Recent content by Epiclightning

Ok yes that was the error. I solved it taking torque about the center of mass and got the correct value of alpha. Then I got the right answer(N = 85mg/118). Thanks for the help

Oh, I think i found a possible error - am I not allowed to apply torque = I*alpha about that point as it is a noninertial frame which is not the instantaneous axis of rotation or the centre of mass

FBD attached

But that would mean the normal is the same as before(no slipping) as Vertical acceleration of lowest pt= Accn.(CM) - component of angular acceleration = (Mg-N)/M - (Rw)(2/√5) =0 This gives N = 17mg/20 as nothing has changed from when there was no slipping. So what am I doing wrong?

The vertical acceleration of the lowermost point should be 0, then at this instant?

Homework Statement A hemispherical bowl of mass m and radius R is placed on a rough horizontal surface. Initially the centre C of the bowl is vertically above the point of contact with the ground(see figure). Now the bowl is released from rest. Find the normal force acting on the hemisphere...

Why don't you try to find the relationship between the numbers whose digits have been increased by one and the originals ( for both 2 and 4 digits?) Take specific examples to help if you want to.

Thanks for the help :)

That makes a lot more sense, thank you. I found the net capacitance as bε((K-1)x +l) /d If I assume the dielectric is at x distance within the capacitor. I found U = 1/2 CV^2 and evaluated -dU/DX to get the force on the capacitor at that instant. Could you explain how the force was suddenly...

Oh, alright. But I don't think I understand. I get that q= CV is the same for each, but if all the charges are zero, then there is no potential change at the capacitors. The answer is given as -10.3 volts

but the voltages are 12 volts each, so how can the voltage on the left be double of the last?

Homework Statement Consider the situation shown in the figure. The width of each plate is b. the capacitor plates are rigidly clamped in the lab and connected to a battery of emf E. All surfaces are frictionless. Calculate the value of M for which the dielectric will stay in equilibrium...

yes, i think: For (b): VA +12 - q/4 -12 = VB VA - VB = q/4 Also VA - VB = -q/2 if I go the other way but this gives q = 0... The answer is given as difference in potential = -8 volts

okay but i will have to assume charges to use kirchhoffs laws. how can the charges exist if the batteries don't give a current...unless the capacitors are charged already? To clarify, the circuit is not physically possible, right? what exactly is the purpose of the extra batteries?

so the capacitors are charged already? how can there be potential difference without current?