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OP warned about not using the homework template
Can any one help me on this problem from this video starting from 8:03 , I've been working on it for a week ,and I couldn't find a solution.
Correct. What would happen if he didn'tI think he used his timer to apply force to the ruler
##Mv^2/l=mg\cos x-n\sin x\ ##: in ##\ mv^2\over 2\ ## I see some energy. In ##\ mgl\cos x\over 2\ ## also. But the other one ? Perhaps it's good to set up a list of known/unknown variables and a few more equations -- all of these important constituents of the template !!!Describing normal force from the floor.
Do you have a ruler and a reasonably smooth surface nearby ?Maybe, ruler wouldn't slide , correct?
What is ##nl\sin x\over 2 ## ?can you explain further?
and did you observeYes , I have made that experiment twice.
as correct or as proven wrong ?Maybe, ruler wouldn't slide , correct?
gravity is one componentwhat makes the centripetal force to the center of the mass
How do you now ?It wouldn't slide if he didn't put the timer.
'it' being 'blocked the foot of the ruler from sliding to the left' (while the ruler is falling to the right) ?And it slided when i did it
NoIt wouldn't slide if he didn't put the timer. And it slided when i did it
Yes. So you reproduced Walters result.I pushed the ruler a bit and it started to fall then it's gone away from the block i put behind it , and slided until it lied at the surface
Yes, Walter said it was more complicated. And no, there does not seem to be a worked out solution within google range. We have to think for ourselvesThis problem is more trickier for me , I even tried to search about it but i didn't find anything helpful
In my case the bottom end of the ruler (that is falling to the right) slides to the left, already when theta is about 30 degrees.It slided also but only when it lied at the ground
It certainly can. That's why the block is needed -- for the horizontal componentcan a force which is applied from the pivot point make a radial acceleration?
That's the vertical component and the normal force works outwards, but: yesi.e the normal force from the surface that touches the pivot point
No. The ruler is not just translating but also rotating. So more complicated than the girl on the igloo: there is a rotational energy term too.So Mv^{2}/l=mg cos(x) ?
And does normal force from the block does torque?I don't understand why we need opposite torque