Question

A weightless thread can bear tension upto 3.7 kg weight. a stone of mass 500 gram is tied at its one end and revolved in a vertical circular path of radius 4 m. if g = 10 m/s2, then the maximum angular velocity of the stone is (radians/sec) will be

Answer 1

When a body moves in circular path, a centripetal force is required to keep the body along its path.

As per the question, the string is tied to a stone and the stone is moving in a vertical circular path.

Here the tension of the string will provide the necessary centripetal force.

The maximum tension T = 3.7 kg weight

                                        = 3.7×g       [g is the acceleration due to gravity]

                                        = 3.7×10 N

                                        =37 N

The mass of the stone [m]= 500 gram i.e 0.5 kg.

The radius of circular path is 4m.

We are asked to calculate the maximum angular velocity.

The centripetal force is calculated as -

                                     $F=\frac{mv^2}{r}$

                                     $F=mr\omega^{2}$     $[v=\omega r]$

$Here\ v\ is\ the\ tangential\ velocity\ and\ \omega\ is\ the\ maximum\ angular\ velocity\ regarding\ to\ v$

       Here, centripetal force T = F

                                             $Hence\ T\ =mr\omega^2$

                                             $\omega^2\ =\frac{T}{mr}$

                                             $=\frac{37}{0.5*4}$

                                             $\omega^2=18.5$

                                             $\omega=\sqrt{18.5}\ rad/s$

                                             $\omega=4.3012\ rad/s$  [ans]

                                     

Answer 2

Maximum tension that string can bear = 37 N

tension at lowest point of vertical loop =

$mg \: + \: mv {}^{2} r$

$= 0.5 \times 10 \times 0.5 \times v \times 4 \\ \\ = 5 + 2v {}^{2} \\ \\ 37 = 5 + 2v {}^{2} \\ \\ v \: = \: 4 \: rad$