Question

A 2.4 kh mass is connected to a spring with a spring constant k= 170 N/m and unstretched length 16cm. The pair are mouted on a frictionless air table, with the free end of the spring attached to a frictionless pivot. the mass is set into circular motion at 1.5 m/s.

Find the radius of its path

Answer 1

spring force, F = kx

But the force is provided by the centripetal force of the mass. And the equation for centripetal force is F = mv²/r, where m is the mass, v the speed and r the radius of the circle.  

But the radius of the circle is the total length of the srping. In metres this is 0.16 +x if x is also in metres.

   Since the force is provided by the centripetal force, that must mean the centripetal force is equal to the force given by Hooke's Law, so

   kx = mv²/(0.16+x). Multiplying both sides by 0.16+x gives

kx(x+0.16) = mv², and this can be written as   kx² + 0.16kx - mv² = 0,

   x = [-b ± √(b²-4ac)]/2a.  

where   a = k = 170 b = 0.16k = 0.16*170 = 27.2 c = -mv² = -2.4*1.5² = -5.4. Putting these numbers into the quadratic formula gives   

Evaluating this gives x =0.1155 Therefore the radius of the circle is 0.16 + x = 0.16 + 0.1155= 0.2755 m