Cray VA ework 20-SP17 x C Secure h ps://www.webassign.n /web/Student/Assignment-
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Question
Cray VA ework 20-SP17 x C Secure h ps://www.webassign.n /web/Student/Assignment-Responses/last?dep 5321263 Q8 6. -16 points MI4 .2.020 My Notes Calculate the angular momentum for a rotating disk, sphere, and rod (a) A uniform disk of mass 15 kg, thickness 0.5 m, and radius 0.3 m is located at the origin, oriented with its axis along the y axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the +y axis and look toward the origin at the disk). The disk makes one complete rotation every 0.7 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? Express your answer for rotational angular momentum in vector form.) kg m /s (b) A sphere of uniform density, with mass 27 kg and radius 0.5 m, is located at the origin and rotates around an axis parallel with the x axis. If you stand somewhere on the +x axis and look toward the origin at the sphere, the sphere spins counterclockwise. One complete revolution takes 0.8 s. What is the rotational angular momentum of the sphere? What is the rotational kinetic energy of the sphere? (Express your answer for rotational angular momentum in vector form (c) A cylindrical rod of uniform density is located with its center at the origin and its axis along the z axis. Its radius is 0.02 m, its length is 0.7 m, and its mass is 2 kg. It makes one revolution every 0.09 s. If you stand on the +x s and look toward the origin at the rod, the rod spins clockwise. What is the rotational angular momentum of the rod? What is the rotational kinetic energy of the rod? (Express your answer for rotational angular momentum in vector form.) m3/s Additional Materials Section 11.2 2.48 PM Ask me anything 4/2/2017Explanation / Answer
a) The rotational angular momentum is Lrot = I*w
The moment of inertia of the disk about its axis is = I = 0.5*m*R^2
And the angular frequency = w = 2pi/T where T is the period.
Therefore,
Lrot = 0.5*m*R^2*2pi/T = pi*m*R^2/T = 6.058 kg.m^2/s
Krot = 0.5*I*w^2 = 75.53 J
b) The rotational angular momentum is Lrot = I*w
The moment of inertia of the sphere about its axis is = I = 2/5*m*R^2
And the angular frequency = w = 2pi/T where T is the period.
Therefore,
Lrot = 2/5*m*R^2*2pi/T = 4pi*m*R^2/5T = 21.206 kg.m^2/s
Krot = 0.5*I*w^2 = 83.3 J
c) The rotational angular momentum is Lrot = I*w
The moment of inertia of the rod about x axis is = I = (mL^2/12) + (mR^2/4)
And the angular frequency = w = 2pi/T where T is the period.
Therefore,
Lrot = [(mL^2/12) + (mR^2/4)]*2pi/T = 5.715 kg.m^2/s