If a 2000kg coaster is moving at 10 m/s on the top of the loop, what is the magnitude of the normal force that the track applies to the coaster?
Normal force = mg cos theta
But if theta is 90 degrees, in this case (I'm assuming, because the coaster is perpendicular to the top) then the normal force is zero...
What should it be?
track=mv^2/r -mg
so you need the radius of the loop
Radius is 5m...
Does that mean the answer would be 20,400?
Remember centripetal acceleration.
I assume the coaster is inside the loop and upside down.
If it goes too slow it will fall.
The force down on car is m g + force down exerted by track, call it T
then T + m g = m v^2/R
If T is zero or less, it falls
It the car is on top
then T is up on car
mg - T = m v^2/R
if T is zero, the car flies off the track
Now you tell me.
Ok, still getting 20,400 because I know the car is upside-down in the loop.
That's right - thank you! :)
In this case, the normal force will not be zero. The normal force is the force exerted by a surface perpendicular to the surface. Even though the roller coaster is on the top of the loop and appears to be perpendicular to the track, gravity still acts vertically downwards.
To determine the magnitude of the normal force, you can start by analyzing the forces acting on the coaster. There are two primary forces acting on the coaster at the top of the loop:
1. Gravitational force (mg): This force pulls the coaster downwards towards the center of the loop.
2. Centripetal force (mv^2 / r): This force keeps the coaster moving in a circular path, directed towards the center of the loop.
At the top of the loop, the direction of the net force should be towards the center of the loop to maintain circular motion.
To find the magnitude of the normal force, you can calculate the net force in the vertical direction and equate it to the sum of the gravitational force and the centripetal force:
Net force in the vertical direction = mg + mv^2 / r
Since the coaster is on the top of the loop, the net force should be equal to zero. Therefore, you can set the equation equal to zero and solve for the magnitude of the normal force (N):
mg + mv^2 / r = 0
N = -mg - mv^2 / r
The negative sign indicates that the normal force is acting in the opposite direction of the gravitational force. The magnitude of the normal force will be equal to the absolute value of the expression -mg - mv^2 / r.
Please note that if you provide additional information, such as the radius of the loop or any other relevant parameters, a more accurate calculation can be performed.