Question

1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The length of a windmill blade is 36 m. If the windmill blade takes 50 spins to come to a stop: a. What is the magnitude of the rotational acceleration at the tip of a blade while it is slowing down? b. What is the magnitude of the rotational acceleration at the midpoint of the blade (r = 18 m)? Before you begin your calculations state whether you expect the answer to be larger or smaller and explain why. c. While the windmill spins at 20 times per minute, what is the centripetal acceleration of a repair person desperately holding on at the end of the blade? And at the halfway point? d. How would the centripetal acceleration at these two points change if the angular velocity of the blades were cut in half? First predict whether it will get larger or small and explain how you know, then do the calculation.

Answer 1

Answer:

a) $\alpha = -0.233 rad/s^{2}$

b) the rotational acceleration will remain the same,  $\alpha = -0.233 rad/s^{2}$

c) When r = 36 m, $a_{c} = 157.25 m/s^{2}$

   When r = 18 m, $a_{c} = 78.63 m/s^{2}$

d) When r = 36 m, $a_{c} = 39.31 m/s^{2}$

   When r = 36 m, $a_{c} = 19.66 m/s^{2}$

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60  = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

$\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}$

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, $a_{c} = w^{2} r$

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

$a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}$

At the halfway point, r = 18 m

$a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}$

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

$a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}$

When r = 18 m

$a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}$