Recent content by kekpillangok

Can we explain this from the fact that, at each instant, there are acceleration vectors of equal magnitude along the directions of the velocities relative to plate 1 and plate 2?

The two forces on the block are ##\vec{f }_{1 } ## and ##\vec{f }_{2 } ##. The net force in the x-direction, therefore, is ## f _{1 }\cos \left( \theta _{1 }\right) \, -f _{2 }\cos \left( \theta _{2 }\right) \, ##. So, by Newton's second law, ## f _{1 }\cos \left( \theta _{1 }\right) \, -f...

I'm a bit puzzled. If the particle starts from rest, it seems to me that, at each instant, the acceleration and velocity vectors should be aligned. I'm not sure if this is right. At the very first instant, however, this is certainly the case: both acceleration and velocity make an angle of 45°...

Thanks for replying! I found the following expressions for the derivatives: ##\frac{\, \mathrm{d } \, v _{1 } \, }{\, \mathrm{d } \, t \, }=\frac{f }{m }{\left( \cos \left( \theta _{1 }\right) \, -\cos \left( \theta _{2 }\right) \, \right) } ## ##\frac{\, \mathrm{d } \, v _{2 } \, }{\...

Here's what I've tried. ##\vec{v } ## is the velocity of the brick after it has stopped changing. I've been able to conclude that this must happen when the relative velocity vectors ## \vec{v }-\vec{u }_{1 }## and ## \vec{v }-\vec{u }_{2 }## lie along the line connecting the tip of ##\vec{u...

I'm having a hard time understanding the conditions under which the string will become slack in this problem. Maybe I'm just bad at visualising the situation and playing the film in my head, but the only situation I could imagine, at first, where I was 100% sure the string would become slack is...

I can't understand what is going on here. My intuition tells me that the traction pulling on the block should be equal to F. Apparently, however, the problem would have me believe that this is not the case, as it gives me a 30° angle at which the force acts and a 6-8-10 triangle as ways to...

The first image shows the rings and the absorbing disc from the front; the second shows them from the side. I am told that, to solve this problem, I have to imagine a light source located at ##r _{1 } ##. The light passing through the "open" rings (the ones that let light through, located at...

Here's what I've tried. First of all, I assume that q is positive. For particle A, then, I can write $$q E -k {\left( x _{A }-x _{B }\right) }=m \ddot{x }_{A }, $$ where ##x _{A } ## and ##x _{B } ## are the coordinates of the particles relative to their equilibrium positions from the point of...

I was taught to solve this problem by first finding the velocity of the body (of mass ##m ##) relative to the block of mass ##M ##. One way of doing this is as follows: first write $$ {v _{m _{B }}}^{2 }={v _{mx _{B }}}^{2 }+{v _{my }}^{2 } (I)...

That would be ##R =R _{0 }/{\left( 1 -\frac{m {\omega }^{2 }}{k }\right) } ##, where ##R _{0 } ## is the natural length of the spring. I found this value by setting ## k x =m {\omega }^{2 }{\left( R _{0 }+x \right) }##. ##x ## here is measured from the natural length. This results in ## x...

Thanks for all the answers, guys! Ok, here's what I've tried. To find the equilibrium position, I set ##m \omega {\left( R +x \right) }-k x =0 ## and solved for ##x ##, which gave me ##x =-\frac{m {\omega }^{2 }R }{m {\omega }^{2 }-k } ##. This, as I understand it, is the amount by which the...

Thanks a lot! I've thought about the signs and came up with the following: $$-k x +m {\omega }^{2 }x =m \ddot{x }. $$ Now, when the mass is within the circle of radius ##R ##, ##x < 0 ## and the elastic and centrifugal forces combine in the outward direction; and when the mass is outside the...

Thanks for your reply! From the rotating frame of reference, there is the centrifugal force and the spring force. The spring force points now inwards, now outwards, whereas the centrifugal force always points outwards, with its magnitude changing with the mass's radial position according to $$F...

If I understand the problem correctly, I need to find the angular frequency of the mass's oscillations about the radius R, which, I think, should be the length of the spring when the mass is merely rotating with angular speed ω (and not oscillating along the radial direction). I was able to find...