What drawing? Where is point 3?
We are not clairvoyant.
For the normal force in the drawing to have the same magnitude at all points on the vertical track, the stunt driver must adjust the speed to be different at different points. Suppose, for example, that the track has a radius of 4.84 m and that the driver goes past point 1 at the bottom with a speed of 16.6 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?
4 answers
How do I attach photos?
It is located on the top
I figured it out.
2mg=mv3^2/r-mv1^2/r
V3 being the speed at the top and V1 being the speed given at the bottom
set them equal: V3^2=2gr+V1^2
sqrt(2*9.8*4.84+16.6^2)
V3= 19.2459 m/s
2mg=mv3^2/r-mv1^2/r
V3 being the speed at the top and V1 being the speed given at the bottom
set them equal: V3^2=2gr+V1^2
sqrt(2*9.8*4.84+16.6^2)
V3= 19.2459 m/s
1 answer