Question

a. An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle $u$. Find an expression for the angular velocity$v$.

b. A student ties a $500g$ rock to a$1.0-m$-long string and swings it around her head in a horizontal circle. At what angular speed, in rpm, does the string tilt down at a$10°$ angle?

Short answer

a). The expression of the angular velocity is $\omega =\sqrt{\frac{g}{L\sin \theta }}$.

b). The angular velocity of the rock is $72\mathrm{rpm}$.

Step-by-step solution

Given Information (Part a)

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 $\theta$. We also know that the acceleration due to gravity is $g$.

Explanation (Part a)

We have to find out the expression for the angular velocity $\omega$.

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

$r=L\cos \theta$

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

$T\sin \theta =mg$

$T=\frac{mg}{\sin \theta }$

Explanation (Part a)

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

$T\cos \theta =m{\omega }^{2}r$

$\frac{mg}{\sin \theta }\cos \theta =m{\omega }^{2}L\cos \theta$

Now solving the above equation for $\omega$ we have

$\omega =\sqrt{\frac{g}{L\sin \theta }}$

Final Answer (Part a)

The expression of the angular velocity is $\omega =\sqrt{\frac{g}{L\sin \theta }}$.

Given Information (Part b)

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 $\theta =10°$, and the gravitational acceleration is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Explanation (Part b)

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

$\omega =\sqrt{\frac{g}{L\sin \theta }}$

Substituting the values we have,

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

$=7.52\mathrm{rad}/\mathrm{s}$

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

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

Final Answer (Part b)

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