An amusement park ride consists of a large vertical cylinder that spins about its axis fast 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/s2 .Given g = 9.8 m/s2, the coefficient μ = 0.2 of static friction between a person and the wall, and the radius of the cylinder R = 6.5 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*pi)*radius)/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 48 kg person from falling down.
Answer in units of s
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To determine the maximum rotation period T of the cylinder that would prevent a 48 kg person from falling down, we need to calculate the minimum speed required for the person to stay on the wall.
First, let's determine the maximum static friction force (Ff_max) between the person and the wall. This can be found using the formula Ff_max = μN, where N is the normal force.
The normal force (N) can be calculated as the weight of the person, which is given by the formula N = mg, where m is the mass of the person and g is the acceleration due to gravity.
N = (48 kg)(9.8 m/s²) = 470.4 N
Now, we can find the maximum static friction force:
Ff_max = (0.2)(470.4 N) = 94.08 N
The maximum static friction force is equal to the centripetal force required to keep the person stuck to the wall.
The centripetal force can be calculated using the formula Fc = m(v²/R), where Fc is the centripetal force, m is the mass of the person, v is the speed, and R is the radius of the cylinder.
Fc = 48 kg(v²/6.5 m)
Since we want to find the maximum rotation period T, we can substitute v = ((2π)(radius))/T into the formula for Fc:
Fc = 48 kg((((2π)(6.5 m))/T)²/6.5 m)
Simplifying:
94.08 N = 48 kg(((2π(6.5 m))²/6.5 m)/T)²
Now, we can solve this equation to find T:
94.08 N = 48 kg(((2π(6.5 m))²/6.5 m)/T)²
Rearranging:
T² = (((48 kg)(2π(6.5 m)²))/94.08 N)²
T² = (0.16895131 s²)
Taking the square root of both sides:
T ≈ 0.411 s
Therefore, the maximum rotation period T of the cylinder that would prevent a 48 kg person from falling down is approximately 0.411 seconds.
To solve this problem, we need to consider the forces acting on the person when they are stuck to the wall of the spinning cylinder.
The force of gravity pulling the person downward is given by:
F_gravity = m*g
Where m is the mass of the person (48 kg) and g is the acceleration due to gravity (9.8 m/s^2).
The force of static friction between the person and the wall is given by:
F_friction = μ*N
Where μ is the coefficient of static friction (0.2) and N is the normal force acting on the person. In this case, the normal force is equal to the force of gravity since the person is stuck to the wall and not moving away from it.
Now, let's consider the centripetal force required to keep the person stuck to the wall, which is provided by the frictional force:
F_centripetal = m*a
Where a is the centripetal acceleration, which is given by:
a = v^2 / R
v is the speed of the person along the circular path, which can be calculated using the formula you provided:
v = (2*pi*R) / T
R is the radius of the cylinder (6.5 m) and T is the rotation period of the cylinder (the time to complete a full circle).
From Newton's second law, we know that the sum of the forces in the vertical direction is zero because the person is not falling down:
F_gravity - F_friction = 0
Substituting the expressions for F_gravity and F_friction, we get:
m*g - μ*N = 0
Simplifying further, we have:
m*g - μ*m*g = 0
Solving for μ, we find:
μ = g / (g - a)
Substituting the expression for a, we get:
μ = g / (g - v^2 / R)
Now, we can substitute the given values into this equation and solve for the rotation period T:
0.2 = 9.8 / (9.8 - ((2*pi*R) / T)^2 / R)
Simplifying this equation and solving for T will give us the maximum rotation period that prevents the person from falling down.
Please note that since the equation involves a trigonometric function and complex calculations, it is best to use a computer program or calculator to solve for T.