91Ó°ÊÓ

We know that the mass of the object is$m$, the length of the string is $L$, and the angle of the string with the vertical is $\mathrm{θ}$. We also know that the acceleration due to gravity is $g$.

We have to find out the expression for the angular velocity $\mathrm{Ó\neg }$.

Consider the tension in the string is $T$and the radius of the circle is $r$. Then we have

$r=L\cos \mathrm{θ}$

Now the from the figure we can see that the vertical component of the tension will balance the weight. Hence, we have

$T=\frac{mg}{\sin \mathrm{θ}}$

And the horizontal component of the tension will balance the centrifugal force. Hence, we have

$\frac{mg}{\sin \mathrm{θ}}\cos \mathrm{θ}=m{\mathrm{Ó\neg }}^{2}L\cos \mathrm{θ}$

Now solving the above equation for $\mathrm{Ó\neg }$ we have

The expression of the angular velocity is $\mathrm{Ó\neg }=\sqrt{\frac{g}{L\sin \mathrm{θ}}}$.

We know that the mass of the rock is $m=500\mathrm{g}=0.500\mathrm{kg}$, the length of the string is $L=1.0\mathrm{m}$, the angle with horizontal is $\mathrm{θ}=10°$, and the gravitational acceleration is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

We have to calculate the angular velocity of the rock.

Solution

We will use the expression derived in the previous section, that is the angular velocity is given by

$\mathrm{Ó\neg }=\sqrt{\frac{g}{L\sin \mathrm{θ}}}$

Substituting the values we have,

$\mathrm{Ó\neg }=\sqrt{\frac{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{(1.0\mathrm{m})\sin \left(10°\right)}}$

$=(7.52\mathrm{rad}/\mathrm{s})\frac{(60\mathrm{s}/\min )}{(2\mathrm{Ï€}\mathrm{rad}/\text{revolution})}$

$=72\text{revolution}/\min$

The angular velocity of the rock is$72rpm$.