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 the conservation of mechanical energy, as the bead slides without friction and is acted upon by conservative forces (gravity and spring force).

Let's define the initial and final states of the bead. We will set the reference level for gravitational potential energy ($h=0$) at the bottom of the hoop.

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

• Height ($h_1$): The hoop has radius $R$. If the bottom is at $h=0$, the center is at $h=R$, and the top is at $h=2R$. So, $h_1 = 2R$.
• Initial Speed ($v_1$): The bead is released with negligible zero initial speed, so $v_1 = 0$.
• Spring Length ($L_1$): The spring is attached to the bottom of the hoop. When the bead is at the top, the distance between the top of the hoop and the bottom of the hoop is the diameter, $2R$. So, $L_1 = 2R$.
• Spring Extension ($x_1$): The equilibrium length of the spring is $R$. The extension is $x_1 = L_1 - R = 2R - R = R$.

The total mechanical energy in the initial state ($E_1$) is the sum of kinetic energy ($KE_1$), gravitational potential energy ($PE_{g1}$), and spring potential energy ($PE_{s1}$).

$E_1 = KE_1 + PE_{g1} + PE_{s1}$

$E_1 = \frac{1}{2}mv_1^2 + mgh_1 + \frac{1}{2}kx_1^2$

$E_1 = \frac{1}{2}m(0)^2 + mg(2R) + \frac{1}{2}k(R)^2$

$E_1 = 2mgR + \frac{1}{2}kR^2$

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

• Spring Length ($L_2$): We are given that the final length of the spring is $L_2 = R$.

• Spring Extension ($x_2$): Since the equilibrium length of the spring is also $R$, the extension is $x_2 = L_2 - R = R - R = 0$. This means the spring is at its natural length.

• Height ($h_2$): To find the height of the bead when the spring length is $R$, let's set up a coordinate system. Let the center of the hoop be the origin $(0,0)$. The bead is at a point $(x,y)$ on the hoop, so $x^2 + y^2 = R^2$. The bottom of the hoop is at $(0, -R)$. The distance between the bead $(x,y)$ and the bottom of the hoop $(0,-R)$ is the spring length $L_2 = R$.

  Using the distance formula:

  $L_2^2 = (x-0)^2 + (y - (-R))^2$

  $R^2 = x^2 + (y+R)^2$

  Substitute $x^2 = R^2 - y^2$ (from the hoop equation):

  $R^2 = (R^2 - y^2) + (y+R)^2$

  $R^2 = R^2 - y^2 + y^2 + 2yR + R^2$

  $R^2 = 2R^2 + 2yR$

  $0 = R^2 + 2yR$

  $2yR = -R^2$

  $y = -R/2$

  The y-coordinate of the bead is $-R/2$ relative to the center of the hoop. Since the bottom of the hoop is at $y=-R$ (relative to the center), and we set $h=0$ at the bottom, the height of the bead from the bottom is $h_2 = R + y = R + (-R/2) = R/2$.

• Final Speed ($v_2$): Let the velocity of the bead at this point be $v_2$.

The total mechanical energy in the final state ($E_2$) is:

$E_2 = KE_2 + PE_{g2} + PE_{s2}$

$E_2 = \frac{1}{2}mv_2^2 + mgh_2 + \frac{1}{2}kx_2^2$

$E_2 = \frac{1}{2}mv_2^2 + mg(R/2) + \frac{1}{2}k(0)^2$

$E_2 = \frac{1}{2}mv_2^2 + \frac{1}{2}mgR$

3. Conservation of Mechanical Energy:

Since there is no friction, the total mechanical energy is conserved: $E_1 = E_2$.

$2mgR + \frac{1}{2}kR^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}mgR$

Now, solve for $v_2$:

$\frac{1}{2}mv_2^2 = 2mgR - \frac{1}{2}mgR + \frac{1}{2}kR^2$

$\frac{1}{2}mv_2^2 = \left(2 - \frac{1}{2}\right)mgR + \frac{1}{2}kR^2$

$\frac{1}{2}mv_2^2 = \frac{3}{2}mgR + \frac{1}{2}kR^2$

Multiply the entire equation by 2:

$mv_2^2 = 3mgR + kR^2$

Divide by $m$:

$v_2^2 = 3gR + \frac{kR^2}{m}$

Take the square root:

$v_2 = \sqrt{3gR + \frac{kR^2}{m}}$

Comparing this result with the given options, it matches option (1).