Solution to Question 34
Let the bead describe a horizontal circle of radius $r$ at a depth $z$. The tension $T$ has a vertical component balancing gravity and a horizontal component providing centripetal force. $$ T \cos \theta = mg $$ $$ T \sin \theta = \frac{m v^2}{r} $$ Dividing these gives the relationship between geometry and velocity: $$ \tan \theta = \frac{r}{z} = \frac{v^2}{rg} \implies z = \frac{r^2 g}{v^2} $$
Initially, $r = R$ and velocity is $v_0$. $$ z_1 = \frac{R^2 g}{v_0^2} $$
As the thread is pulled up slowly, there is no torque about the vertical axis passing through the hole. Thus, angular momentum is conserved. $$ m v r = \text{constant} $$ $$ v_0 R = v_2 r_2 \implies v_2 = \frac{v_0 R}{r_2} $$
The problem states the depth is reduced to half: $z_2 = z_1 / 2$. Using the equilibrium condition $z = \frac{r^2 g}{v^2}$ for the final state: $$ z_2 = \frac{r_2^2 g}{v_2^2} $$ Substitute $v_2 = v_0 R / r_2$: $$ z_2 = \frac{r_2^2 g}{(v_0 R / r_2)^2} = \frac{r_2^4 g}{v_0^2 R^2} $$ We know $z_2 = \frac{z_1}{2} = \frac{R^2 g}{2 v_0^2}$. Equating the two expressions for $z_2$: $$ \frac{r_2^4 g}{v_0^2 R^2} = \frac{R^2 g}{2 v_0^2} $$ $$ r_2^4 = \frac{R^4}{2} \implies r_2 = \frac{R}{2^{1/4}} $$
The work done by the pulling force equals the change in total mechanical energy (Kinetic + Potential). $$ W = \Delta KE + \Delta PE $$ Change in Kinetic Energy: $$ KE_i = \frac{1}{2} m v_0^2 $$ $$ KE_f = \frac{1}{2} m v_2^2 = \frac{1}{2} m \left( \frac{v_0 R}{r_2} \right)^2 = \frac{1}{2} m v_0^2 \left( \frac{R^2}{R^2 / \sqrt{2}} \right) = \frac{1}{2} m v_0^2 \sqrt{2} $$ $$ \Delta KE = \frac{1}{2} m v_0^2 (\sqrt{2} – 1) $$ Change in Potential Energy: (Taking table top as $z=0$) $$ PE_i = -mg z_1 $$ $$ PE_f = -mg z_2 = -mg \frac{z_1}{2} $$ $$ \Delta PE = PE_f – PE_i = mg \frac{z_1}{2} $$ Substitute $z_1 = \frac{R^2 g}{v_0^2}$: $$ \Delta PE = \frac{m g^2 R^2}{2 v_0^2} $$
$$ W = \frac{m v_0^2}{2} (\sqrt{2} – 1) + \frac{m g^2 R^2}{2 v_0^2} $$ Factoring out $\frac{m v_0^2}{2}$: $$ W = \frac{m v_0^2}{2} \left[ (\sqrt{2} – 1) + \frac{g^2 R^2}{v_0^4} \right] $$