The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.2 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?
To determine the largest value that the radius r can have for the car to remain in contact with the circular track, we need to analyze the forces acting on the car at the highest point of the loop-the-loop.
At the highest point, the car experiences the following forces:
1. Gravitational force (mg) acting downward.
2. Normal force (N) acting perpendicular to the track.
3. Centripetal force (F_c) acting inward toward the center of the circular path.
For the car to remain in contact with the track, the sum of these forces at the highest point must be equal to or greater than zero.
The gravitational force and the normal force can be calculated as follows:
Gravitational force: mg,
Normal force: N = mg.
The centripetal force acting inward is given by:
F_c = m(v^2 / r),
where m is the mass of the car, v is the car's velocity, and r is the radius of the circular track.
By setting the sum of these forces equal to zero, we can solve for the maximum radius r:
mg + mg - m(v^2 / r) ≥ 0.
Combining like terms, we get:
2mg - m(v^2) / r ≥ 0.
Simplifying further:
2mg ≥ m(v^2) / r.
The mass m cancels out from both sides, leaving us with:
2g ≥ v^2 / r.
Rearranging the equation to solve for r, we get:
r ≥ v^2 / (2g),
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values given:
v = 4.2 m/s,
g = 9.8 m/s^2,
We can calculate the largest value of r:
r ≥ (4.2^2) / (2 * 9.8).
Evaluating the equation:
r ≥ 1.764 / 19.6.
Calculating the result:
r ≥ 0.09 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 0.09 meters.