Physics - W.E.P. Question with Solution | TestHub

Frequency of revolution, $n=30\text{ rpm}=\frac{30}{60}\text{ rev/s}=0.5\text{ Hz}$

Angular velocity, $\omega =2\pi n=2\pi \times 0.5=\pi \text{ rad/s}$

Radius of the vertical circle, $r=49\text{ cm}=0.49\text{ m}$

The tension in the string at the lowermost point of the vertical circle is given by:

$T=mg+mr{\omega }^2$

Substituting the given values ($g=10{\text{ m/s}}^2$ and ${\pi }^2=10$):

$T=m(10)+m(0.49)({\pi }^2)$

$T=10m+m(0.49)(10)$

$T=10m+4.9m=14.9m\text{ N}$

Therefore, the tension is nearly $15m\text{ N}$.

Answer: $(15\mathrm{m})\mathrm{N}$