A ceiling fan consists of a small cylindrical disk with 5 thin rods coming from

Question

A ceiling fan consists of a small cylindrical disk with 5 thin rods coming from the center. The disk has mass md = 2.8 kg and radius R = 0.26 m. The rods each have mass mr = 1.3 kg and length L = 0.74 m.

1) What is the moment of inertia of each rod about the axis of rotation? kg-m2

2) What is the moment of inertia of the disk about the axis of rotation? kg-m2

3) What is the moment of inertia of the whole ceiling fan? kg-m2

4) When the fan is turned on, it takes t = 3.7 s and a total of 14 revolutions to accelerate up to its full speed.

What is the magnitude of the angular acceleration? rad/s2

5) What is the final angular speed of the fan? rad/s

6) What is the final rotational energy of the fan? J

7) Now the fan is turned to a lower setting where it ends with half of its rotational energy as before. The time it takes to slow to this new speed is also t = 3.7 s.

What is the final angular speed of the fan? rad/s

8) What is the magnitude of the angular acceleration while the fan slows down? rad/s2

I got the answers for 1, 2, 3, and 4. I need 5, 6, 7 and 8.

Explanation / Answer

1) I1 = (1/3)*M*L^2 = (1/3)*1.3*0.74^2 = 0.237 kg-m^2

2) I2 = (1/2)*M*r^2 = 0.5*2.8*0.26^2 = 0.09464 kg-m^2

3) I = I1+I2 = 0.237+0.09464 =0.33164 kg-m^2

4) theta = 14*2*3.142 = 87.976 rad

but theta = (wi*t)+(0.5*alpha*t^2)

87.976 = 0+(0.5*alpha*3.7^2)

alpha = 12.85 rad/s^2

5) final angular speed is wf = wi+(alpha*t) = 0+(12.85*3.7) = 47.545 rad/sec

6) final rotational energy is KE = 0.5*I*w^2 = 0.5*0.33164*(47.545^2) = 374.84 J

7) Now rotational energy is (374.84/2) = 187.42 J

0.5*I*wf^2 = 187.42

0.5*0.33164*wf^2 = 187.42

wf = 33.61 rad/sec is the final angular speed

8) alpha = (wf-wi)/t = (33.61-47.545)/3.7 =-3.76 rad/s^2

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