Solution to Question 25
Let the hand move in a circle of radius $r$ with speed $v$. The angular velocity of the system is: $$ \omega = \frac{v}{r} $$ The stone moves in a concentric circle of radius $R$ with the same angular velocity $\omega$. The speed of the stone is: $$ v_s = \omega R = \frac{v}{r} R $$
The forces acting on the stone are the tension $T$ in the cord and the air resistance $f$. Since gravity is neglected, the vertical equilibrium is not considered.
1. Geometry:
Consider the triangle formed by the center of the path $O$, the hand position $A$, and the stone position $B$. The sides are $OA = r$, $OB = R$, and $AB = l$. Let $\phi$ be the angle between the radius vector of the stone ($R$) and the cord ($l$).
Using the Law of Cosines on $\triangle OAB$:
$$ r^2 = R^2 + l^2 – 2Rl \cos \phi $$
$$ \cos \phi = \frac{R^2 + l^2 – r^2}{2Rl} $$
From this, we can find $\tan \phi$:
$$ \tan \phi = \sqrt{\frac{1}{\cos^2 \phi} – 1} = \sqrt{\left(\frac{2Rl}{R^2 + l^2 – r^2}\right)^2 – 1} $$
2. Dynamics:
Resolving the tension $T$ into radial and tangential components relative to the stone’s circular path:
- Radial Component: Provides the necessary centripetal force. $$ T \cos \phi = m \omega^2 R = m \left(\frac{v}{r}\right)^2 R $$
- Tangential Component: Balances the air resistance $f$ (since speed is uniform). $$ f = T \sin \phi $$
Substituting the geometric expression for $\tan \phi$: $$ f = \frac{m v^2 R}{r^2} \sqrt{\frac{4R^2 l^2}{(R^2 + l^2 – r^2)^2} – 1} $$