Answer to Question #192886 in Mechanics | Relativity for Jordan Yung

Question

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.

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

1. The radius of rotation R of body A.

2. What is the stretching of the wire connecting the bodies.

3.Determine, without calculating which of the two bodies exerts a greater equivalent force? 

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

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

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

Answer 1

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 $= Fx=mR\omega^2\times 0.5=4\times 1\times (3)^2\times 0.5=18 J$