released to roll straight down. Assuming energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom

of the incline. Which pieces of data, if any, are unnecessary for calculating the solution?

- Relevant Equations
- vf=vi+at

K=1/2*I*w^2+1/2*m*v^2

PE=mgh

v=rw

a=2/3*g*sin(B)

a=2/3*g*sin(25*(pi/180))=>a=2.8507 m/s^2

vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s

So the translational motion of the cylinder is 4.2760 m/s.

4.2760=R*w

w=134.04 rad/s

PE=mgh=>PE=215*9.8*.108=>PE=227.56 J

PE = KE at the end of the roll because of energy conservation.

227.56 = 1/2*I*w^2+1/2*m*v^2

227.56=0.5*I*(134.04)^2+0.5*215*(4.276)^2

I=-0.1931 kg*m^2

The answer is 1.21 *10^4 kg*m2

How is this solved and why is my approach wrong? I think I did all the calculations correctly, so I must have gone wrong in applying the physics.

vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s

So the translational motion of the cylinder is 4.2760 m/s.

4.2760=R*w

w=134.04 rad/s

PE=mgh=>PE=215*9.8*.108=>PE=227.56 J

PE = KE at the end of the roll because of energy conservation.

227.56 = 1/2*I*w^2+1/2*m*v^2

227.56=0.5*I*(134.04)^2+0.5*215*(4.276)^2

I=-0.1931 kg*m^2

The answer is 1.21 *10^4 kg*m2

How is this solved and why is my approach wrong? I think I did all the calculations correctly, so I must have gone wrong in applying the physics.