A boy ties a stone of mass 100   g to the end of a 2   m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80   N . If the maximum speed with which the stone can revolve is K π   rev   min - 1 . The value of K is :…

Question

A boy ties a stone of mass 100 &#160; g to the end of a 2 &#160; m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 &#160; N . If the maximum speed with which the stone can revolve is K &#960; &#160; rev &#160; min - 1 . The value of K is : (Assume the string is massless and un-stretchable)

Aicharya · Accepted Answer

The maximum angular velocity of the stone is $ω = r p m \times \frac{2 π}{60} = \frac{K}{30} rad s^{- 1}$

The tension in the string will provide the required centripetal force. Therefore,

$T = m ω^{2} l \Rightarrow 80 = 0.1 \times {(\frac{K}{30})}^{2} \times 2 \Rightarrow K^{2} = 360000 \Rightarrow K = 600$

A boy ties a stone of mass 100   g to the end of a 2   m long string and whirls it around in a…

Solution

A boy ties a stone of mass 100 &#160; g to the end of a 2 &#160; m long string and whirls it around in a…

Solution

A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a…