Suggestions for Problem 6/175
Note that this problem is very similar to that of Problem 6/176, a problem for which a video solution is posted. Please review the solution of 6/176 prior to working this problem.
There are two primary changes that you need to make for Problem 6/175:
- The force "P" in 6/176 is to be replaced by -m*g*sin(theta) in the summation of forces in the x-direction, where theta = arctan(0.1) and "x" is along the inclined surface. (The red is a correction made to the original post.)
- For Problem 6/175, the initial velocity of O is +2 m/sec in the x-direction; in Problem 6/176, the initial velocity of O is zero.
All other aspects of the solution are the same, including Step 3 - Kinematics.
Please post any questions as comments.
9 comments:
I am curious how the problem changes since it moves up the incline, stops, and then moves down. Should i just carry out the signs with omega_1 in the negative k and omega_2 in the positive k through out my equations and hope the math works itself out?
why is it theta = sin^-1 (0.1) ? isnt't theta supposed to be sin^-1 (1/root of 101) ?
I think he meant arctan(1/10) not arcsin(1/10)
Sorry...To find theta you use arctan(1/10) and then you use -mgsin(theta) as the force in the x direction. Remember that the coordinate axis is along the incline with +X up the incline.
i got my final equation to be v2= (g*sin(theta)(t)-2) / -1-(k^2/r^2) this doesn't work. anybody know where i went wrong.
i couldn't get the answer too, i dunno why... seems the answer is way off from the answer given in the book
--to mark--
You are correct ... I goofed. The trig is theta = arctan(0.1), NOT arcsin(0.1). (For this small of an angle, the two results actually give very close results).
Sorry for the error.
--to jonbrueck--
You are close to the correct result. You made a small error in algebra right at the end. The answer should read:
v2=-2+ (g*sin(theta)(t)) / (-1-(k^2/r^2))
The "-2" does NOT get divided by -1-(k^2/r^2).
--to Mark (first comment)--
The entire process of the wheel moving up the incline, stopping and then moving back down is governed by the same equation. Just set up the problem using a consistent set of sign conventions, as described in the video solution for 6/176, and the math will take care of itself. No need to fix up anything to account for the reversal of direction in motion.
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