Question

1. Starting from rest, a CD takes 3.0 s to reach its operating angular velocity of 450 rpm. The mass of a CD is 17 g and its diameter is 12 cm. You may assume that the small opening at the center of the CD is unimportant when calculating the rotational inertia. Assume that the angular acceleration is constant.

a. What is the rotational kinetic energy of the CD after it has completely spun up?

b. How high off the floor would you need to hold the CD in order to have an equivalent amount of gravitational potential energy?

c. How many revolutions does the CD make before reaching full speed?

d. How much net torque is applied to the CD? e. If the torque is applied by friction at the edge of the center opening, what frictional force is required? The center opening has a diameter of 1.4 cm.

2. When an airplane lands, the wheels skid until the speed of the outer radius of the wheel is the same as the speed of the airplane. Friction on the wheels provide the torque makes the wheels spin; they are not rotating before the plane lands. An average landing speed of a commercial airplane is 70 m/s. Each wheel has a mass of approximately 100 kg and a radius of 0.7 m; the wheels can be modeled as a uniform disk.

a. What is the angular speed of the wheel when the outer radius of the wheel has a tangential speed of 70 m/s?

b. How much torque is applied by friction to each wheel as it is skidding? The coefficient of friction is about 0.5. Assume the mass of the airplane is approximately 300,000 kg and the weight of the airplane is supported by 3 wheels.

c. What is the angular acceleration of the wheel during this time? d. How long does it take the wheel to reach the speed where it stops skidding?

Answer 1

1)
given
wi = 0
at t = 3.0s, wf = 450 rpm = 450*2*pi/60 = 47.1 rad/s
m = 17 g = 0.017 kg
d = 12 cm
r = d/2 = 12/2 = 6 cm = 0.06 m

moment of inertic of the disk, I =0.5*m*r^2

= 0.5*0.017*0.06^2

= 3.06*10^-5 kg.m^2

a) Rotational kinetic energy, KE_rotational = (1/2)*I*wf^2

= (1/2)*3.06*10^-5*47.1^2

= 0.034 J

b) Apply, KE_rotational = m*g*h

h = KE_rotaional/(m*g)

= 0.034/(0.017*9.8)

= 0.20 m

c)) angular acceleration of the Cd, alfa = (wf - wi)/t

= (47.1 - 0)/3

= 15.7 rad/s^2

angular displacement, theta = (wf^2 - wi^2)/(2*alfa)

= (47.1^2 - 0^2)/(2*15.7)

= 70.65 rad

= 70.65/(2*pi)

= 11.2 revolutions.

d) Torque = I*alfa

= 3.06*10^-5*15.7

= 4.8*10^-4 N.m

e) Torque = F*R

F = Torque/R

= 4.8*10^-4/0.014

= 0.034 N