Question
Find the maximum revolution per minute of a body of 500 gram tied to a 1.5 m long string, if the string can bear a maximum tension of 40 N .
✓
Answer
$\begin{array}{l}
m=\text { mass of body }=500 g=0.5 kg \\
R=\text { radius of circle }=1.5 m \\
F_c=\text { centripetal force }=40 N
\end{array}$
Now,
$\begin{aligned}
F_c & =\text { mass } \times \text { acceleration } \\
F_c & =m a_r \\
& =\frac{m v^2}{R}=\frac{m}{R}(R \omega)^2 \quad \because v=R \omega \\
& =\frac{m R^2 \omega^2}{R}=m R \omega^2 \\
& =m R(2 \pi f)^2=4 \pi^2 m R f^2 \\
f^2 & =\frac{F_c}{4 \pi^2 m R} \\
f & =\sqrt{\frac{F_c}{4 \pi^2 m R}}=\sqrt{\frac{40}{4 \times(3.14)^2 \times 0.5 \times 1.5}} \\
& =\sqrt{1.351}=1.162 revolution / sec \\
& =1.162 \times 60=69.74 rev / min
\end{aligned}$
Hence, the body can revolve at rate of 69 revolution/ minute without breakage of string.
m=\text { mass of body }=500 g=0.5 kg \\
R=\text { radius of circle }=1.5 m \\
F_c=\text { centripetal force }=40 N
\end{array}$
Now,
$\begin{aligned}
F_c & =\text { mass } \times \text { acceleration } \\
F_c & =m a_r \\
& =\frac{m v^2}{R}=\frac{m}{R}(R \omega)^2 \quad \because v=R \omega \\
& =\frac{m R^2 \omega^2}{R}=m R \omega^2 \\
& =m R(2 \pi f)^2=4 \pi^2 m R f^2 \\
f^2 & =\frac{F_c}{4 \pi^2 m R} \\
f & =\sqrt{\frac{F_c}{4 \pi^2 m R}}=\sqrt{\frac{40}{4 \times(3.14)^2 \times 0.5 \times 1.5}} \\
& =\sqrt{1.351}=1.162 revolution / sec \\
& =1.162 \times 60=69.74 rev / min
\end{aligned}$
Hence, the body can revolve at rate of 69 revolution/ minute without breakage of string.
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