A particle is connected to a string of length l whose one end is fixed , the particle has velocity root 3.5 gl at the lowermost point. Then, what will be the minimum velocity of the particle in subsequent motion ?

To determine the minimum velocity of the particle in its subsequent motion after reaching the lowermost point, we can analyze the situation using principles of energy conservation and circular motion. The particle is attached to a string and moves in a vertical circular path. At the lowest point, it has a certain kinetic energy, and as it moves upward, this kinetic energy will convert into potential energy.

Understanding the Initial Conditions

At the lowest point of the swing, the particle has a velocity of \( v = \sqrt{3.5gl} \). Here, \( g \) represents the acceleration due to gravity, and \( l \) is the length of the string. The kinetic energy (KE) at this point can be calculated using the formula:

• KE = \( \frac{1}{2} mv^2 \)

Substituting the velocity into the kinetic energy formula gives:

• KE = \( \frac{1}{2} m (\sqrt{3.5gl})^2 = \frac{1}{2} m (3.5gl) = 1.75 mgl \)

Potential Energy at the Highest Point

As the particle moves upward, it gains potential energy (PE) while losing kinetic energy. The potential energy at a height \( h \) is given by:

• PE = \( mgh \)

At the highest point of its motion, the height \( h \) can be expressed as \( h = l \) (the length of the string). Therefore, the potential energy at the highest point is:

• PE = \( mgl \)

Energy Conservation Principle

According to the conservation of mechanical energy, the total mechanical energy at the lowest point must equal the total mechanical energy at the highest point. Thus, we can set up the equation:

• KE (lowest point) = PE (highest point) + KE (highest point)

Substituting the known values:

• 1.75 mgl = mgl + KE (highest point)

Now, we can solve for the kinetic energy at the highest point:

• KE (highest point) = 1.75 mgl - mgl = 0.75 mgl

Finding the Minimum Velocity

The kinetic energy at the highest point can also be expressed in terms of velocity:

• KE = \( \frac{1}{2} mv^2 \)

Setting this equal to our expression for kinetic energy at the highest point gives:

• \( \frac{1}{2} mv^2 = 0.75 mgl \)

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

• \( \frac{1}{2} v^2 = 0.75 gl \)

Multiplying both sides by 2 leads to:

• v^2 = 1.5 gl

Taking the square root gives us the minimum velocity:

• v = \( \sqrt{1.5gl} \)

Conclusion

Thus, the minimum velocity of the particle in its subsequent motion, particularly at the highest point of its trajectory, is \( \sqrt{1.5gl} \). This analysis illustrates how energy conservation plays a crucial role in understanding the dynamics of the particle's motion in a vertical circular path.