Question

A bead of mass 'm' slides without friction on the wall of a vertical circular hoop of radius 'R' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is 'R'. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes 'R', would be (spring constant is 'k', g is acceleration due to gravity)

• A. $\sqrt{3Rg + \frac{kR^2}{m}}$
• B. $2\sqrt{gR + \frac{kR^2}{m}}$
• C. $\sqrt{2Rg + \frac{kR^2}{m}}$
• D. $\sqrt{2Rg + \frac{4kR^2}{m}}$

Answer 1

Answer: (A)

$\sqrt{3Rg + \frac{kR^2}{m}}$

Answer 2

The problem involves a bead sliding without friction on a vertical circular hoop under the influence of gravity and a spring force. Since friction is absent and both gravity and spring forces are conservative, the total mechanical energy of the system is conserved.

We define the initial and final states of the bead and apply the principle of conservation of mechanical energy:

$E_{initial} = E_{final}$

$KE_{initial} + PE_{gravity,initial} + PE_{spring,initial} = KE_{final} + PE_{gravity,final} + PE_{spring,final}$

Let's set the reference level for gravitational potential energy ($PE_g = 0$) at the bottom of the hoop. The radius of the hoop is $R$. The natural length of the spring ($L_0$) is $R$.

1. Initial State (Bead at the top of the hoop):

• Position: The bead is at the very top of the hoop.
• Height: Relative to the bottom of the hoop, the height ($h_i$) is the diameter, so $h_i = 2R$.
• Speed: The bead is released with negligible zero initial speed, so $v_i = 0$.
• Spring Length: The spring is attached to the bottom of the hoop. When the bead is at the top, the distance between the bead and the attachment point (bottom of the hoop) is the diameter, $L_i = 2R$.
• Spring Extension: The extension of the spring is $\Delta L_i = L_i - L_0 = 2R - R = R$.

Therefore, the initial energies are:

• Kinetic Energy: $KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}m(0)^2 = 0$
• Gravitational Potential Energy: $PE_{g,i} = mgh_i = mg(2R)$
• Spring Potential Energy: $PE_{s,i} = \frac{1}{2}k(\Delta L_i)^2 = \frac{1}{2}kR^2$

Total initial energy: $E_i = 0 + 2mgR + \frac{1}{2}kR^2$

2. Final State (When the length of the spring becomes 'R'):

• Position: The spring is attached to the bottom of the hoop (let's call this point B). The bead is at some point P on the hoop. We are given that the length of the spring $BP = R$.

  • Let C be the center of the hoop. The distance from the center to the bottom of the hoop is $CB = R$ (radius).
  • Since the bead P is on the hoop, the distance from the center to the bead is $CP = R$ (radius).
  • Thus, we have a triangle BCP with sides $CB=R$, $CP=R$, and $BP=R$. This means triangle BCP is an equilateral triangle.
  • In an equilateral triangle, all angles are $60^\circ$. So, the angle between the vertical line CB and the radius CP is $60^\circ$.
  • Let's find the height of the bead P from the bottom of the hoop. If we consider the center C at height $R$ from the bottom, the vertical distance of P from C is $R \cos(60^\circ) = R/2$. Since the angle is measured from the downward vertical (CB), the bead is $R/2$ above the horizontal line passing through the center.
  • So, the height of the bead P from the bottom of the hoop is $h_f = R + R \cos(60^\circ) = R + R/2 = 3R/2$.
  • Wait, let's re-verify the height. If the bottom of the hoop is at $(0,0)$, the center is at $(0,R)$. The bead P is at a point $(x,y)$ on the circle $(x-0)^2 + (y-R)^2 = R^2$. The spring is attached at $(0,0)$. The length of the spring is $L = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2+y^2}$.
  • Alternatively, using the equilateral triangle property, if the center is at $(0,0)$, the bottom is at $(0,-R)$. The bead P is at a position such that the vector CP makes an angle of $60^\circ$ with the negative y-axis (vector CB). So, the coordinates of P are $(R\sin(60^\circ), -R\cos(60^\circ)) = (R\sqrt{3}/2, -R/2)$.
  • The height of the bead from the bottom of the hoop (which is at $y=-R$) is $h_f = y - (-R) = -R/2 + R = R/2$. This is the correct height.

• Speed: Let the final speed be $v$.

• Spring Length: The problem states the spring length becomes $R$, so $L_f = R$.

• Spring Extension: Since the natural length of the spring is $L_0 = R$, the final extension is $\Delta L_f = L_f - L_0 = R - R = 0$.

Therefore, the final energies are:

• Kinetic Energy: $KE_f = \frac{1}{2}mv^2$
• Gravitational Potential Energy: $PE_{g,f} = mgh_f = mg(R/2)$
• Spring Potential Energy: $PE_{s,f} = \frac{1}{2}k(\Delta L_f)^2 = \frac{1}{2}k(0)^2 = 0$

Total final energy: $E_f = \frac{1}{2}mv^2 + \frac{1}{2}mgR + 0$

3. Apply Conservation of Mechanical Energy:

$E_i = E_f$

$2mgR + \frac{1}{2}kR^2 = \frac{1}{2}mv^2 + \frac{1}{2}mgR$

Now, solve for $v$:

Multiply the entire equation by 2 to clear fractions:

$4mgR + kR^2 = mv^2 + mgR$

Rearrange to solve for $mv^2$:

$mv^2 = 4mgR - mgR + kR^2$

$mv^2 = 3mgR + kR^2$

Divide by $m$:

$v^2 = \frac{3mgR + kR^2}{m}$

$v^2 = 3gR + \frac{kR^2}{m}$

Take the square root:

$v = \sqrt{3gR + \frac{kR^2}{m}}$

Comparing this result with the given options:

(1) $\sqrt{3Rg + \frac{kR^2}{m}}$ (2) $2\sqrt{gR + \frac{kR^2}{m}}$ (3) $\sqrt{2Rg + \frac{kR^2}{m}}$ (4) $\sqrt{2Rg + \frac{4kR^2}{m}}$

The calculated velocity matches option (1).