How far does a rolling hoop travel up an incline with known variables?

[therayne]

Sorry, the title is wrong, it's hoop rolling UP an incline

Homework Statement

In a circus performance, a large 3.8 kg hoop
with a radius of 1.8 m rolls without slipping.
If the hoop is given an angular speed of
5.5 rad/s while rolling on the horizontal and is
allowed to roll up a ramp inclined at 17◦ with
the horizontal, how far (measured along the
incline) does the hoop roll? The acceleration
of gravity is 9.81 m/s2 .
Answer in units of m.

Known variables:

M=3.8 kg
R=1.8 m
wi = 5.5 rad/s
wf=0 rad/s
vi= wR
Angle A=17 degrees
h= d*sin(A)
d=?

*im using v and w at the center of mass, same for inertia

Homework Equations

I = MR^2
rotational KE= (1/2)Iw^2 + (1/2)Mv^2

The Attempt at a Solution

KEf + Uf = KEi + Ui
0 + Mgh = (1/2)Iw^2 + (1/2)Mv^2 + 0
Mgh = (1/2)Iw^2 + (1/2)Mv^2

substitute:
h=d*sin(A)
w= v/R
I=MR^2

Mgd*sin(A) = (1/2)(MR^2)(v^2 / R^2) + (1/2)Mv^2
Mgd*sin(A) = (1/2)Mv^2 + (1/2) Mv^2
Mgd*sin(A) = Mv^2

d = v^2 / (g*sin(A))
using v=wR = 9.9

d= 234.17

Which is wrong.

I wasn't sure if i was supposed to include friction, does it even make sense for an object to roll without friction? I'm not sure, and I'm not sure how to even do the problem with friction. that would mean i would have to use torque right?
Thanks for the help.

 

[ehild]

Your derivation is correct, but I do not understand how you get the final result.

sin 17o = 0.292

$$d=\frac{9.9^2}{9.8 \cdot 0.292} =?$$

About friction: yes, static friction is present, the loop could not roll otherwise, but static friction does not do any work (the point where it acts is in rest.)

ehild

 

[therayne]

I ended with
Mgd*sin(A) = Mv^2

I divided both sides by Mgsin(A), so the M's cancel and I'm left wit
d = v^2 / gsin(A)

oops! I made a type when i got d=234.17
it's just d = 34.17 m

AH!
I just checked the website where i submit my answers and 34.17 is right. I thought i had already entered that answer and got it wrong, but i had something else the first time.

Well thanks for your help, i was a little confused about the friction but I think I got it now.

Thanks a million, this website is great, I've been using it for help for the past few weeks but only just registered.