A uniform solid cylindrical log begins rolling without slipping down a ramp that rise

1. Apr 15, 2012

Mdhiggenz

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

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0 above the horizontal. After it has rolled 4.20 m along the ramp, the magnitude of its linear acceleration is closest to

2. Relevant equations

3. The attempt at a solution

What I did was

mgh=1/2mv^2+1/2Iω^2

mgh=1/2mv^2+1/2[1/2mr^2]ω^2

v=rω

mgh=1/2mv^2+1/4mv^2
√4/3gh=v

solved for v and got 7.41m/s

I then used the constant acceleration equations

vx^2=vo+2aΔx

solving for a and got 6.54 which is the incorrect answer.

Where did I go wrong

Thank you!

2. Apr 15, 2012

billVancouver

Re: A uniform solid cylindrical log begins rolling without slipping down a ramp that

Hi Mdhiggenz,

The problem is in your first calculation, where you come up with 7.41m/s. Your manipulation of the algebra looks correct to me (double check!), but at the very last step you seem to have used h=4.20m. The problem says that the log moved 4.20m along the ramp, which means it would have descended 4.20m*sin(28deg), which is the value you should use for h. Does this fix your problem?

Hope this helps,
Bill Mills

Last edited by a moderator: Apr 16, 2012