Recent content by ShaunPereira

Ok. Thank You.

Right so if I understood you correctly then what you essentially mean is: The sign depends on who is doing the work IF it is by an external agent who lifts the brick, both the the force and the displacement are in the same direction and hence the work done by it is positive and the work done by...

I encountered a problem regarding the appropriate sign needed to be taken for the work done on a dipole when it rotates in a uniform electric field and would appreciate some help. The torque on a dipole can be defined as τ=PEsinθ The work done on a dipole to move it from an angle ##\theta_0##...

Ok thanks.

Yes. and the mass here is proportional to ##R^3## which when we put in our equation gives us the answer. The inverse square law takes into account both mass and distances. Since both change we have to take both into account Right?

Right. So does that mean that the force between the two spheres obeys the inverse square law when the masses are fixed and the distances vary. However when the sizes of the two spheres vary keeping density constant the force is no more proportional to ##R^2## but rather proportional to ##R^4##...

I recently encountered this problem on a test where the solution for the above problem was given as follows: $$F= \frac{Gm_1m_2} {r^2} $$ (1) but $$ m=\frac{4}{3}\pi R^3 $$ substituting in equation (1) $$F= \frac{{G(\frac{4}{3}\pi R^3\rho})^2 }{2R^2} $$ where r=radii of the two spheres m=mass...

Thank you!

This is exactly what I was trying to imply. So the force acting on a planet which is gravity, being an inverse square force has a required tangential velocity for a circular orbit to be formed. Anything less or more would lead to an elliptical orbit. However the force acting on a small ball...

Right, so if I have understood you correctly there is no way to conserve angular momentum and also retain the original shape of the orbit. In order to do so we would first have to accelerate the Earth thus creating an elliptical orbit and then wait till we reach apoapsis and burn prograde until...

Well what I wanted was an explanation as to why the mathematics(equations) turn out the way they did. To change a planets orbit we would indeed have to change the angular momentum and as you pointed out it would have to be done twice to achieve the desired orbit. However I have encountered...

The classic way to go about this problem would be to use Kepler's laws and thus find the new time period of earth. However I encountered this question in a test on rotational motion which deals with conservation of angular momentum. The equation used here would be I1ω1= I2ω2 Replacing I with MR2...

Thank you.

Oh! So do we have to find the force at any instant, which in this case must also be the maximum value ? The force due to elemental mass dm would be 2λgx. But since we need the maximum value of force it would be 2λgL which means that the instant an elemental mass dm which was originally at height...

Yes. But that would solve only part of the problem. I have stated in the previous post that the force exerted due to change in momentum of elemental mass is 2λgx But this is only the force on a small mass dm and not on the entire chain How do I find the total force exerted on the chain due to...