Answer:
928.34 N
Explanation:
Radius=r=9.5 m
Time period of revolution=T=7.6 s
Mass of rider=m=57 kg
We have to find the rate of change of the rider's momentum at the bottom of the ride
According to Newton's second law of motion
[tex]F_{net}=ma_c[/tex]
[tex]N-mg=mr\omega^2[/tex]
[tex]N=mg+mr(\frac{2\pi}{T})^2[/tex]
Where [tex]\pi=3.14,g=9.8 m/s^2[/tex]
Using the formula
[tex]N=57\times 9.8+57\times 9.5(\frac{2\times 3.14}{7.6})^2[/tex]
[tex]N=928.34 N[/tex]
Hence, the rate of change of the rider's momentum=928.34 N
