Question

A particle is thrown horizontally with the speed of 20m/s from the top of a high tower .What is the radius of the curvature of the particle 2 seconds later?​

Answer 1

Radius of curvature

• any surface can be defined as part of some circle with some defined radius.
• more the curvature of surface less will be its radius.

Radius of curvature (when path is not physically present)

$\implies \bf R = \dfrac{ V_{net}^2}{a_{\bot}}$

(here we will take the component of acceleration perpendicular to net velocity)

Solution:

First draw the path of resulting projectile.

net velocity after 2secs

$\rightarrow\rm V_x = U_x$

$\rightarrow\bf V_x = 20 \hat{i}\: ms^{-1}$

$\rightarrow\rm V_y = U_y - gt$

$\rightarrow\rm V_y = 0 - 10(2)$

$\rightarrow\bf V_y = - 20\hat{j}$

$\implies\bf V_{net }= 20\sqrt{2}$

Perpendicular component of g(acceleration)

• as the angle of velocities are 45°
• perpendicular component = $\rm gcos\theta$

$\implies\bf a_{\bot} = \dfrac{10}{\sqrt{2}}$

Finding ROC

$\rightarrow\rm R = \dfrac{ V_{net}^2}{a_{\bot}}$

$\rightarrow\rm R = \dfrac{ (20\sqrt{2})^2}{\dfrac{10}{\sqrt{2}}}$

$\rightarrow\rm R = \dfrac{ 20\times 20\times 2 \times \sqrt{2}}{10}$

$\rightarrow\rm R = \dfrac{ 800\sqrt{2}}{10}$

$\implies\bf R = 80\sqrt{2}$

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Answer 2

Concept:

There will be zero deceleration in horizontal direction, and g in vertical direction.

Given:

Initial horizontal velocity = 20 m/s

Initial vertical velocity = 0 m/s

time = 2 seconds

Find:

Radius of curvature of particle after 2 seconds

Solution:

Refer to the diagram attached for more clarity

Velocity in horizontal direction after 2 seconds

Vx = 20 m/s as no deceleration

Velocity in vertical direction after 2 seconds

Vy = Uy + gt

Vy = gt

Vy = 20 m/s

Find the value of theta θ

Tan θ = Vy/Vx = 20/20

Tan θ = 1

θ = 45°

Net velocity after 2 seconds

Vnet = √(Vx² + Vy²)

Vnet = √(20² + 20²)

Vnet = 20√2 m/s

Acceleration perpendicular to it will be

a( perpendicular to Vnet ) = g sinθ = 10 sinθ

a( perpendicular to Vnet ) = 10 sin45°

a( perpendicular to Vnet ) = 10 × 0.7

a( perpendicular to Vnet ) = 7 m/s²

Formula of Radius of curvature will be = Vnet / a ( perpendicular to Vnet )

Formula of Radius of curvature will be = 20√2 / 7

Formula of Radius of curvature will be = 4 m (approximately)

Hence, The radius of curvature after 2 seconds will be 4 m.

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