Question

Mass $m_1$on the frictionless table of FIGURE EX8.13 is connected by a string through a hole in the table to a hanging mass $m_2$. With what speed must $m_1$ rotate in a circle of radius $r$if $m_2$is to remain hanging at rest?

FIGURE EX8.13

Short answer

The speed must $m_1$rotate in a circle of radius $r$ if $m_2$ is to remain hanging at rest is

$v=\sqrt{\left(\frac{m_2}{m_1}\right)rg}$1

Step-by-step solution

Given information

Given in the question that, Mass $m_1$on the frictionless table of FIGURE EX8.13 is connected by a string through a hole in the table to a hanging mass $m_2$.

FIGURE EX8.13

Explanation

Here,

Here

$F_c=F_w$

$F_c=\frac{m_1{v}^2}{r}$

$F_W=m_{2}g$

$\frac{m_1{v}^2}{r}=m_{2}g$

From the above equation we get,

$v=\sqrt{\left(\frac{m_2}{m_1}\right)rg}$

This is the speed must $m_1$rotate in circle of radius $r$if $m_2$is to remain handing at rest

$v=\sqrt{\left(\frac{m_2}{m_1}\right)rg}$

Final answer

The speed must $m1$rotate in a circle of radius $r$if $m_2$ is to remain hanging at rest is

$v=\sqrt{\left(\frac{m_2}{m_1}\right)rg}$