Question

A spring of spring constant k is compressed by an amount of x and a mass m is just placed in contact with it and released. Calculate minimum value of x for which the mass m completes the vertical circle in the circular track.

• A. $\sqrt{2gR}\sqrt{\frac{m}{k}}$
• B. $\sqrt{\frac{5gR}{2}}\sqrt{\frac{m}{k}}$
• C. $\sqrt{5gR}\sqrt{\frac{m}{k}}$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{5gR}\sqrt{\frac{m}{k}}$

Answer 2

To solve this problem, we need to apply the principle of conservation of mechanical energy and the condition for completing a vertical circular motion.

Step 1: Energy stored in the spring and initial velocity of the mass. When the spring is compressed by an amount 'x', the potential energy stored in it is given by: $PE_{spring} = \frac{1}{2}kx^2$

When the mass 'm' is released, this potential energy is converted into kinetic energy of the mass as it leaves the spring at the natural length position. Let the velocity of the mass at this point (which is also the bottom of the circular track) be $v_{bottom}$. By conservation of energy: $\frac{1}{2}kx^2 = \frac{1}{2}mv_{bottom}^2$

This gives us the relationship between 'x' and $v_{bottom}$: $v_{bottom}^2 = \frac{kx^2}{m}$ (Equation 1)

Step 2: Condition for completing a vertical circle. For a mass to complete a vertical circle of radius 'R' on a smooth track, the minimum velocity at the highest point of the circle (the top of the loop) must be $\sqrt{gR}$. Let this velocity be $v_{top}$. So, $v_{top} = \sqrt{gR}$

Step 3: Conservation of energy in the circular track. Now, we apply the conservation of mechanical energy between the bottom of the circular track and the top of the circular track. Let the bottom of the track be the reference level for potential energy ($PE = 0$). At the bottom: Kinetic Energy ($KE_{bottom}$) = $\frac{1}{2}mv_{bottom}^2$ Potential Energy ($PE_{bottom}$) = $0$ Total Energy at bottom ($E_{bottom}$) = $\frac{1}{2}mv_{bottom}^2$

At the top: The height of the top of the circle from the bottom is $2R$. Kinetic Energy ($KE_{top}$) = $\frac{1}{2}mv_{top}^2$ Potential Energy ($PE_{top}$) = $mg(2R)$ Total Energy at top ($E_{top}$) = $\frac{1}{2}mv_{top}^2 + mg(2R)$

By conservation of mechanical energy ($E_{bottom} = E_{top}$): $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$

Step 4: Substitute $v_{top}$ and solve for $v_{bottom}$. Substitute $v_{top} = \sqrt{gR}$ into the energy conservation equation: $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}m(\sqrt{gR})^2 + mg(2R)$ $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mgR + 2mgR$ $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mgR + \frac{4}{2}mgR$ $\frac{1}{2}mv_{bottom}^2 = \frac{5}{2}mgR$ Multiply by 2: $mv_{bottom}^2 = 5mgR$ $v_{bottom}^2 = 5gR$ (Equation 2)

Step 5: Equate the expressions for $v_{bottom}^2$ and solve for x. From Equation 1, we have $v_{bottom}^2 = \frac{kx^2}{m}$. From Equation 2, we have $v_{bottom}^2 = 5gR$. Equating these two expressions: $\frac{kx^2}{m} = 5gR$ Now, solve for x: $x^2 = \frac{5mgR}{k}$ $x = \sqrt{\frac{5mgR}{k}}$ This can be written as: $x = \sqrt{5gR} \sqrt{\frac{m}{k}}$

Comparing this result with the given options: (1) $\sqrt{2gR}\sqrt{\frac{m}{k}}$ (2) $\sqrt{\frac{5gR}{2}}\sqrt{\frac{m}{k}}$ (3) $\sqrt{5gR}\sqrt{\frac{m}{k}}$ (4) None of these

The calculated value of x matches option (3).

The final answer is $\boxed{\text{(3)}}$

Explanation of the solution:

1. Energy Conversion (Spring to Kinetic): The potential energy stored in the compressed spring ($\frac{1}{2}kx^2$) is entirely converted into kinetic energy ($\frac{1}{2}mv_{bottom}^2$) of the mass as it leaves the spring at the bottom of the track. This gives $v_{bottom}^2 = \frac{kx^2}{m}$.
2. Vertical Circle Condition: For the mass to complete the vertical circle, its minimum speed at the highest point of the loop (height $2R$) must be $v_{top} = \sqrt{gR}$.
3. Energy Conservation (Bottom to Top of Loop): Apply conservation of mechanical energy between the bottom of the loop and the top of the loop. The initial kinetic energy at the bottom ($\frac{1}{2}mv_{bottom}^2$) is converted into kinetic energy ($\frac{1}{2}mv_{top}^2$) and gravitational potential energy ($mg(2R)$) at the top.
4. Solve for $v_{bottom}$: Substituting $v_{top} = \sqrt{gR}$ into the energy conservation equation yields $\frac{1}{2}mv_{bottom}^2 = \frac{5}{2}mgR$, which simplifies to $v_{bottom}^2 = 5gR$.
5. Solve for x: Equating the two expressions for $v_{bottom}^2$ (from step 1 and step 4) gives $\frac{kx^2}{m} = 5gR$. Solving for x, we get $x = \sqrt{5gR} \sqrt{\frac{m}{k}}$.

Answer: The minimum value of x is $\sqrt{5gR}\sqrt{\frac{m}{k}}$. The correct option is (3).