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.
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?
3 answers
but you don't know hf or ho, how do you you use the conservation of energy equation?
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