Question 82: Work, Energy and Power & Circular Motion

Question

A stone of mass $1 \mathrm{~kg}$ is tied to end of a massless string of length $1 \mathrm{~m}$. If the breaking tension of the string is $400 \mathrm{~N}$, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :

Accepted Answer

$$
\begin{aligned}
& T \sin 	heta=\frac{m v^{2}}{l \sin 	heta} \\
& \cos 	heta=\frac{m g}{T} ........(1) \\
& \sin^2 	heta=\frac{m v^{2}}{T l} ........(2)
\end{aligned}
$$
From (1) and (2),
$1=\left(\frac{m g}{T}ight)^{2}+\frac{m v^{2}}{T l}$
$$ \Rightarrow $$ $1=\left(\frac{10}{400}ight)^{2}+\frac{v^{2}}{400}$
$$ \Rightarrow $$ $v^{2}=399.78$
$$ \Rightarrow $$ $v=20 \mathrm{~m} / \mathrm{s}$

Answer

$20 \mathrm{~ms}^{-1}$

Answer

$40 \mathrm{~ms}^{-1}$

Answer

$400 \mathrm{~ms}^{-1}$

Answer

$10 \mathrm{~ms}^{-1}$

Work, Energy and Power & Circular Motion

A stone of mass 1 \mathrm{~kg} is tied to end of a massless string of length 1 \mathrm{~m}. If the breaking tension of the string is 400 \mathrm{~N}, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :

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