A roller coaster ride includes a circular loop with radius R = 13.3 m. What minimum speed must the car have at the top to stay in contact with the tracks?
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Basically, the centripetal acceleration has to offset gravity, 9.8 m/s².
Since the centripetal acceleration is given by v²/r, so we equate the two quantities to get
v²/r = 9.8
v=sqrt(9.8*r) m/s
= sqrt(9.8*13.3) m/s
= 11.4 m/s.
Well, to stay in contact with the tracks, the car needs to have enough speed so that gravity won't pull it off the track. Let me calculate that for you.
Fetching my handy-dandy calculator... Beep boop beep...
Okay, according to my calculations, you'd need a minimum speed of at least 20.7 meters per second at the top of the loop to defy gravity and stay in contact with the tracks.
So hold on tight, and make sure to bring some extra laughter for the ride!
To determine the minimum speed the car must have at the top of the circular loop to stay in contact with the tracks, we can use the concept of centripetal force.
The centripetal force is given by the equation:
F = m * a
Where F is the centripetal force, m is the mass of the car, and a is the centripetal acceleration.
In this case, the centripetal force required to keep the car in contact with the tracks is the gravitational force acting on the car at the top of the loop.
F = m * g
Where g is the acceleration due to gravity.
To find the centripetal acceleration, we can use the equation:
a = (v^2) / R
Where v is the velocity of the car at the top of the loop, and R is the radius of the loop.
Since we want to find the minimum velocity, we can assume that the car is just about to lose contact with the tracks. Therefore, the gravitational force is equal to the centripetal force.
m * g = m * [(v^2) / R]
Simplifying the equation, we can cancel out the mass:
g = (v^2) / R
Rearranging the equation to solve for v^2, we get:
v^2 = g * R
Now we can substitute the known values:
g = 9.8 m/s^2 (acceleration due to gravity)
R = 13.3 m (radius of the loop)
v^2 = 9.8 * 13.3
v^2 = 130.34
Taking the square root of both sides of the equation, we get:
v ≈ 11.41 m/s
Therefore, the minimum speed the car must have at the top of the circular loop to stay in contact with the tracks is approximately 11.41 m/s.
To determine the minimum speed the car must have at the top of the circular loop, we need to consider the forces acting on the car. At the top of the loop, the car is subjected to two forces: gravity (mg) and the normal force (N). The net force acting towards the center of the loop is given by:
net force = N - mg
For the car to stay in contact with the tracks, the net force must be greater than or equal to zero. This means that the normal force must be greater than or equal to the force due to gravity:
N - mg ≥ 0
The normal force can be calculated as:
N = m * v² / R
where m is the mass of the car, v is the speed of the car at the top of the loop, and R is the radius of the loop.
Substituting N in the inequality above, we have:
m * v² / R - mg ≥ 0
Simplifying the inequality:
v² / R ≥ g
Solving for v, we find:
v ≥ √(g * R)
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Now, we can substitute the given value of R:
v ≥ √(9.8 * 13.3)
v ≥ √(129.54)
v ≥ 11.38 m/s
Therefore, the minimum speed the car must have at the top of the loop to stay in contact with the tracks is approximately 11.38 m/s.