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Note that is problem is suited for a POLAR kinematics description. Since e_r always points perpendicularly outward from the rotation axis, it is as shown in the figure above (NOT parallel to the edge of the cone on which the block rests).
For the MINIMUM rotation rate, the block will have an impending slipping direction of DOWN the inside of the cone (therefore, the friction force is up the incline).
For the MAXIMUM rotation rate, the block will have an impending slipping direction of UP the inside of the cone (therefore, the friction force is down the incline).
You will need to work the problem twice: once of the minimum rotation rate and again for the maximum rotation rate.
10 comments:
i've used all different coordinate planes and i keep getting the same wrong answer. i was wondering if you could set the y-axis perpendicular to the cone to solve for the normal force. making the normal = mgcos30
you have to consider a_r as a vector so one part of the a_r (i would say y direction) should be added to N and the other one should be added (in x direction) with friction to find w. you will get close answer for minimum part without plugging in a_r part in the equation, but for maximum, it will be totally different.
i forgot to add, after adding y component of a_r, to gcos 30, you can find the friction.
Is the acceleration in the e_theta direction 0? does the FBD for the block just have 3 forces: normal, friction, and weight? How do you solve for the normal force here?
In the homework hint it says you have to work this problem twice. I found it to be alot easier to just specify my friction as +-f. It gave me both answers and took half the time!
Good idea. Thanks for pointing out this.
if we're solving for N, there is a mass component in there. How and when do we cancel that out with the other unknown m's ?
Good question that jon brueck has here. Anyone have an answer or a suggestion for him?
Every term of your equations(force and energy), have mass in them. If it doesn't, then there is probably a mistake somewhere. So, to answer your question, I would leave the mass term in the equations as "m" and then when you have all the equations combined, you'll see that they will cancel.
My suggestion would be to choose one set of axes that is natural for the kinematics and stick with it all the way through. Less confusing that way.
I have added FBD's in the figure for the original post. Say we are looking first for the minimum speed (using FBD on the far right). From this we get two force component equations:
e_r: f*cos(30°) - N*sin(30°) = m*a_r
k: f*sin(30°) + N*cos(30°) - m*g = m*a_z
The kinematics are:
a_r = r_dot_dot - r*omega^2 = -r*omega^2
a_z = 0 (moves in circle with constant height z)
Use these kinematics along with f = mu_s*N (impending slip) to solve the force equations for the two unknowns of N and omega.
As kul points out, N is NOT equal to m*g*cos(30°) (this is not statics!).
Repeat for the maximum speed case using the other FBD (with f pointing DOWN the inside surface of the cone).
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