Question

A wind turbine is rotating counterclockwise at 0.5 rev/s and slows to a stop in 10 s. Its blades are 20 m in length. (a) What is the angular acceleration of the turbine? (b) What is the centripetal acceleration of the tip of the blades at t = 0 s? (c) What is the magnitude and direction of the total linear acceleration of the tip of the blades at t = 0 s?

Answer 1

The angular acceleration of the turbine is -0.05 rev/s^2. The centripetal acceleration of the tip of the blades at t = 0 s is -6.2 m/s^2. The magnitude and direction of the total linear acceleration of the tip of the blades at t = 0 s are 6.26 m/s^2 counterclockwise.

Step-by-step explanation:

(a) The angular acceleration of the turbine can be calculated using the formula α = (final angular velocity - initial angular velocity) / time taken. In this case, the initial angular velocity is 0.5 rev/s, the final angular velocity is 0 rev/s (since it comes to a stop), and the time taken is 10 s. Therefore, the angular acceleration is (α) = (0 - 0.5 rev/s) / 10 s = -0.05 rev/s2.

(b) The centripetal acceleration of the tip of the blades at t = 0 s can be calculated using the formula ac = rα, where r is the length of the blades and α is the angular acceleration. In this case, r = 20 m and α = -0.05 rev/s2. Converting rev/s2 to rad/s2, we get α = -0.05 x (2π rad/rev) = -0.31 rad/s2. Therefore, the centripetal acceleration is ac = (20 m) x (-0.31 rad/s2) = -6.2 m/s2.

(c) The total linear acceleration of the tip of the blades at t = 0 s is the vector sum of the centripetal acceleration and the tangential acceleration. Since the blades are initially rotating counterclockwise, the direction of the centripetal acceleration is towards the center of rotation (inwards), and the direction of the tangential acceleration is in the direction of the velocity (counterclockwise). Therefore, the magnitude of the total linear acceleration is the square root of the sum of the squares of the centripetal acceleration and the tangential acceleration. Given that ac = -6.2 m/s2 and the angular velocity at t = 0 s is 0.5 rev/s, the tangential acceleration (at) can be calculated using the formula at = rα, where r is the length of the blades and α is the angular acceleration. Substituting the values, at = (20 m) x (-0.05 rev/s2) = -1 m/s2. Therefore, the magnitude of the total linear acceleration is |atotal| = √((-6.2 m/s2)2 + (-1 m/s2)2) = 6.26 m/s2. The direction of the total linear acceleration is counterclockwise.

Answer 2

-0.314 rad/s²

197.39 m/s²

197.49 m/s² and 1.822° counterclockwise.

Step-by-step explanation:

`\omega_f` = Final angular velocity

`\omega_i` = Initial angular velocity

`\alpha` = Angular acceleration

`\theta` = Angle of rotation

t = Time taken

R = Radius = 20 m

`\omega_i=0.5* 2\pi\\\Rightarrow \omega_i=\pi`

`\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=(\omega_f-\omega_i)/(t)\\\Rightarrow \alpha=(0-\pi)/(10)\\\Rightarrow a=-0.314\ rad/s^2`

Angular acceleration of the turbine is -0.314 rad/s²

Centripetal acceleration

`a_c=\omega_i^2R\\\Rightarrow a_c=(\pi)^2* 20\\\Rightarrow a_c=197.39\ m/s^2`

The centripetal acceleration of the tip is 197.39 m/s²

Tangential acceleration

`a_t=\alpha R\\\Rightarrow a_t=-0.314* 20\\\Rightarrow a_t=-6.28\ m/s^2`

Acceleration

`a=√(a_c^2+a_t^2)\\\Rightarrow a=√(197.39^2+(-6.28)^2)\\\Rightarrow a=197.49\ m/s^2`

`\theta=tan^(-1)(|a_t|)/(|a_c|)\\\Rightarrow \theta=tan^(-1)(6.28)/(197.39)\\\Rightarrow \theta=1.822^(\circ)`

Magnitude and direction of the total linear acceleration of the tip of the blades is 197.49 m/s² and 1.822° counterclockwise.