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jarptica [38.1K]
3 years ago
9

A uniform disk with mass m = 8.54 kg and radius R = 1.35 m lies in the x-y plane and centered at the origin. Three forces act in

the +y-direction on the disk: 1) a force 334 N at the edge of the disk on the +x-axis, 2) a force 334 N at the edge of the disk on the –y-axis, and 3) a force 334 N acts at the edge of the disk at an angle θ = 30° above the –x-axis.
a. What is the magnitude of the torque on the disk about the z axis due to F1?
b. What is the magnitude of the torque on the disk about the z axis due to F2?
c. What is the magnitude of the torque on the disk about the z axis due to F3?
d. What is the x-component of the net torque about the z axis on the disk?
e. What is the y-component of the net torque about the z axis on the disk?
f. What is the z-component of the net torque about the z axis on the disk?
g. What is the magnitude of the angular acceleration about the z axis of the disk?
h. If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.6 s?
Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

A) τ_1 = 450.9 N.m

B) τ_2 = 0 N.m

C) τ_3 = 390.47 N.m

D) There is no x-component.

E) There is no y-component.

F) τ_z = 60.43 N.m

G) α = 7.765 rad/s²

H) K = 600.6J

Explanation:

A) τ_1 = F1•r1•sinθ1

τ_1 = 334 x 1.35 x sin90

τ_1 = 450.9 N.m

B) τ_2 = F2•r2•sinθ2

τ_2 = 334 x 1.35 x sin180

τ_2 = 0 N.m

C) τ_3 = F3•r3•sinθ3

τ_3 = 334 x 1.35 x sin60

τ_3 = 390.47 N.m

D) Torque from F2 is zero and both torques from F1 and F3 are in z direction, thus, there is no x-component.

E) Torque from F2 is zero and both torques from F1 and F3 are in z direction, thus, there is no y-component.

F) τ_1 is in the positive z direction

τ_2 is zero

τ_3 is in the negative z direction

Hence, τ_z = 450.9 - 390.47 = 60.43 N.m

G) moment of inertia is given as ;

I = (1/2)MR²

Thus ;

I = (1/2) x 8.54 x 1.35² = 7.782 N.m

Angular acceleration is;

α = τ_z/I

α = 60.43/7.782 = 7.765 rad/s²

H) angular velocity is;

ω = αt = 7.765 x 1.6 = 12.424 rad/s

Rotational energy, K = (1/2)Iω²

K = (1/2) x 7.782 x (12.424)² = 600.6J

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