A CYLINDER OF RAIUS R LYING ON HORIZONTAL PLANE.WHAT IS THE MINIMUM VELOCITY OF A PARTICLE PROJECTED AT A CERTAIN ANGLE SO THAT IT JUST PASSE THE CYLINDER????AND ALSO FIND ANGLE OF PROJECTION??

To determine the minimum velocity required for a particle projected at a certain angle to just pass over a cylinder lying on a horizontal plane, we need to consider the geometry of the situation and the physics of projectile motion. Let's break this down step by step.

Understanding the Geometry

Imagine a cylinder with radius \( R \) positioned on a flat surface. For a particle to just clear the top of the cylinder, it must reach a height equal to the cylinder's radius at the point directly above the cylinder. This height is crucial for our calculations.

Projectile Motion Basics

When a particle is projected at an angle \( \theta \) with an initial velocity \( v_0 \), its motion can be analyzed using the equations of projectile motion. The vertical and horizontal components of the initial velocity are given by:

• Vertical component: \( v_{0y} = v_0 \sin(\theta) \)
• Horizontal component: \( v_{0x} = v_0 \cos(\theta) \)

Finding the Required Height

To just clear the cylinder, the particle must reach a height of \( R \) at a horizontal distance equal to the radius of the cylinder. The time \( t \) it takes to reach this horizontal distance can be calculated using the horizontal motion equation:

Distance = Velocity × Time, or \( R = v_{0x} t \), which gives us:

\( t = \frac{R}{v_0 \cos(\theta)} \)

Vertical Motion Equation

Now, we can use the vertical motion equation to find the height at time \( t \):

Height = \( v_{0y} t - \frac{1}{2} g t^2 \)

Substituting for \( t \) and the vertical component:

\( R = v_0 \sin(\theta) \left( \frac{R}{v_0 \cos(\theta)} \right) - \frac{1}{2} g \left( \frac{R}{v_0 \cos(\theta)} \right)^2 \)

Rearranging the Equation

After simplifying, we get:

\( R = R \tan(\theta) - \frac{g R^2}{2 v_0^2 \cos^2(\theta)} \)

Rearranging this gives:

\( \frac{g R^2}{2 v_0^2 \cos^2(\theta)} = R \tan(\theta) \)

Which simplifies to:

\( v_0^2 = \frac{g R}{2 \cos^2(\theta) \tan(\theta)} \)

Finding the Minimum Velocity

To minimize \( v_0 \), we can express \( \tan(\theta) \) in terms of \( \sin(\theta) \) and \( \cos(\theta) \):

\( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

Substituting this back into our equation allows us to find the optimal angle. The minimum velocity occurs when \( \theta = 45^\circ \), as this angle maximizes the range of projectile motion.

Final Calculation

At \( \theta = 45^\circ \), \( \tan(45^\circ) = 1 \) and \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \). Plugging these values into our equation gives:

\( v_0^2 = \frac{g R}{2 \cdot \frac{1}{2}} = \frac{g R}{1} \)

Thus, the minimum velocity \( v_0 \) required is:

\( v_0 = \sqrt{g R} \)

Summary

In summary, the minimum velocity of the particle projected at an angle of \( 45^\circ \) to just pass over the cylinder is \( v_0 = \sqrt{g R} \). This analysis combines the principles of projectile motion with geometric considerations to ensure the particle clears the cylinder effectively.