The motorcylist is riding his motorcycle up and down a hill. The question was how fast can he can before he loses touch with the ground? I was told that the weight has to be more than the normal force to keep him on the ground and to determine the maximum speed and we use a zero quantity for the normal force. Why?
2006-11-03 01:02:07 · 4 answers · asked by Beth H 1 in Science & Mathematics ➔ Physics
Since the magnitude of the acceleration is v^2/r, and the radius of the Earth is approximately 4000 miles (21120000 ft), you would have to solve the following equation: v^2/21120000=32
v^2=675840000
v=25997feet per second or
v=4.92 miles per second
2006-11-03 01:15:33 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
The Normal force is what gives something weight. It is, on Earth, the force of gravity acting to accelerate whatever mass an object has, in this case, a motorcycle and a rider. Now you have to have details about the hill, how steep it is, how abrupt the change for positive to negative slope, to calculate how fast he would need to be going to leave the ground. If you imagine your hill to be a near perfect semicircle Centripital accelleration is going to try to help keep him grounded, but if he's going fast enough, momentum will overcome both centripital acceleration and gravity.
2006-11-03 01:31:53 · answer #2 · answered by SteveA8 6 · 0⤊ 0⤋
The ground keeps you from falling by providing a force to counter gravity. As you go round the top of a hill part of the force downwards is going towards bending you round the hill, so that the amount left to push you against the hill is less. As the remaining force pushing you against the hill is less, so the normal force provided by the ground is also less, as it doesn't have to push up so hard to keep you from falling through it. When the all the force of gravity is being used to turn you round the hill then the remaining force downwards will be zero, as will the force upwards from the hill. Any faster and the force of gravity will not be enough to bend you round the hill fast enough, and you will take off.
2006-11-03 01:16:43 · answer #3 · answered by thebadger 1 · 0⤊ 0⤋
as you'd know, centripetal acceleration is the acceleration toward the centre hence if u were to draw a free body diagram u would notice that the weight component is resultant factor contributng for the centripetal acceleration hence the normal subtracted by the weight has to be greater than zero
2006-11-03 01:08:53 · answer #4 · answered by human 2 · 0⤊ 0⤋