A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers (Fig. P7.26). (a) If the vehicle has a speed of 20.0 m/s at point A , what is the force of the track on the vehicle at this point? (b) What is the maximum speed the vehicle can have at point B in order for gravity to hold it on the track?
(a) The force of the track on the vehicle at point A is equal to the mass of the vehicle multiplied by its acceleration. Since the vehicle has a speed of 20.0 m/s at point A, its acceleration is zero. Therefore, the force of the track on the vehicle at point A is zero.
(b) The maximum speed the vehicle can have at point B in order for gravity to hold it on the track is equal to the square root of the acceleration due to gravity multiplied by the distance between points A and B. Assuming the distance between points A and B is 10 m, the maximum speed the vehicle can have at point B is 14.1 m/s.
(a) To calculate the force of the track on the vehicle at point A, we can use the equation:
Force = Mass * Acceleration
At point A, the vehicle is moving with a constant speed of 20.0 m/s, which means its acceleration is zero. Therefore, the force of the track on the vehicle at point A is:
Force = Mass * Acceleration
Force = 500 kg * 0 m/s²
Force = 0 Newtons
So, the force of the track on the vehicle at point A is 0 Newtons.
(b) To determine the maximum speed the vehicle can have at point B in order for gravity to hold it on the track, we need to consider the centripetal force required to keep the vehicle moving in a circle at that point. At the topmost point of the roller coaster (point B), the gravitational force acting on the vehicle must provide the necessary centripetal force to keep it on the track.
The centripetal force is given by the equation:
Centripetal Force = Mass * Velocity^2 / Radius
Since the vehicle is moving in a circle, the net force acting on it must be the centripetal force. At the topmost point, the net force is the difference between the gravitational force and the force of tension (exerted by the track), which acts in the opposite direction to gravity. So we have:
Net Force = Gravitational Force - Force of Tension
At the maximum speed, the force of tension is zero, as the track is barely able to hold the vehicle on the track. Therefore, the net force is equal to the gravitational force:
Net Force = Gravitational Force
In this case, the gravitational force is the weight of the vehicle, given by:
Gravitational Force = Mass * Gravitational Acceleration
Gravitational Acceleration is approximately equal to 9.8 m/s^2.
Now, we can equate the centripetal force and the gravitational force:
Centripetal Force = Gravitational Force
Mass * Velocity^2 / Radius = Mass * Gravitational Acceleration
Canceling out the mass on both sides of the equation, we can solve for the maximum velocity:
Velocity^2 / Radius = Gravitational Acceleration
Velocity^2 = Gravitational Acceleration * Radius
Taking the square root of both sides:
Velocity = √ (Gravitational Acceleration * Radius)
Given the radius of the roller coaster track, you can substitute it into the equation to find the maximum velocity the vehicle can have at point B in order for gravity to hold it on the track.
To solve this problem, we need to understand the physics principles involved.
(a) To find the force of the track on the vehicle at point A, we need to consider the forces acting on the vehicle. At this point, there are two forces: the force of gravity pulling the vehicle downwards and the normal force exerted by the track pushing the vehicle upwards.
The force of gravity is given by the equation: F_gravity = m * g
where m is the mass of the vehicle and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force exerted by the track is equal in magnitude but opposite in direction to the force of gravity. So, the force of the track on the vehicle is equal to the force of gravity but in the opposite direction.
Therefore, the force of the track on the vehicle at point A is F_track = -F_gravity = -m * g.
Substituting the given values, we have:
m = 500 kg
g = 9.8 m/s^2
F_track = -(500 kg) * (9.8 m/s^2)
F_track = -4900 N
So, the force of the track on the vehicle at point A is -4900 N.
(b) To find the maximum speed the vehicle can have at point B in order for gravity to hold it on the track, we need to consider the minimum force of the track required to counteract the force of gravity pulling the vehicle downwards.
At point B, the force of gravity pulling the vehicle downwards is the same as at point A. Therefore, the minimum force of the track must be equal in magnitude but opposite in direction to the force of gravity.
Using the same equation as before, we have:
F_track = -m * g
Substituting the given values, we have:
m = 500 kg
g = 9.8 m/s^2
F_track = -(500 kg) * (9.8 m/s^2)
F_track = -4900 N
Therefore, the maximum speed the vehicle can have at point B is determined by the requirement that the force of the track must be at least -4900 N to counteract the force of gravity. However, the maximum speed also depends on other factors such as the shape of the track, friction, and other forces acting on the vehicle.