Answer:
Part a)
[tex]F_n = 306 N[/tex]
Part b)
[tex]v = 12.1 m/s[/tex]
So this speed is independent of the mass of the rider
Explanation:
Part a)
By force equation on the rider at the position of the hump we can say
[tex]mg - F_n = ma_c[/tex]
now we will have
[tex]mg - F_n = \frac{mv^2}{R}[/tex]
[tex]F_n = mg - \frac{mv^2}{R}[/tex]
now we have
[tex]F_n = 100(9.81) - \frac{100(9^2)}{12}[/tex]
[tex]F_n = 981 - 675[/tex]
[tex]F_n = 306 N[/tex]
Part b)
At the top of the loop if the minimum speed is required so that it remains in contact so we will have
[tex]F_n + mg = ma_c[/tex]
[tex]F_n = 0[/tex] at minimum speed
[tex]mg = \frac{mv^2}{R}[/tex]
[tex]v = \sqrt{Rg}[/tex]
[tex]v = \sqrt{15 \times 9.81}[/tex]
[tex]v = 12.1 m/s[/tex]
So this speed is independent of the mass of the rider