Recent content by nerdvana101

First of all, thank u both for pondering over this. And using haruspex's idea of taking angular momentum of the ball at the point of contact, I have got the correct answer. Thanks a lot man!

Hi, I have another query related to rolling motion. Lets say a disc is rolling purely with a velocity vo (this being the vcom). It encounters a step.lets say that the step has a height h, where h < r, i.e., the step has a height less than that of the height of the center of the disc. SO...

thx, just worked it out. got the right answer. Anyway, I was thinking that if the wall were shorter than the sphere itself? let's say that the wall has a height h, where h < r, i.e., the wall has a height less than that of the height of the center of the sphere. SO, only a corner of that wall...

So, taking the torque about the COM and then finding the angular acceleration, what happens to the linear velocity? I think the friction force will act in it's opposite direction and stop it from sliding. so that a = f/m. Then, I cud equate it as - v-at = r(ω - αt) and get the time?

Hi, I have a problem which I can't figure out.:confused: Let us take a sphere which is rolling purely at a constant velocity vo. Now, if the sphere were to collide inelastically with a wall, with coeff. of restitution = e. Then what is the time after which the sphere starts pure rolling...

isn't there any option to select the best answer?

Thx tiny-tim. What I forgot to add was the basics of angular momentum - L at the point of contact = Icomxω + rcomxmxvcom. THX a lot!

HI, I am having some trouble understanding the effects of an impulse on rolling motion, particularly without slipping. For eg, if we take a sphere of radius R and mass M, then what is the point h above the centre C at which when an impulse is imparted, will cause pure rolling motion...