The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of v = 4.7 m s-1, what is the largest value that the radius r can have if the car is to remain in contact with the track at all times?
To find the largest value that the radius r can have for the car to remain in contact with the track at all times, we need to consider the condition for the car's acceleration at the top of the loop to be equal to or greater than the acceleration due to gravity.
Let's start by analyzing the forces acting on the car at the topmost point of the loop. At this point, the car is moving in a circular path with its weight acting downward and the normal force acting perpendicular to the track.
The net force acting on the car at the top of the loop is the difference between these two forces. The net force can be calculated using the following formula:
Net Force = Centripetal Force - Weight
The centripetal force is given by the equation:
Centripetal Force = (mass of the car) × (centripetal acceleration)
The centripetal acceleration can be calculated using the equation:
Centripetal Acceleration = (velocity squared) / (radius of the loop)
Now, the weight of the car is given by:
Weight = (mass of the car) × (acceleration due to gravity)
Since we want the car to remain in contact with the track, the net force acting on the car at the top of the loop should be greater than or equal to zero. Therefore:
Net Force ≥ 0
Substituting the equations above, we have:
(mass of the car) × (centripetal acceleration) - (mass of the car) × (acceleration due to gravity) ≥ 0
Rearranging the equation gives:
(mass of the car) × (centripetal acceleration) ≥ (mass of the car) × (acceleration due to gravity)
Canceling out the mass of the car on both sides of the equation gives:
centripetal acceleration ≥ acceleration due to gravity
Substituting the equation for centripetal acceleration, we have:
(velocity squared) / (radius of the loop) ≥ acceleration due to gravity
Now, we can plug in the given values. The initial speed of the car, v, is 4.7 m/s, and the acceleration due to gravity, g, is approximately 9.8 m/s^2.
Substituting the values, we get:
(4.7^2) / r ≥ 9.8
Simplifying further:
22.09 / r ≥ 9.8
To find the largest value for the radius, we want the right side of the equation to be as small as possible. So, let's assume the right side is equal to zero:
22.09 / r = 9.8
Now, solve for r:
r = 22.09 / 9.8
Calculating this, we find:
r ≈ 2.25 meters
Therefore, the largest value that the radius r can have for the car to remain in contact with the track at all times is approximately 2.25 meters.