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A 75kg passenger is riding the cliffhanger barrel ride at the fair. If the barrel radius is 4m and the coefficient of friction between the rider and the wall is 0.5, what is normal force the rider experiences and the minimum speed the ride can turn so that the rider doesn't slide down the wall?

Normal force = ___ N

Min speed = ____ m/s (round to the nearest tenth)

A 75kg passenger is riding the cliffhanger barrel ride at the fair. If the barrel-example-1
User Mcherm
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Answer: The normal force the rider experiences is the sum of the gravitational force acting on the rider and the friction force acting between the rider and the wall of the barrel.

Normal force = gravitational force + friction force

The gravitational force acting on the rider can be calculated using the weight of the rider, which is the mass of the rider multiplied by the acceleration due to gravity:

Weight = mass * acceleration due to gravity

Weight = 75 kg * 9.8 m/s^2 = 735 N

The friction force acting between the rider and the wall of the barrel can be calculated using the coefficient of friction and the normal force:

Friction force = coefficient of friction * normal force

Solving for the normal force, we get:

Normal force = Weight / (1-coefficient of friction) = 735 N / (1-0.5) = 1470 N

To find the minimum speed the ride can turn so that the rider doesn't slide down the wall, we can use the formula

velocity = square root of ( (Normal force * radius ) / (mass))

Minimum speed = √(1470 N * 4 m / 75 kg) = 14.5 m/s (round to the nearest tenth)