track=mv^2/r -mg
so you need the radius of the loop
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?
6 answers
Radius is 5m...
Does that mean the answer would be 20,400?
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
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! :)
2 answers