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

Answer:

0.025 m/sec

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

$F_c=\frac{mv^2}{r}=\frac{0.25\times 0.05^2}{0.5}=0.00125 N$

now again using this equation for finding the final velocity

$0.00125=\frac{mv_{final}^2}{r}=\frac{0.25v_{final}^2}{0.125}$

$v_{final}=\sqrt{\frac{0.00125\times 0.125}{0.25}}=0.025\ m/sec$

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