Part A

Each blade on a wind turbine is engineered to produce an effective force through its center of mass that is perpendicular to its long axis. (Part A figure) The magnitude of the force produced by each blade in a wind of speed v is F=(40.0~{\rm kg/s})v. If the center of mass of each blade is r_1 = 11.0 m, what is U_1, the total work done by the wind turbine's blades, in a wind of speed v_1 = 3.90 m/s after 10 revolutions?

Part B
A brake system is tested by rotating a tire and measuring the number of rotations required for the brake system to bring the tire to a stop. (Part B figure) The tire's radius is R_2 = 50.0 cm and the brake system's radius is r_2 = 16.7 cm. A moment of M_1 = 18.3 N \cdot m is applied to the tire for 5 rotations before the brake system is applied. The brake system is composed of two pads that are pushed out against the drum with a force that increases as the tire rotates and is described by F=(10.0\,\theta)~{\rm N}. If the coefficient of kinetic friction between the brake pads and the outer ring of the brake system is mu_1 = 0.550, how many rotations, n, will the tire go through before coming to a stop?

Part A:

To find the total work done by the wind turbine's blades, we need to calculate the work done by each blade and then multiply it by the number of blades.

1. Work done by one blade: The work done is given by the formula W = F * d * cos(theta), where F is the force, d is the distance moved, and theta is the angle between the force and the displacement.

In this case, the force produced by each blade is F = (40.0 kg/s) * v, where v is the wind speed. The distance moved by each blade is the circumference of the circle covered in one revolution, which is 2 * pi * r1. The angle between the force and displacement is 0 degrees since the two are perpendicular.

So, the work done by one blade is W1 = (40.0 kg/s) * v * 2 * pi * r1.

2. Total work done by all blades: Since there are multiple blades, we need to multiply the work done by one blade by the number of blades. In this case, the number of revolutions is given as 10.

So, the total work done by the wind turbine's blades is U1 = n * W1, where n is the number of revolutions and W1 is the work done by one blade.

Plug in the given values:
v1 = 3.90 m/s
r1 = 11.0 m
n = 10

U1 = 10 * [(40.0 kg/s) * v1 * 2 * pi * r1]

Calculate the value of U1 using the given values.

Part B:

To find the number of rotations before the tire comes to a stop, we need to calculate the work done by the moment applied to the tire and the work done by the brake system. When the brake system brings the tire to a stop, the work done by the brake system is equal to the work done by the moment applied to the tire.

1. Work done by the moment: The work done by the moment is given by the formula W = M1 * theta, where M1 is the moment applied and theta is the angle in radians.

In this case, the moment applied to the tire is M1 = 18.3 N · m, and the number of rotations before the brake system is applied is given as 5. Since each rotation is 2 * pi radians, the total angle is 5 * 2 * pi.

So, the work done by the moment is W_moment = M1 * (5 * 2 * pi).

2. Work done by the brake system: The work done by the brake system is given by the formula W = F * d * mu, where F is the force, d is the distance moved, and mu is the coefficient of friction.

In this case, the force applied by the brake system is F = (10.0 * theta) N. The distance moved is the circumference of the circle covered in each rotation, which is 2 * pi * r2. The coefficient of friction is given as mu1 = 0.550.

So, the work done by the brake system is W_brake = (10.0 * theta) * (2 * pi * r2) * mu1.

3. Equating the work done by the moment and the work done by the brake system:

W_moment = W_brake

M1 * (5 * 2 * pi) = (10.0 * theta) * (2 * pi * r2) * mu1

Simplify the equation and solve for the number of rotations n.

Plug in the given values:
R2 = 50.0 cm
r2 = 16.7 cm
M1 = 18.3 N · m
mu1 = 0.550

Calculate the value of n using the given values.