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a = | a | = k v y = k v 0

Motion of a balloon

The acceleration of the balloon has two components in mutually perpendicular directions.

Thus, we see that total acceleration is not only one dimensional, but constant as well. However, this does not mean that component accelerations viz tangential and normal accelerations are also constant. We need to investigate their expressions. We can obtain tangential acceleration as time rate of change of the magnitude of velocity i.e. the time rate of change of speed. We, therefore, need to first know an expression of the speed. Now, speed is :

v = k y 2 + v 0 2

Differentiating with respect to time, we have :

a T = đ v đ t = 2 k 2 y 2 k 2 y 2 + v 0 2 x d y d t

a T = d v d t = k 2 y v 0 k 2 y 2 + v 0 2

In order to find the normal acceleration, we use the fact that total acceleration is vector sum of two mutually perpendicular tangential and normal accelerations.

a 2 = a T 2 + a N 2

a N 2 = a 2 a T 2 = k 2 v 0 2 k 4 y 2 v 0 2 k 2 y 2 + v 0 2

a N 2 = k 2 v 0 2 { 1 k 2 y 2 k 2 y 2 + v 0 2 }

a N 2 = k 2 v 0 2 { k 2 y 2 + v 0 2 k 2 y 2 k 2 y 2 + v 0 2 }

a N 2 = k 2 v 0 4 k 2 y 2 + v 0 2

a N = k v 0 2 k 2 y 2 + v 0 2

Nature of motion

Problem : The coordinates of a particle moving in a plane are given by x = A cos(ωt) and y = B sin (ωt) where A, B (<A) and “ω” are positive constants of appropriate dimensions. Prove that the velocity and acceleration of the particle are normal to each other at t = π/2ω.

Solution : By differentiation, the components of velocity and acceleration are as given under :

The components of velocity in “x” and “y” directions are :

đ x đ t = v x = - A ω sin ω t

đ y đ t = v y = B ω cos ω t

The components of acceleration in “x” and “y” directions are :

đ 2 x đ t 2 = a x = - A ω 2 cos ω t

đ 2 y đ t 2 = a y = - B ω 2 sin ω t

At time, t = π 2 ω and θ = ω t = π 2 . Putting this value in the component expressions, we have :

Motion along elliptical path

Velocity and acceleration are perpendicular at the given instant.

v x = - A ω sin ω t = - A ω sin π / 2 = - A ω

v y = B ω cos ω t = B ω cos π / 2 = 0

a x = - A ω 2 cos ω t = - A ω 2 cos π / 2 = 0

a y = - B ω 2 sin ω t = - b ω 2 sin π / 2 = - B ω 2

The net velocity is in negative x-direction, whereas net acceleration is in negative y-direction. Hence at t = π 2 ω , velocity and acceleration of the particle are normal to each other.

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Problem : Position vector of a particle is :

r = a cos ω t i + a sin ω t j

Show that velocity vector is perpendicular to position vector.

Solution : We shall use a different technique to prove as required. We shall use the fact that scalar (dot) product of two perpendicular vectors is zero. We, therefore, need to find the expression of velocity. We can obtain the same by differentiating the expression of position vector with respect to time as :

v = đ r đ t = a sin ω t i + a cos ω t j

To check whether velocity is perpendicular to the position vector, we take the scalar product of r and v as :

r . v = a cos ω t i + a sin ω t j . - a sin ω t i + a cos ω t j

r . v = - a sin ω t cos ω t + a sin ω t cos ω t = 0

This means that the angle between position vector and velocity are at right angle to each other. Hence, velocity is perpendicular to position vector.

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Displacement in two dimensions

Problem : The coordinates of a particle moving in a plane are given by x = A cos(ω t) and y = B sin (ω t) where A, B (<A) and ω are positive constants of appropriate dimensions. Find the displacement of the particle in time interval t = 0 to t = π/2 ω.

Solution : In order to find the displacement, we shall first know the positions of the particle at the start of motion and at the given time. Now, the position of the particle is given by coordinates :

x = A cos ω t

and

y = B sin ω t

At t = 0, the position of the particle is given by :

x = A cos ω x 0 = A cos 0 = A

y = B sin ω x 0 = B sin 0 = 0

At t = π 2 ω , the position of the particle is given by :

x = A cos ω x π / 2 ω = A cos π / 2 = 0

y = B sin ω x π / 2 ω = a sin π / 2 = B

Motion along an elliptical path

The linear distance equals displacement.

Therefore , the displacement in the given time interval is :

r = A 2 + B 2

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Questions & Answers

if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
can anyone tell who founded equations of motion !?
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n=a+b/T² find the linear express
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Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
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Torque
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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