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?
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.
but you don't know hf or ho, how do you you use the conservation of energy equation?
To find the largest value that the radius r can have for the car to remain in contact with the circular track, we need to consider the minimum velocity required for the car to stay on the track at the top of the loop.
To do this, we can use the concept of centripetal force. At the top of the loop, the only forces acting on the car are gravity (mg) and the normal force (N) from the track. The centripetal force required for circular motion is provided by the normal force.
To stay on the track, the centripetal force must be equal to or greater than the gravitational force. Mathematically, this can be represented as:
N ≥ mg
Next, we can equate the gravitational force (mg) to the centripetal force (mv²/r), where m is the mass of the car, v is its velocity, and r is the radius of the circular track.
mg = mv²/r
Now, we can rearrange the equation to solve for the radius r:
r ≥ v²/g
Given that the initial velocity v is 4.5 m/s and the acceleration due to gravity g is approximately 9.8 m/s², we can substitute these values into the equation:
r ≥ (4.5 m/s)² / (9.8 m/s²)
Calculating this, we find:
r ≥ 2.06 m
Therefore, the largest value that the radius r can have for the car to remain in contact with the circular track at all times is approximately 2.06 meters.
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