Suggestions:- This problem is well-suited for a polar description of kinematics (you are given the r_dot and theta_dot components of velocity).
- When you draw the FBD of the particle, the particle will be in contact with one side of the tube only (either top or bottom, not both). Make an assumption of contact at either top or bottom surface when drawing the FBD.
- Your answer in the end will tell you if your assumption on contact surface was correct. If N > 0, then your assumption was correct. Otherwise, your assumption was incorrect. You might want to view the discussion of this in the solution video for Problem 3/86 on the website. A link to this is found on the blog page.
- You are given that theta_dot is constant. This tells you the value of theta_dot_dot. You do NOT have any given information on r_dot_dot.
- In the end, you will have two equations (from Newton) with two unknowns: N and r_dot_dot.
- After you solve these equations, think about the influence of the direction of motion of the particle on the normal contact force N -- how would your answer for N change if the sign of r_dot is changed?
14 comments:
This question isn't related to this problem, but i just wanted to make sure the assignments due fri are still 3/9 and 3/44?
Yes, that is correct. This hint is for the next homework assignment. I posted this last night while it was fresh on my mind.
what is r_dot? i see theta_dot is 3 rad/sec, we are given v_relative to O and slide, theta=30 degrees and a force of .2 lb for slide.
Your e_R vector should point from O to the particle P along the tube. Therefore, r is the distance between O and P. Since P is moving inward with a relative speed of 4 ft/ sec, then r_dot is - 4 ft/sec (r_dot is the speed of P that one would see if they were riding along on the tube).
I ended up with the wrong answer, and I'm not really sure why. I assumed N acted on the bottom surface and got the equation N = -mgcos(theta) - 2m*r_dot*theta_dot. Is this equation wrong?
ebcox, I had the same equation but 2m*r_dot*theta_dot is positive because the e_theta vector of the acceleration points in the direction of increasing theta.
I understand everything except what is r ?? I dont know how to get it and is the e_theta part of the velocity 0 , is that how you get r but then r is just 0... so Im lost
In terms of the correct signs:
* Be sure to draw your FBD. Write down a set of e_r and e_theta components of Newton's 2nd Law.
* Write down the expression for acceleration in terms of polar components.
Signs will take care of themselves if you do the above correctly.
For the question on the value for "r":
Work through the problem leaving "r" as an unknown. Since we are not given a value for "r", it is likely that you answer will be INDEPENDENT of the value of "r". If you find that you need to know "r" to get the answer, then you are probably doing something wrong. Let me know in that case and email me a scan of your solution.
how many forces does the free body diagram have? will it be easier to solve this problem setting the x and y axis in the e_r and e_theta directions?
The contact surfaces are smooth; therefore, no friction.
The particle is in contact with only one surface; therefore, there is only one normal contact force.
With this normal contact force and the weight force, there are only two forces in the FBD.
Forgot to address the last part of the last comment.
You can choose either a polar description (as was done on the similar problem in the Video Solution) or the moving reference xyz (as was done in lecture).
Both approaches give the same answer, of course. Your choice will depend on with which description you are most comfortable.
the equation i keep ending up with is: -N-mg*sin(60) = .2/32.2 * (-24). this is wrong and i don't understand why
From your equation, it looks like you have assumed that the particle is in contact with the tube on its top surface.
If that is correct, then so is your equation. It also appears to give the correct answer. Why do you think that it is wrong?
I just bought a TI-89 and i don't know how to use it. I got it to work though. thanks.
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