A ceiling fan with 67-cm-diameter blades is turning at 54 rpm. Suppose the fan coasts to a stop 27.3 s after being turned off. What is the speed of the tip of a blade 9.6 s after the fan is turned off?

1.228 m/s

Explanation:

In rotational kinematics, we have the following equations;

ω_f = ω_i + αt - - - - (1)

Where;

ω_f is final angular velocity

ω_i is initial angular velocity

α is angular acceleration

t is time taken

Also; instantaneous velocity is given by;

v = rω - - - - - (eq2)

Where r is the distance from the angle of rotation

We are given;

ω_i = 54 rpm = 54 x 0.10472 rad/s =

5.655 rad/s

Time required for it to stop running after turning it; t = 27.3 s

Diameter(D) = 67cm = 0.67m

Thus, let's find the angular acceleration by plugging in the relevant values into eq 1

Since it comes to rest after 27.3s,thus ω_f = 0

0 = 5.655 + 27.3α

-27.3α = 5.655

α = -5.655/27.3 = -0.2071 rad/s²

Now, we want to find the speed after 9.6s

Thus,

ω_f = 5.655 + 9.6(-0.2071)

ω_f = 5.655 - 1.988 = 3.667 rad/s

Now, since the diameter is 0.67m,then the distance between the tip and the angle of rotation is;

r = d/2 = 0.67/2 = 0.335m

So, for the final velocity of the tip of a blade 9.6 s after the fan is turned off; let's plug in the relevant values into eq 2.

v = rω = 0.335 x 3.667 = 1.228 m/s

prolly a beacuse the momentuem the car has increases while going down hill