when the body enters the curve we can assume that: mgcos(θ) - N = mu^2/R, where R is the radious of its curved trajectory. Since it doesn't loose contact must be: u0^2 < gRcos(θ). From conservation of energy: u^2 = = u0^2 + 2gR(1 - cos(θ)) and ux = ucos(θ) for the horizontal velocity. Now its...
You are wrong. The vx velocity while the body moves along the curve is larger than v0, (increasing to an amount while going down and decreasing to its original value v0 while going up). So the second body travels the horizontal distance of the gap faster than the first body.
First of all this happens in a normal g-field. This is not homework or coursework. Its a multiple choise question from a national competition of Physics which deals with basic physics laws and i don't like the way it is written. Now the two bodies will not reach at the same time to the wall...
The two bodies shown at the picture are moving along the surfaces marked as A and B with no friction and no rotation and with the same initial velocity. Which one will arrive first at the wall? We may assume that the second body never looses contact with surface B. Note that this question was...
That's right. It has to do with the position of the CM of the body. As far as momentum and energy conservation laws, none of them are violated since this is not an isolated system and since the CM is higher than normal we have gravitational energy converting to kinetic. Thank you.
Hello everybody. I have a rather strange question regarding the rotational motion of a solid object across an incline plane. I wonder if it is possible for a solid object to move upwards the plane without the application of any external force, other than its own weight and the friction with the...