# Motorcycle Rotational problems

1. Mar 7, 2009

### hellothere123

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

1) A 320 kg motorcycle includes two wheels, each of which is 52 cm in diameter and has rotational inertia 2.1 kg·m2. The cycle and its 73 kg rider are coasting at 82 km/h on a flat road when they encounter a hill. If the cycle rolls up the hill with no applied power and no significant internal friction, what vertical height will it reach?

3. The attempt at a solution

for 1) I did the total kinetic energy(translational and rotational) to equal mgh, and that didnt work am i missing something?
so i had .5(393)(82*(1000/3600))^2 + .5(2.1)(82*(1000/3600)/.26)^2 = mgh

Any help would be greatly appreciated, i would like to learn how to do this..

Last edited: Mar 7, 2009
2. Mar 7, 2009

### Staff: Mentor

Hint: You are given the rotational inertia of each wheel. (Your basic approach is fine.)

3. Mar 7, 2009

### hellothere123

so i thought i would multiply everything by 2 for each wheel, but still get it wrong.. am i doing anything else wrong?

Last edited: Mar 7, 2009
4. Mar 7, 2009

### Staff: Mentor

If you show your corrected equation we can check. What are you using for m?

5. Mar 7, 2009

### hellothere123

i will change the numbers slightly to match a different problem that i have the solution for so i can see where i went wrong.
not much difference: A 320 kg motorcycle includes two wheels, each of which is 52 cm in diameter and has rotational inertia 2.1 kg·m2. The cycle and its 75 kg rider are coasting at 85 km/h on a flat road when they encounter a hill. If the cycle rolls up the hill with no applied power and no significant internal friction, what vertical height will it reach?

from .5mv^2 + Iw^2 = mgh

with numbers:
.5(395)(85*(1000/3600))^2 + 2.1([85*(1000/3600)]/.26)^2 = (395)(9.8)(h+.26)

6. Mar 7, 2009

### Staff: Mentor

The only part I would question is where you put "h + .26". Just use "h", which will give you the height increase of the motorcycle. (I suspect that's all they want.)

7. Mar 7, 2009

### hellothere123

yea.. you are right. but to my understanding.. isnt it to the height of the center of mass? so i would have thought i should include the .26 but i guess not.. thanks.