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Suggestions:
4 comments:
i dont really understand teh kinematic steps we need to take to get tehe solution on this problem, anyone with any advice it would be greatly appreciated.
Recall that you do not know at the beginning if the cylinder rolls without slipping or slips as it rolls down the incline.
If NO SLIP:
In this case the contact point C does not slip as the cylinder rolls. This says that aC = aC*j (j is the unit vector perpendicular to the incline). Also, you know that G moves only along x, or aG = aG*i. From this, you can write:
aG = aC + alpha x rG/C - omega^2*rG/C
OR
aG*i = aC*j + (alpha*k) x (R*j) - omega^2*(R*j)
Using the x-components:
aG = - alpha*R
If the cylinder SLIPS:
In this case, there are no usable kinematics from the cylinder (aG is NOT equal to -R*alpha). However, you use f = muk*N to solve.
Good question.
Does this help?
so assuming no slip is it correct to have N -mgcos = 0 since Agy is 0?
Nick i think that u can write N-mgcos40=0 anyway because the disk is only sliding on the ramp supposedly.
If i m wrong please someone correct me.
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