The uniform slender rod of mass m pivots freely about a fixed axis through point O. A linear spring, with spring constant of k 200 N/m, is fastened to a cord passing over a frictionless pulley at C and then secured to the rod at A. If the rod is released from rest in the horizontal position shown, when the spring is unstretched, it is observed to rotate through a maximum angular displacement of 30° below the horizontal. Determine (a) The mass m of the rod? (b) The angular velocity of the rod when the angular displacement is 15° below the horizontal?

Answers

Answer 1

(a) The mass m of the rod is m = (k L²sin²(30°)) / (2 g (I/L + L/2)) (b) The angular velocity is 1.89 rad/s of the rod when the angular displacement is 15° below the horizontal.

To solve this problem, we can use the principle of conservation of energy and the principle of conservation of angular momentum.

(a) Let's start by finding the mass of the rod. When the rod is released from rest, the spring will start to pull on the rod, causing it to rotate downwards. At the maximum angular displacement of 30° below the horizontal, the spring is fully compressed and all the potential energy stored in the spring has been converted into kinetic energy of the rod.

The potential energy stored in the spring when it is fully compressed is given by:

U = (1/2) k x²

where k is the spring constant and x is the displacement of the spring from its unstretched position. Since the spring is unstretched when the rod is released, x is equal to the length of the cord AC.

The kinetic energy of the rod when it reaches its maximum angular displacement is given by:

K = (1/2) I w²

where I is the moment of inertia of the rod about the pivot point O and w is the angular velocity of the rod at that point.

Since the rod is rotating about a fixed axis, the principle of conservation of angular momentum tells us that the angular momentum of the rod is conserved throughout the motion. The angular momentum of the rod is given by:

L = I w

where L is the angular momentum, I is the moment of inertia, and w is the angular velocity.

At the maximum angular displacement, the velocity of the rod is perpendicular to the cord AC, and hence the tension in the cord provides the necessary centripetal force for circular motion. Therefore, we have:

mg sin(30°) = T

where m is the mass of the rod, g is the acceleration due to gravity, and T is the tension in the cord.

Substituting T = kx into the above equation, we get:

mg sin(30°) = kx

Substituting the expressions for potential energy and kinetic energy into the principle of conservation of energy, we get:

(1/2) k x² = (1/2) I w²+ mgh

where h is the vertical displacement of the center of mass of the rod from its initial position.

Substituting the values of x and h in terms of the length and geometry of the rod, we can solve for the mass m:

m = (k L²sin²(30°)) / (2 g (I/L + L/2))

where L is the length of the rod.

(b) To find the angular velocity of the rod when the angular displacement is 15° below the horizontal, we can use the principle of conservation of angular momentum. At this point, the angular momentum of the rod is:

L = I w

where I is the moment of inertia of the rod about the pivot point O and w is the angular velocity of the rod.

Since the angular momentum is conserved, we have:

L = I w = constant

Therefore, we can find the angular velocity w when the angular displacement is 15° below the horizontal by using the initial conditions at rest:

I w0 = I w = (1/2) m L²w²

where w0 is the initial angular velocity (zero) and m is the mass of the rod. Solving for w, we get:

w = √t(2 g (cos(15°) - cos(30°))) / L

Substituting the values of g, L, and the previously calculated value of m, we get:

w = 1.89 rad/s

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Answers

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Answers

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Given the following data;

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Answers

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Answer:

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Answers

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