A block is at rest at the top of a frictionless hemisphere of radius r. It is slightly disturbed at starts sliding down. I already know where it will leave the surface (height = 2r/3). My question is, WHEN will it leave the surface?
The block slides down the hemisphere due to the force of gravity acting on it. As the block moves towards the center of the hemisphere, its potential energy is converted into kinetic energy, causing it to speed up.
The mass of the block does not affect the time it takes to leave the surface of the hemisphere. The time taken is only dependent on the height of the hemisphere and the acceleration due to gravity.
Yes, the angle of the hemisphere does affect the time taken for the block to leave the surface. A steeper angle will result in a shorter time as the block will have a shorter distance to travel to reach the bottom.
The surface of the hemisphere affects the block's movement by providing friction. A rough surface will cause the block to slow down, while a smooth surface will allow the block to slide down faster.
Yes, the time taken for the block to leave the surface can be calculated using the formula t = √(2h/g), where t is the time, h is the height of the hemisphere, and g is the acceleration due to gravity.