Question

A slider of mass 0.25 kg on a string, 0.5 m long is rotating around a pivot on a frictionless table. The velocity of the slider is initially 0.05 m/s. When the string is pulled into a radius of 0.125 m how fast is the mass spinning?

Answer 1

0.025 m/sec

Step-by-step explanation:

we have given m =0.25 kg

velocity=0.05 m/sec

radius =0.5 meter

the centrifugal force produced due to rotational motion

`F_c=(mv^2)/(r)=(0.25* 0.05^2)/(0.5)=0.00125 N`

now again using this equation for finding the final velocity

`0.00125=(mv_(final)^2)/(r)=(0.25v_(final)^2)/(0.125)`

`v_(final)=\sqrt{(0.00125* 0.125)/(0.25)}=0.025\ m/sec`

so the final speed of mass spring will be 0.025 m/sec