Troubleshooting Rolling Log Acceleration on a Ramp

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A uniform solid cylindrical log rolls down a ramp inclined at 28 degrees, and the goal is to determine its linear acceleration after rolling 4.20 m. The initial calculations for velocity were based on incorrect height, using the full distance instead of the vertical descent calculated as 4.20 m multiplied by the sine of 28 degrees. This error led to an incorrect acceleration result of 6.54 m/s². Correcting the height in the calculations should yield the accurate acceleration. Adjusting the height to account for the ramp's angle is crucial for solving the problem correctly.
Mdhiggenz
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Homework Statement



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

Homework Equations





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!
 
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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
 
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