An amusement park ride consists of a large vertical cylinder that spins about it

Question

An amusement park ride consists of a large vertical cylinder that spins about its axis fastenough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s2. Given g = 9.8 m/s 2 , the coefficient µ = 0.436 of static friction between a person and the wall, and the radius of the cylinder R = 5.3 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is v = 2?R/ T where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum rotation period T of the cylinder which would prevent a 64 kg person from falling down. Answer in units of s. dooley (gd8465) - HW6- Circular Motion - fabi (3 enough that a person inside is stuck to the wall and does not slide down when the floor drops away The acceleration of gravity is 9.8 m/s Given g 9.8 m/s2, the coefficient 0.436 of static friction between a person and the wall, and the radius of the cylinder R = 5.3 m. For simplicity, neglect the person's depth and assume he or she is just a physical point on the wall. The person's speed is 2T R where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum rotation period T of the cylinder which would prevent a 64 kg person from falling down. Answer in units of s. 1D Motion-s... dt HW3-2D Motion-....pdf HW4- Newtor 155 17 32

Explanation / Answer

Using Force balance on the person

Fn = Fc

N = m*V^2/R

given that

V = 2*pi*R/T, So

N = 4*pi^2*m*R/T^2

Now Person would not fall as long as the upward frictional force can balance downward gravitational force, So

Ff >= W

uk*N >= m*g

Using above value of normal force

uk*4*pi^2*m*R/T^2 >= m*g

T^2 =< 4*pi^2*uk*R/g

T =< 2*pi*sqrt (uk*R/g)

Now Using givne values:

T =< 2*pi*sqrt (0.436*5.3/9.8)

T =< 3.05 sec

Longest rotation period will be

T = 3.05 sec

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