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?
The largest value that the radius r can have is determined by the centripetal force, which is equal to the mass of the car multiplied by its velocity squared, divided by the radius. Therefore, the largest value for the radius r is equal to the mass of the car multiplied by its velocity squared, divided by the centripetal force. In this case, the mass of the car is 1 kg, the velocity is 4.5 m/s, and the centripetal force is 9.8 N. Therefore, the largest value for the radius r is 1 kg x (4.5 m/s)^2 / 9.8 N, which is equal to 0.23 m.
To determine the largest value of the radius 'r' for the car to remain in contact with the circular track at all times during the loop-the-loop trick, we need to consider the minimum speed required for the car to maintain contact with the track at the topmost point of the loop.
At the topmost point of the loop, the centripetal force should be equal to the gravitational force acting on the car.
The gravitational force can be calculated using the formula:
Fg = m * g
Where:
Fg is the gravitational force
m is the mass of the car
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Now, the centripetal force at the topmost point can be calculated using the formula:
Fc = (m * v^2) / r
Where:
Fc is the centripetal force
m is the mass of the car
v is the speed of the car at the topmost point of the loop
r is the radius of the circular track
Since the centripetal force should be equal to the gravitational force, we can equate the two equations:
(m * v^2) / r = m * g
Simplifying the equation, we have:
v^2 = r * g
Now, substituting the given values:
v = 4.5 m/s (initial speed)
g = 9.8 m/s^2 (acceleration due to gravity)
(4.5 m/s)^2 = r * 9.8 m/s^2
Solving for 'r', we have:
r = (4.5 m/s)^2 / 9.8 m/s^2
r ≈ 2.07 meters
Therefore, the largest value that the radius 'r' can have for the car to remain in contact with the circular track is approximately 2.07 meters.
To find the largest radius that the car can have and still remain in contact with the circular track, we need to consider the forces acting on the car at the highest point of the loop-the-loop.
At the highest point of the loop-the-loop, the car is momentarily at rest for an instant, which means that the net force acting on it must be directed towards the center of the circle.
The two forces acting on the car at the highest point are the weight (mg) of the car acting downwards, and the normal force (N) acting upwards. The net force can be calculated as:
net force = N - mg
For the car to remain in contact with the circular track at all times, the net force must be greater than or equal to zero. This condition can be expressed as:
N - mg ≥ 0
Simplifying this expression, we get:
N ≥ mg
Since the normal force N is the force supplying the centripetal force, we can relate it to the car's velocity and the radius of the loop-the-loop using the following equation:
N = m * v^2 / r
where m is the mass of the car, v is its velocity, and r is the radius of the circular track.
Combining the above two equations, we get:
m * v^2 / r ≥ m * g
Now, we can plug in the given values:
m = mass of the car (not provided)
v = initial speed of the car = 4.5 m/s
g = acceleration due to gravity ≈ 9.8 m/s^2
Therefore, the largest value that the radius (r) can have for the car to remain in contact with the circular track at all times can be found by rearranging the above inequality:
r ≤ (m * v^2) / (m * g)
Since the mass of the car is not provided, we cannot determine the numerical value of the largest radius (r).