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The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.5 m/s, what i...Asked by Jordan
The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.5 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?
Answers
Answered by
drwls
There is no drawing.
Require that the centripetal force at the top of the loop be equal to M g. Solve for the r that satisfies that requirement, with V determined by conservation of energy.
Require that the centripetal force at the top of the loop be equal to M g. Solve for the r that satisfies that requirement, with V determined by conservation of energy.
Answered by
jamie
but you don't know hf or ho, how do you you use the conservation of energy equation?
Answered by
JJ
This is what I found....
When the car is at the top of the track the centripetal
force consists of the full weight of the car.
mv2/r = mg
Applying the conservation of energy between the bottom and the top of the track gives
(1/2)mv^2 + mg(2r) = (1/2)mv0^2
Using both of the above equations
v0^2 = 5gr
so
r = v0^2/(5g) = (4.5 m/s)^2/(49.0m/s^2) =
Hope that helps
When the car is at the top of the track the centripetal
force consists of the full weight of the car.
mv2/r = mg
Applying the conservation of energy between the bottom and the top of the track gives
(1/2)mv^2 + mg(2r) = (1/2)mv0^2
Using both of the above equations
v0^2 = 5gr
so
r = v0^2/(5g) = (4.5 m/s)^2/(49.0m/s^2) =
Hope that helps
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