Any questions?
Note added by CMK in response to comments added by Fady and Brandon:
Fady asks a very good question: why does the block slip outward when there are not forces acting outward?
Brandon added a good comment where he used a "centripetal force" argument. As Brandon describes, the centripetal force is not a true force but is rather a kinematic force constraint.
Click here to see a variation on Brandon's argument based on Newton's laws. Let me know if this helps to better understand this problem.
Again, very good question and comment.
11 comments:
is theta-dot the angular velocity omega?
assuming that my hypothesis is correct.
My solution has the sum of forces in theta direction which becomes zero (theta-double dot=0 and r-dot =zero). Thus, I only have left with two equations; the sum of forces in r direction and k direction. I figured that I can divide these two equations in order to find theta-dot, but my final answer is off by .31 when the cone is moving at its maximum speed.
Does that mean my hypothesis is wrong or i had sum of the forces incorrectly? or there may be something i never have thought of in this problem?
-GC
I made the mistake of assuming theta_dot is equal to omega as well. The omega_min is reasonable, but the omega_max is completely incorrect. i dont have an answer to solve this one yet but im headed off to the help room...
I propose that we solve this using e_n and e_t vectors. e_n points to the axis of rotation and e_t points into the page. k points up.
We know that v is ~constant and therefore v_dot is 0. We also know that v=r*omega. They give r=0.2m in the picture.
This method works. It does involve many cos() and sin(), I'm not sure if the other method is easier.
Personal Note: If you draw your e_n like the e_r shown in the picture above, it doesn't matter because we're looking for omega's magnitude, you'll just get the wrong signs.. like I did.
Yes, doing this in path coordinates (i.e., using the e_t and e_n unit vectors) is a perfectly fine alternative. Be sure to draw the e_n unit vector INWARD (e_n = - e_R); this negative sign avoids the sign problem that j biberstine talks about above.
Good point.
what is the sum of the forces in the k direction equal to? (m * a_k)? then what is a_k?
since the box only moves in the e_n direction, the sum of the forces in the k direction = 0
This is more of a general question.
The only forces acting on the block are weight, friction and normal force. None of these three forces act up the incline, so what causes the block to move up the incline when the dish is rotating at a high speed?
The centripetal force is the external force required to make a body follow a curved path. Hence centripetal force is a kinematic force requirement, not a particular kind of force, like gravity or electromagnetic force. Hope this helps
I was just wondering, to keep the block fixed on the cone, will it experience any friction opposite the direction (-e_t) of its movement, and if so, will it at any point experience any slipping horizontally? I guess since it's not accelerating, there shouldn't be any force, but it seems counterintuitive that it doesn't have friction holding it on.
--to Fady and Brandon--
Please see the notes that I added to the original post (I am not very good with HTML and did not know how to add a link to a PDF file here in the Comments text box.
--to Sheng--
Good question.
When the block does not slip on the cone (R_dot = 0), the component of friction force in the e_theta (or equivalently, in the e_t) direction is zero. To see this, sum forces in the e_theta direction:
f_theta = m*(R*omega_dot + 2*R_dot*omega)
Since R_dot = omega_dot = 0, f_theta = 0.
However, once you exceed the maximum rotation rate for no slip, you will then develop an R_dot > 0. From above, you see that now:
f_theta = 2*m*R_dot*omega > 0
In that case, friction in the theta direction is non-zero.
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