Recent content by mrdoe

Thanks, it makes sense now. Intuitively the initial tug of gravity on mass m2 would stretch the spring and move m1 at the same time, but I guess we live in a world full of friction so that isn't really accurate.

This is a problem I came up with myself, so it's not homework and so I didn't post in that forum. Suppose a mass m1 rests on a frictionless table. m1 is directly connected to a massless ideal spring at equilibrium length of spring constant k which is connected to a string going over a pulley...

Well \frac{v^2}{r} = \frac{v^2}{10} > 5g\cos\theta. Sorry Then later on we have \frac{v^2}{10} = \frac{490.5(1+\cos\theta)}{25} and 1+\cos\theta>2.5\cos\theta so 1.5\cos\theta < 1 so \cos\theta < 2/3 and we have \theta = 48.19 degrees.

OK I'll do it the way I think it should be done. Correct me if there are mistakes. We have mg\cos\theta as the centripetal force. At some velocity v, \frac{v^2}{20} > 5g\cos\theta. Then \frac{v^2}{100} > g\cos\theta and using KE/PE, we know that setting the base of the ball to be PE = 0, we...

The normal force goes away when the block loses contact. By that time, the only acceleration is due to gravity. Am I correct? However, assume T is the time at which the block loses contact. Then at T-e where e is infinitely small, the block still has the normal force. So it wouldn't have...

The real problem is that the normal force counteracts any centripetal force.

Are you talking about the gravity between the sphere and the block? I honestly don't think that's a force to consider in this problem.

There are two forces acting on a block on an inclined plane (the plane tangent to the sphere at the tangent point of the block to the sphere): gravity and normal. The net force is down the inclined plane, the magnitude of this is mg sin(theta) where theta is the inclination of the plane...

Bump, I really need help on this one.

Wait: there is no centripetal force. For angle of theta with respect to vertical, there's only a force of mg\sin\theta, which is parallel to the inclined plane (or the sphere's tangent). Therefore, the block should fly off the sphere at 0 degrees. What am I getting wrong here..

The centripetal acceleration has to be v^2/r and thus force must be v^2/20...

The centripetal net force is not large enough to keep the block in "orbit"?

Is it just that at some angle theta, the velocity is no longer in balance with the pseudocentripetal force and the ball therefore leaves the surface?

The problem is that the pseudo-"centripetal" force is entirely canceled out by the normal force. Am I right? So it has something to do with a circle/centripetal pseudoforce but I'm not getting it.

Yes it's circular in shape. So you mean I should consider the component of the forces acting on the block perpendicular to the tangent?