Recent content by palaphys

ok that makes sense. but I have a question. IF there was no external torque (hypothetical) then is it valid to conserve momentum as I have said?

I have another question: If angular momentum is conserved, then it would be conserved in all directions. But in the horizontal direction, the initial angular momentum is zero, and the final angular momentum is non zero (## L_{aboutcm}## of the vertical disc). Why is this? Am I right in the first...

Consider the diagram shown above. I came up with two ideas to solve this: i) I do not see any external torques about the vertical axis in this problem. so, angular momentum is conserved about this axis. (is it?) ii)the standard way, that is, writing newton's second law for rotation, and applying...

so are you implying, that the value of F I got using force balance, IS a valid value for some a force to get the block moving? but it's not the maximum value. Did I get you right?

but when velocity is maximum, there is a momentary equilibrium, isn't it? this means forces at that point of time will be balanced. is it not?

why is this? why not the eqb. position?

When the acceleration is maximum, the velocity will be zero, and vice verca. I'm aware of this. But how is it going to help in this case?

Well, force balance leads to the answer: ##F= μg(m_1 +m_2)##using work done (energy balance equation) leads to ##F= μg(m_2 + 2m_1)/2##, which is given as the right answer. So yes, you are right. Force balance is giving a solution which is more than the required force.

When F is large enough to move m1, m1 will start with some acceleration, such that F>kx+friction where x is the elongation in the spring. At one instant, when the mass has maximum velocity, F=kx+ friction , after which the mass m1 will begin to decelerate, i.e its velocity will decrease. This...

The diagram above describes the situation. Say the spring constant is k. The main goal will be to apply the condition that when the spring force exactly balances the limiting friction, m2 will begin to slide. i.e ##kx=μm_2g## I came up with two approaches for this one. i) Assume that m1 is in...

EDIT- yes, I meant to say that it is ##\arccos (\frac 23) ## with the vertical

Yes, this is my understanding Here is my working: yes, I meant to say that it is ##\arccos (\frac 23) ## with the horizontal. thanks for the help

okay, so you are saying that the string can only reach the top slack, with the given speed in the question, that too in a hypothetical case. Whereas a rigid massless rod given the same speed would reach the top always, as tension would prevent it from collapsing at the bottom. (please correct me...

I will try to use this fact to proceed with the problem

not understanding what you mean by compression. is it stress/ tension?