A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.What is the ratio of the car's apparent weight to its true weight?
2006-10-14 18:22:34 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
At critical speed
F = W = mg = m(v)^2/r,
and the apparent weight of the car is 0
At twice critical speed, then,
F = m(2v)^2)/r, and
F-W = m(2v)^2)/r - m(v)^2/r = 2^2 - 1
2006-10-14 18:51:35 · answer #1 · answered by Helmut 7 · 2⤊ 0⤋
1 to 1
2006-10-14 18:37:40 · answer #2 · answered by charley128 5 · 0⤊ 1⤋
Ac = V^2 / r
You are doubling the velocity, so the centripetal acceleration with increase by a factor of 4. This is also the factor that the car's apparent weight will change by, since weight is the gravitational force plus the force due to centripetal acceleration.
2006-10-14 18:35:58 · answer #3 · answered by resurrection_of_t_o 2 · 0⤊ 0⤋
I am noy sure what you mean by "critical speed", but if it what I imagine it is the ratio would be -1. If "critical speed" is the speed to balance the gravitational force exactly, twice that would give the force of gravity upward.
2006-10-14 18:31:04 · answer #4 · answered by sofarsogood 5 · 0⤊ 0⤋
Rotary action is a particular case of round action in which the axis of rotation lies in the body. for this reason on your case, at the same time as the disc reports a rotary action, for the fly sitting on it, its a round action (until eventually it is sitting accurate on the middle).
2016-12-04 20:30:21 · answer #5 · answered by Anonymous · 0⤊ 0⤋
half the centrifical force
2006-10-14 18:24:19 · answer #6 · answered by gussie r 3 · 0⤊ 0⤋
lolly d why don't you just listen in class instead of expecting us to help you cheat.
2006-10-14 18:24:02 · answer #7 · answered by Spadesboffin 3 · 1⤊ 3⤋