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


As we know-

kx=mrω2m=72×0.532×1=4kgkx=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=mRω2×0.5=4×1×(3)2×0.5=18J= Fx=mR\omega^2\times 0.5=4\times 1\times (3)^2\times 0.5=18 J

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