SOLUTION: A mass of 0.675 kg, on a frictionless table, is attached to a string which passes through a hole in the table. The hole is at the center of a horizontal circle in which the mass mo
Question 1056855: A mass of 0.675 kg, on a frictionless table, is attached to a string which passes through a hole in the table. The hole is at the center of a horizontal circle in which the mass moves with constant speed. The radius of the circle is 0.530 m and the speed of the mass is 12.0 m/s. It is found that drawing the string down through the hole and reducing the radius of the circle to 0.370 m has the effect of multiplying the original tension in the string by 4.63. Compute the total work (in joules) done by the string on the revolving mass during the reduction of the radius.
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A mass of 0.675 kg, on a frictionless table, is attached to a string which passes through a hole in the table.
The hole is at the center of a horizontal circle in which the mass moves with constant speed.
The radius of the circle is 0.530 m and the speed of the mass is 12.0 m/s. It is found that drawing the string down
through the hole and reducing the radius of the circle to 0.370 m has the effect of multiplying the original tension
in the string by 4.63. Compute the total work (in joules) done by the string on the revolving mass during the reduction of the radius.
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You should use two conservation laws.
First, use the Angular Momentum conservation law to find the rotational speed of the mass after reducing the radius of the circle from 0.503 m to 0.370 m.
=
and calculate = = in meters per second.
Second, calculate the change of the kinetic energy of the mass
E = - .
Now, use the Energy conservation law, which says that the change of the kinetic energy is equal to the work
of the tension force on the distance d = .
So, the total work under the question is exactly the change of the kinetic energy E.
