A particle is projected horizontally from the top of a tower with a ve...
Let the particle be projected horizontally with an initial velocity u. Let further its velocity be v at a point P after some time making an angle α with the horizontal. The horizontal velocity does not change with time. It is u at P. We have
vcosα=u⇒cosα=uv
Now g is the acceleration due to gravity acting downward. Its component gcosα is perpendicular to instantaneous velocity at P and hence it is directed towards the centre of the local curvature of the trajectory at P. Thus this is the centripetal acceleration at that point. Hence
gcosα=v2r
Substituting the value of cosα from equation (1), we have
guv=v2r
Or r=v3ug⇒r∝v3
This question is part of UPSC exam. View all NEET courses
A particle is projected horizontally from the top of a tower with a ve...
Introduction:
When a particle is projected horizontally from the top of a tower with an initial velocity v0, it follows a curved path under the influence of gravity. At any instant, the velocity of the particle is represented by v. The question asks us to determine the relationship between the radius of curvature of the particle's path and its velocity at any instant.
Explanation:
To understand the relationship between the radius of curvature and velocity, let's consider the motion of the particle.
1. Initial Velocity:
- The particle is projected horizontally, which means its initial velocity is entirely in the horizontal direction.
- Since there is no initial vertical velocity, the particle falls vertically under the influence of gravity.
2. Effect of Gravity:
- As the particle falls, its vertical velocity increases due to the acceleration caused by gravity.
- The horizontal velocity remains constant throughout the motion.
3. Path of the Particle:
- The resulting motion of the particle is a projectile motion, which is a combination of horizontal motion (constant velocity) and vertical motion (uniformly accelerated motion).
- The path followed by the particle is a parabolic curve.
4. Radius of Curvature:
- The radius of curvature is a measure of how sharply a particle is turning at a particular point on its path.
- In this case, the particle is moving horizontally, so its radius of curvature is determined by the vertical motion.
- The radius of curvature is inversely proportional to the acceleration in the vertical direction.
5. Acceleration in the Vertical Direction:
- The only force acting in the vertical direction is gravity, which causes the particle to accelerate downwards.
- The acceleration due to gravity is constant and equal to g (acceleration due to gravity).
- The vertical acceleration remains constant throughout the motion.
6. Velocity and Centripetal Acceleration:
- At any instant, the particle has both horizontal and vertical components of velocity.
- The centripetal acceleration, which keeps the particle moving in a curved path, is provided by the vertical component of velocity.
- The centripetal acceleration is given by the equation: ac = v^2 / r, where v is the velocity and r is the radius of curvature.
Conclusion:
From the above explanation, we can conclude that the radius of curvature of the particle's path is directly proportional to the square of its velocity at any instant. Therefore, the correct answer is b) v^2.
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.