a 60kg man is on a steadily rotating Ferris wheel. the man is standing on a bathroom scale and the scale reads 45kg at the top of the wheel. take g=10m/s^2
(a) what does the scale read when the man is at the bottom of the wheel?
(b)what does the scale read when the man is level with the hub of the wheel going up? assume their is sufficient friction that the man's feet do not slide off the scale?
(c) what co-efficient of friction is needed to keep the man's feet on the scale in (b)?
i would like the work shown on how to get the answer, i know that the acceleration points towards the center of the Ferris wheel. i can find the acceleration of 2.5 m/s but don't know what to do next.
DIdn't I just do this?
To find the answers to the questions, we need to consider the forces acting on the man at different positions on the Ferris wheel.
(a) When the man is at the bottom of the wheel, the scale will read higher than his actual weight because there will be an additional normal force acting upwards due to the centripetal acceleration.
First, let's find the net force acting on the man when he is at the bottom of the wheel:
Net force = Weight - Normal force
Weight = mass * g (mass = 60 kg, g = 10 m/s²)
Weight = 60 kg * 10 m/s² = 600 N
Since the man is in uniform circular motion, the net force acting on him is given by:
Net force = mass * centripetal acceleration
Centripetal acceleration = v² / r (v is the linear velocity, r is the radius of the Ferris wheel)
Linear velocity (v) at the bottom of the wheel is given by:
v = r * angular velocity (ω)
The angular velocity is given by:
ω = 2π / T (T is the period of the Ferris wheel)
Let's assume the radius of the Ferris wheel is R and the period is T.
So, the linear velocity at the bottom can be written as:
v = R * (2π / T) = 2πR / T
Now, let's calculate the centripetal acceleration at the bottom:
Centripetal acceleration = (2πR / T)² / R
Centripetal acceleration = 4π²R / T²
Now, substituting this value into the net force equation:
Net force = 60 kg * (4π²R / T²)
To find the normal force, we need to subtract this net force from the weight:
Normal force = Weight - Net force
Normal force = 600 N - 60 kg * (4π²R / T²)
The scale reading will be the magnitude of the normal force. Therefore, the scale will read:
Scale reading = |Normal force|
(b) When the man is level with the hub of the wheel and moving upwards, the direction of the net force changes. Now, the net force has two components: the weight acting downwards, and the normal force acting upwards. The scale reading will be the sum of these forces.
Net force = Weight + Normal force
To calculate the centripetal acceleration, the same formulas are used as before.
(c) To find the coefficient of friction needed to keep the man's feet on the scale, we need to consider the friction force acting between the man's feet and the scale. The maximum friction force can be given by:
Friction force = coefficient of friction * Normal force
In part (b), the scale reading will be the magnitude of the sum of the weight and normal force. Therefore, the maximum friction force should be equal to the sum of the weight and normal force.
Friction force = Weight + Normal force
Coefficient of friction * Normal force = Weight + Normal force
Simplifying the equation, we get:
Coefficient of friction = (Weight + Normal force) / Normal force
Coefficient of friction = (Weight / Normal force) + 1
Now, let's substitute the values we found earlier for part (b) into this equation to calculate the coefficient of friction needed.
I hope this explanation helps you understand how to approach the problem and find the answers.