Answer to Question #200301 in Mechanics | Relativity for Gil Dror

Question #200301

A round and smooth table revolves around an axis that passes through its center at a constant angular velocity w. The two bodies are moving together on the table. Body A whose mass m is attached to the table hinge by means of a spring. The force constant of the spring is k and the length of the spring in the relaxed state is L. Body A is connected to body B whose mass is 2m by A wire, see diagram.


At the time of rotation, the distance of body A from the center is R and the distance of body B from the center is 2R. express using w, L, m, k the

1. The radius of rotation R of body A.

2.the stretching of the wire connecting the bodies.

3. Does the expression you found in section 1 have a physical meaning at any angular velocity? 

4. Calculate the mass of each of the bodies given: w = 3rad / sec; k = 72N / m; L = 0.5m; R = 1m 

5. What is the work of the wire that connects the two bodies while the bodies complete one round. Necrosis.

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1
Expert's answer
2021-05-30T13:32:31-0400

1) The radius of Rotation of A is R.

(2) Stretching of the wire = R+2R=3R


(3) The body A has gretaer equiivalent force since Its radius is small.


(4) No, The expression do not have any physical meaning at any angular velocity.


(5) "\\omega=3 rad\/s, k=72 N\/m, L=0.5m, R=0.1m"


As we know-

"kx=mr\\omega^2\\\\m=\\dfrac{72 \\times 0.5}{3^2\\times 1}=4 kg"


So Mass of the body B is 2m i.e. 8kg


(6) Work done of the wire "= Fx=mR\\omega^2\\times 0.5=4\\times 1\\times (3)^2\\times 0.5=18 J"

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