For the normal force in the drawing to have the same magnitude at all points on

Question

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 3.80 m and that the driver goes past point 1 at the bottom with a speed of 25.2 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?

link to image: https://www.flickr.com/photos/134122729@N02/18389853783/in/dateposted-public/

Explanation / Answer

here,

radius , r = 3.80 m

mass of stunt driver is m

at point 1

speed at point 1 ,V1 = 25.2 m/s

centripital force = N1 - m * g

m * v^2 / r = N1 - m * g

N1 = m ((25.2^2)/ 3.8 + 9.8 )

N1 = 176.91 * m N

at point 3

let the speed be V3,

centripital force = N3 + m * g

m * V3^2 / r = N3 + 9.8 * g

for the force at the top is equal to the magnitude at bottom , N1 = N3

m * V3^2 / 3.8 = 176.91 * m + 9.8 * m

V3 = 26.63 m/s

the speed at the top is 26.63 m/s

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects