The hollow tube assembly rotates about a vertical axis with an angular velocity

Question

The hollow tube assembly rotates about a vertical axis with an angular velocity of w=theta dot=4 rad/s anddouble dot w= double dot theta=-2 rad/s2. A 0.2 kg slider P moves inside the horizontal tube portion under the control ofthe string which passes out of the bottom of the assembly. If r=0.8 m, rdot=-2 m/s and rdoubledot= 4 m/s2,

determine:

a) The tension T in the string

b) The horizontal force F theta exerted on the slider by the tube.

The hollow tube assembly rotates about a vertical axis with an angular velocity of w=theta dot=4 rad/s and double dot w= double dot theta=-2 rad/s2. A 0.2 kg slider P moves inside the horizontal tube portion under the control of the string which passes out of the bottom of the assembly. If r=0.8 m, rdot=-2 m/s and r doubledot= 4 m/s2,

Explanation / Answer

If we explain the whole of the scenario, condition is

Tube is rotating at omega = 4 rad/sec

Tube is slowing down (negative) at rate of alpha = -2 rad/sec^2

Distance of slider from axis r = 0.8 m

It is moving towards axis with velocity v = 2 m/sec

It is slowing down with acceleration a = -4 m/sec^2

The factors that control tension in the string are,

angular velocity of the tube, position of the slider, acceleration a of the slider

(a) Equation written in on the slider in the frame of tube

(m*r*omega^2) - T = m*a

(0.2 * 0.8 * 4*4) - T = 0.2 * 4

T = 2.56 - 0.8

T = 1.76 N

(b)

Force on the slider is accelerating the slider tangentially

tangential acceleration of slider = r*alpha

= 0.8 * 2

= 1.6 m/sec^2

So, force F = m * tangential acceleration

= 0.2 * 1.6

= 0.32 N

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