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 ${K \over \pi }$ rev./min. The value of K is :
(Assume the string is massless and unstretchable)
Solution
<p>$T = m{\omega ^2}r$</p>
<p>$$ \Rightarrow 80 = 0.1 \times {\left( {2\pi \times {K \over \pi } \times {1 \over {60}}} \right)^2} \times 2$$</p>
<p>$\Rightarrow {{800} \over 2} = {{{K^2}} \over {900}}$</p>
<p>$\Rightarrow K = 30 \times 20 = 600$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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