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A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 12 m and the loop has a radius of curvature of 15 m. (a) When going over the hump, the coaster is traveling with a speed of 9.0 m/s. A 100-kg rider is traveling on the coaster. What is the normal force of the rider’s seat on the rider when he is at the peak of the hump? Compare this with the normal force he would experience when the coaster is at rest. (b) What is the minimum speed the coaster must have at the top of the loop in order for the rider to remain in contact with his seat? Is this speed dependent on the mass of the rider?

Respuesta :

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