Question

A ring 'A' of mass 'm' is attached to a stretched spring of force constant K, which is fixed at C on a smooth vertical circular track of radius R. Points A and C are diametrically opposite. When the ring slips from rest on the track to point B, making an angle of 30° with AC. (∠ACB = 30°) spring becomes unstretched. Find the velocity of the ring at B.

• A. $\left[ \frac{KR^2}{2m}(2-\sqrt{3})^2 + gR\sqrt{3} \right]^{1/2}$
• B. $\left[ \frac{KR^2}{m}(2-\sqrt{3})^2 + gR \right]^{1/2}$
• C. $\left[ \frac{2KR^2}{m}(2-\sqrt{3})^2 + gR\sqrt{3} \right]^{1/2}$
• D. $\left[ \frac{KR^2}{2m}(\sqrt{2}-1)^2 + gR \right]^{1/2}$

Answer 1

Answer: (B)

$\left[ \frac{KR^2}{m}(2-\sqrt{3})^2 + gR \right]^{1/2}$

Answer 2

The problem can be solved using the principle of conservation of mechanical energy, as the track is smooth (no friction) and the spring force is conservative.

1. Define Initial and Final States:

   • Initial State (Point A): The ring is at rest ($v_A = 0$). It is at the top of the vertical circular track, diametrically opposite to C.
   • Final State (Point B): The ring is at point B, where $\angle ACB = 30^\circ$. The spring is unstretched at this point. We need to find the velocity $v_B$.

2. Set Reference for Potential Energies:

   • Let's take the level of point C as the reference for gravitational potential energy ($PE_g = 0$ at C).
   • The natural length of the spring is its length when unstretched. At point B, the spring is unstretched, so its length is CB.

3. Analyze Geometry:

   • Let O be the center of the circular track of radius R.

   • Since A and C are diametrically opposite, AC is the diameter, so $AC = 2R$.

   • The angle subtended by the diameter AC at any point on the circumference is $90^\circ$. Therefore, $\angle ABC = 90^\circ$.

   • In the right-angled triangle ABC:

     • The natural length of the spring, $L_0 = CB = AC \cos(\angle ACB) = 2R \cos(30^\circ) = 2R \left(\frac{\sqrt{3}}{2}\right) = R\sqrt{3}$.

4. Calculate Initial Energies (at Point A):

   • Kinetic Energy ($KE_A$): The ring starts from rest, so $KE_A = \frac{1}{2} m v_A^2 = 0$.
   • Gravitational Potential Energy ($PE_{g,A}$): Since A is diametrically opposite to C, and C is at the bottom (relative to A being at the top), the height of A above C is $2R$. $PE_{g,A} = mg(2R)$.
   • Spring Potential Energy ($PE_{s,A}$): At point A, the spring is stretched. Its length is AC = 2R. The natural length is $L_0 = R\sqrt{3}$. The extension of the spring is $\Delta L_A = AC - L_0 = 2R - R\sqrt{3} = R(2 - \sqrt{3})$. $PE_{s,A} = \frac{1}{2} K (\Delta L_A)^2 = \frac{1}{2} K [R(2 - \sqrt{3})]^2 = \frac{1}{2} K R^2 (2 - \sqrt{3})^2$.

5. Calculate Final Energies (at Point B):

   • Kinetic Energy ($KE_B$): Let the velocity of the ring at B be $v_B$. $KE_B = \frac{1}{2} m v_B^2$.
   • Gravitational Potential Energy ($PE_{g,B}$): We need the vertical height of B above C. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. So, $\angle AOB = 2 \times \angle ACB = 2 \times 30^\circ = 60^\circ$. Let's place the center O at the origin $(0,0)$. If C is at $(0, -R)$ (bottom) and A is at $(0, R)$ (top). Point B is $60^\circ$ from A along the circle. Its coordinates are $(R \sin 60^\circ, R \cos 60^\circ) = (R\sqrt{3}/2, R/2)$. The vertical height of B from C is $y_B - y_C = R/2 - (-R) = 3R/2$. $PE_{g,B} = mg(3R/2)$.
   • Spring Potential Energy ($PE_{s,B}$): At point B, the spring is unstretched, so its extension is zero. $PE_{s,B} = 0$.

6. Apply Conservation of Mechanical Energy:

   Total Mechanical Energy at A = Total Mechanical Energy at B $KE_A + PE_{g,A} + PE_{s,A} = KE_B + PE_{g,B} + PE_{s,B}$ $0 + mg(2R) + \frac{1}{2} K R^2 (2 - \sqrt{3})^2 = \frac{1}{2} m v_B^2 + mg(3R/2) + 0$

7. Solve for $v_B$:

   Rearrange the equation to solve for $v_B^2$: $mg(2R) - mg(3R/2) + \frac{1}{2} K R^2 (2 - \sqrt{3})^2 = \frac{1}{2} m v_B^2$ $mg\left(2R - \frac{3R}{2}\right) + \frac{1}{2} K R^2 (2 - \sqrt{3})^2 = \frac{1}{2} m v_B^2$ $mg\left(\frac{4R - 3R}{2}\right) + \frac{1}{2} K R^2 (2 - \sqrt{3})^2 = \frac{1}{2} m v_B^2$ $mg\left(\frac{R}{2}\right) + \frac{1}{2} K R^2 (2 - \sqrt{3})^2 = \frac{1}{2} m v_B^2$

   Multiply the entire equation by 2: $mgR + K R^2 (2 - \sqrt{3})^2 = m v_B^2$

   Divide by $m$: $v_B^2 = \frac{mgR}{m} + \frac{K R^2}{m} (2 - \sqrt{3})^2$ $v_B^2 = gR + \frac{K R^2}{m} (2 - \sqrt{3})^2$

   Take the square root to find $v_B$: $v_B = \left[ gR + \frac{K R^2}{m} (2 - \sqrt{3})^2 \right]^{1/2}$

   This can also be written as: $v_B = \left[ \frac{KR^2}{m}(2-\sqrt{3})^2 + gR \right]^{1/2}$

   Comparing this with the given options, it matches option (B).