Question

Calculate the angular momentum for a rotating disk, sphere, and rod. (a) A uniform disk of mass 16 kg, thickness 0.5 m, and radius 0.9 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.) L tot = (L , kg.mp/s Krot = ** (b) A sphere of uniform density, with mass 24 kg and radius 0.2 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 0.5 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.) L L tot = ( S , C ?) kg-m/s Krot = ] (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 0.05 m, its length is 0.7 m, and its mass is 4 kg. It makes one revolution every 0.08 s. If you stand on the +x axis 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.) , ( 11) kg-m/s Llot = (C Krot = Krot = -12

Answer 1

a)

m = mass of the disk = 16 kg

r = radius of the disk = 0.9 m

I = moment of inertia of the disk = (0.5) m r² = (0.5) (16) (0.9)² = 6.48 kg m²

T = time period of rotation for disk = 0.7 s

w = angular velocity of disk = 2$\pi$/T = 2(3.14)/(0.7) = 8.97 rad/s

Angular momentum is given as

L = I w

L = (6.48) (8.97)

L = 58.13 kg m²/s

$\underset{L_{rot}}{\rightarrow}$ = 0 i - 58.13 j + 0 k

rotational kinetic energy is given as

K_rot = (0.5) I w² = (0.5) (6.48) (8.97)² = 260.7 J

b)

m = mass of the sphere = 24 kg

r = radius of the sphere = 0.2 m

I = moment of inertia of the sphere = (0.4) m r² = (0.4) (24) (0.2)² = 0.384 kgm²

T = time period of rotation for sphere = 0.5 s

w = angular velocity of sphere = 2$\pi$/T = 2(3.14)/(0.5) = 12.56 rad/s

Angular momentum is given as

L = I w

L = (0.384) (12.56)

L = 4.82 kgm²/s

$\underset{L_{rot}}{\rightarrow}$ = 4.82 i + 0 j + 0 k

rotational kinetic energy is given as

K_rot = (0.5) I w² = (0.5) (4.82) (12.56)² = 380.2 J

c)

m = mass of the rod = 4 kg

L = length of the rod = 0.7 m

I = moment of inertia of the rod = (1/12) m L² = (1/12) (4) (0.7)² = 0.163 kgm²

T = time period of rotation for sphere = 0.08 s

w = angular velocity of sphere = 2$\pi$/T = 2(3.14)/(0.08) = 78.5 rad/s

Angular momentum is given as

L = I w

L = (0.163) (78.5)

L = 12.8 kgm²/s

$\underset{L_{rot}}{\rightarrow}$ = - 4.82 i + 0 j + 0 k

rotational kinetic energy is given as

K_rot = (0.5) I w² = (0.5) (0.163) (78.5)² = 502.22 J