What I have is one washer tied on an end of a string and four washers tied on th

Question

What I have is one washer tied on an end of a string and four washers tied on the other end. This string was placed through a plastic tube. I marked off .25 meters and had to spin the single washer at a speed that kept the radius of the single washer string .25 meters outside the top of the plastic tube. Too slow and the radius decreased, to fast and the radius increased. I am stuck on how to figure out the expression it is requesting below:

How is the centripetal force on the revolving washer related to the force of gravity on the hanging washers? Write an expression that equates the centripetal force Fc of the rotating mass to the force of gravity on the hanging mass. Write your expression in terms of m1 (revolving mass), m2 (hanging mass), T, R, and g.

Explanation / Answer

the gravitational force on the hanging mass is F1 =m2g the centripetal force on the revolving mass is F2=mv^2/r where' r ' the radius of revolving mass m1 we know that linear velocity v=r* where' ' is angular velocity in uniform angular motion is constant there fore 'v' is proportional to the radiusa of the circle so radius decreases the linear velocity also decreases ______________________________________________________________________________ let the tention in the string when the revolving mass m1 is at top is =T1 is directed along upwards there fore the down ward force is gravitational force mass m2 =T+m1v^2/r there fore m2g =T+m1v^2/r let the tention in the string when the revolving mass m1 is at bottom is =T2 is directed along downwards m2g+m1v^2/r =T is directed along upwards there fore the down ward force is gravitational force mass m2 =T+m1v^2/r there fore m2g =T+m1v^2/r let the tention in the string when the revolving mass m1 is at bottom is =T2 is directed along downwards m2g+m1v^2/r =T is directed along downwards m2g+m1v^2/r =T

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