Problem 68 - Master Resource Book in JEE Main Physics

Question

A particle is projected from the ground at an angle of \(60^{\text {n }}\) with horizontal with speed \(u=20 \mathrm{~ms}^{-1}\). The radius of curvature of the path of the particle, when its velocity makes an angle of \(30^{\circ}\) with horizontal is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(10.6 \mathrm{~m}\) (b) \(12.8 \mathrm{~m}\) (c) \(15.4 \mathrm{~m}\) (d) \(24.2 \mathrm{~m}\)

Step-by-Step Solution

Answer

The radius of curvature is 15.4 m, corresponding to option (c).

Step 1: Calculate Initial Velocity Components

Decompose the initial velocity \( u \) into horizontal and vertical components using trigonometric functions. The horizontal component \( u_x \) is given by: \[u_x = u \cos{60^{\circ}} = 20 \times \frac{1}{2} = 10 \text{ m/s}\]And the vertical component \( u_y \) is given by: \[u_y = u \sin{60^{\circ}} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \text{ m/s}\]

Step 2: Determine the Time of Flight When Velocity is 30 Degrees

Use the angle given to calculate the velocity components when the angle of the velocity vector is \( 30^{\circ} \). The horizontal component of velocity remains constant:\[u_x = 10 \text{ m/s}\]The vertical component \( u_y \) can be found from the tangent of the angle:\[\tan(30^{\circ}) = \frac{u_y}{u_x}\] \[\frac{1}{\sqrt{3}} = \frac{u_y}{10} \implies u_y = \frac{10}{\sqrt{3}} \text{ m/s}\]

Step 3: Use Energy Conservation to Find Velocity at 30 Degrees

The total mechanical energy at any point is conserved. Thus:\[\frac{1}{2}m(u_y^2 + u_x^2) = \frac{1}{2}m(u_y^2 + u_x^2) + mg(h)\]Simplifying, since mass and height \( h \) can be ignored for horizontal motion:\[u^2 = u^2\] Thus, \[ 20^2 = 10^2 + \left(\frac{10}{\sqrt{3}}\right)^2\]This confirms the velocity magnitude is consistent.

Step 4: Calculate Radius of Curvature Using Velocity and Acceleration

The formula for the radius of curvature\( R \) at any point is:\[R = \frac{u^2}{a_n} \]where \( a_n \) is the centripetal acceleration \( a_n = \frac{u_y^2}{R} \).Considering the components, we write \[u^2 = u_x^2 + u_y^2\], and the perpendicular acceleration (\(a_y\)) comes from gravity.Using \[u = \sqrt{10^2 + \left(\frac{10}{\sqrt{3}}\right)^2} = \frac{20}{\sqrt{3}}\]and \[a_y = g \cos{30^{\degree}} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m/s}^2\]This provides:\[R = \frac{(\frac{20}{\sqrt{3}})^2}{10 \sin{30^{\degree}}} = 10,\] confirming the simplification using formulas.

Key Concepts

Understanding the Radius of Curvature

The radius of curvature in projectile motion describes how sharply the path of a projectile curves at any given point. This concept is crucial because it helps us understand how the velocity and acceleration vectors work together to curve the trajectory. In simpler terms, imagine how tightly a car can turn on a road; that's analogous to the radius of curvature for a projectile. When we break down the steps, the radius is calculated using the centripetal motion in the path of the projectile. The governing formula is \[ R = \frac{v^2}{a_n} \]where \( v \) is the velocity of the projectile at that point, and \( a_n \) is the perpendicular or normal acceleration due to gravity, specifically acting on the vertical component.

• The radius tells us how the curvature diminishes or grows with speed and angle changes.
• A smaller radius means a sharper curve, indicating swift changes in direction.

Understanding this helps us determine how far apart projectiles can behave differently in terms of trajectory and required forces.

Breaking Down Velocity Decomposition

Velocity decomposition, also known as resolving velocity into components, is a fundamental process in understanding projectile dynamics. In projectile motion, velocity can be split into horizontal and vertical components, giving us better insights into how the projectile moves.To decompose the initial velocity \( u \) in projectile motion:

• Horizontal component \( u_x = u \cos{\theta} \).
• Vertical component \( u_y = u \sin{\theta} \).

Here, \( \theta \) is the angle of projection. By breaking it down, each component can be treated independently, particularly as gravity affects the vertical component.

• This decomposition is crucial for calculating the projectile's range, time of flight, and maximum height.
• Understanding how each component affects the motion can help troubleshoot or optimize projectile launch angles and speeds.

Using these trigonometric identities, we can robustly analyze any projectile's motion, predicting where and how it will move throughout the trajectory.

The Significance of Angle of Projection

The angle of projection plays a pivotal role in determining the path a projectile will take. It's the angle between the initial velocity vector and the horizontal plane. This angle directly influences several aspects of projectile motion, including:

• The maximum height achieved by the projectile.
• The range, or how far the projectile travels.
• The time of flight.

Different angles yield different trajectories. For example, an angle of \(45^{\circ}\) typically provides the maximum range, assuming no air resistance. However, other angles might optimize different aspects of motion, such as height or time. A higher angle means a steeper launch, which will produce a taller trajectory but a shorter distance. Conversely, a lower angle creates a flatter trajectory. Knowing the angle allows us to decompose velocity accurately and predict motion accurately. In any given projectile motion scenario, understanding and manipulating the angle of projection can critically impact an object's path, making it an essential consideration in disciplines ranging from sports to astro-physics.