Question

3/194 The assembly of two 5‐kg spheres is rotating freely about the vertical axis at 40 rev/min with θ = 90°. If the force F which maintains the given position is increased to raise the base collar and reduce θ to 60°, determine the new angular velocity ω. Also determine the work U done by F in changing the configuration of the system. Assume that the mass of the arms and collars is negligible.

Answer 1

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Answer:

A) The new angular velocity ω = 3 rad/s

B) work done by F in changing the configuration of the system = 5.27 J

Explanation:

We are given that ωo = 40 rev/min. We need to convert it to rad/s. We know that 1 rev/min = 0.105 rad/s

Thus, ωo = 40 x 0.105 = 4.2 rad/s

Mass of sphere (Ms) = 5kg

For θ = 90°; it's, ro = 0.1 + 2(0.3 cos (90/2)) = 0.1 + 2(0.3 cos 45°)

= 0.1 + (2 x 0.212) = 0.524m

For θ = 60°; it's, r = 0.1+ 2(0.3cos 30°) =

0.1 + 2(0.2598) = 0.62m

Since momentum is conserved; ΔH = 0.

Thus, ro(vo) = rv

Where v is speed. We know that v=ωr; thus,

ro(roωo) = r(ωr)

Which gives;

ro²ωo = r²ω

So making ω the subject to get;

(ro²ωo)/r²

So plugging in the relevant values to get;

ω = (0.524² x 4.2)/0.62² = 3 rad/s

B) Work done (U) is; U1-2 = ΔT + ΔVg = 2(1/2)(m)(v²-vo²) + 2mgΔH

= 2(1/2)(m)(r²ω² - ro²ωo²) + 2mgΔH

But looking at the diagram, we can deduce that;

To get change in ΔH;

[0.1 + 2(0.3 sin 45°)] - [0.1+ 2(0.3cos 30°)] = 0.5243 - 0.4 = 0.1243m

So, U1-2 = (5)((0.62² x 3²) - (0.524² x 4.2²)) + (2 x 5 x 9.81 x 0.1243) = -6.92 + 12.19 = 5.27J