The smooth hollow tube assembly rotates a vertical axis with angukar velocity w

Question

The smooth hollow tube assembly rotates a vertical axis with angukar velocity w = ?· = 3.7 rad/s and w· = ?·· = -1.8 rad/s^2. A small 0.13-kg slider P moves inside the horizontal tube portion under the control of the strong which passes out the bottom of the assembly. If r = 1 m, r· = -2.0 m/s, and r·· = 4.5 m/s^2, determine the tension T in the string and the horizontal force F? exerted on the slider by the tube. T is positive in tension, and F? is positive if in the positive ? direction. NOTE: T is not 2.0475 N. F? is -2.158 N.

Explanation / Answer

The normal centrifugal force purely depends on angular velocity at that instance and radius of curvature and of course on the mass of the object

So, T=mw^2r is the answer

= 0.13*(3.7^2)*1=1.7797

As the centrifugal force is balanced by the tensile force on the string , then T= the centrifugal force only.

Now v=wr. By derivatives we get a=r*theta.. + w*r.

(1*-1.8)+(3.7*-2)= -9.2

Horizontal force = mass*acceleration= 0.13*-9.2=1.196

But the tensile force need to consider here also . By vector addition we get ?{1.196^2+1.7797^2} =2.1442 . As both were negative in direction . Their vector sum would be also negative

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