Recent content by Galileo

Consider the (physics) situation of a sphere of radius R rolling without slipping on a plane. The configuration of the sphere is given by the cooordinates of the center of the sphere and two angles for its orientation. When the sphere rolls over some given (closed) path in the plane, then its...

Sheesh, I just read right over that part, even though it was mentioned explicitly.

In any real situation where the rope is coiled up, energy will indeed be lost. But in the way the problem is posed, the assumption that the rope is coiled is not used at all. The model more appropriately applies to a flexible, nonstretchable, rope that lies stretched out initially. Energy would...

The point is that y(x) is a function that minimizes (or extremizes) the integral. So in y(x)+{\alpha}{\eta}(x)[/itex], the y(x) is the sought after solution and {\alpha}{\eta}(x) is an arbitrary deviation from that solution. Therefore, the minimum occurs at [tex]{\alpha}=0

Generally, if a rigid body is rotating about an arbitrary axis, the angular momentum need not point in the same direction as the rotation axis, as it does when \vec L = I \vec \omega (for rotation about a principal axis). The relation between \vec L and \omega is still linear, and I is...

What you have calculated is the velocity for which the kinetic energy is twice the initial potential energy. It's not that useful or related to your problem. You already mentioned Fc=mv^2/r. How does this play a role in your problem? What supplies the centipetal force and what is the condition...

The power behind conservation of energy is that you can use the equation for E at any point in the objects trajectory and you will always get the same answer. You haven't been given the mass of the object, but the answer will be independent of the objects mass anyway. You have two special...

They look ok. For the second one you could have used the perpendicular axis theorem straight from the answer in a). Answer is ok though

The easiest way would be to use conservation of energy: E = 1/2mv^2+mgh Alternatively, you could use the equation of motion for the object. You know it has a constant acceleration g downwards.

It just seems like a change of variable, but beware your notation. In the first integral you use v both as the integration variable and in the upper limit of integration. In the second, x plays both those roles too. This is confusing. I'd call the integration variable v' in the first one, then...

Here's an argument coming from a physicist. For general polyhedra, there doesn't seem to be any severe restriction on the degrees of freedom required for their description (unless they are sufficiently regular). I.e. for polyhedron with 30 edges. Any of those edges can be lengthened or...

The total momentum is conserved. All the individual momenta are, generally, not or collisions wouldn't change anything!

Yes, that's correct.

The crucial thing to realize is that the pair forces in Newton's third law act on different bodies, unlike in the second law which relates the net force and acceleration of a single body. So the cart will move simple because it is pushed. The man will be pushed back the other way if he doesn't...

The path a freely falling object takes is a straight(est possible) line in spacetime, so you need to think of time in your space as having a coordinate axis too. The easiest way to picture this is with a worldline in a spacetime diagram (usually with time on the vertical axis). The worldline of...