Question

A small steel sphere of mass $m$ is tied to a string of length $r$ and is whirled in a horizontal circle with a uniform angular velocity $2ω$ . The string is suddenly pulled, so that radius of the circle is halved. The new angular velocity will be:

• A. $2ω$
• B. $4ω$
• C. $6ω$
• D. $8ω$

Answer 1

Answer: (D)

Given : Mass of sphere $=m$
Initial radius of path $r_1=r$
Final radius of path $r_2=\frac{r}{2}$
The initial M.I. of a particle of mass $m$ rotating in a horizontal circle is given by
$=m{r}_1^2=m{r}^2$
Similarly, the final M.I. of a particle rotating in a circle is
$=m{r}_2^2=m{\left(\frac{r}{2}\right)}^2$
$=\frac{m{r}^2}{4}$
Applying law of conservation of momentum
$m{r}^2×2ω=\frac{m{r}^2}{4}×{ω}_2$
$∴{ω}_2=4×2ω=8ω$