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The velocity of the object at the point, where motion changes direction, is zero and force is acting downwards. This situation is same as the case 1. The object is at rest. Hence, object moves in the direction of force i.e. in the downward direction.

In the figure above, the vectors drawn at various points represent the direction and magnitude of force ( F ), acceleration ( a ) and velocity ( v ) during the motion. Note that both force and acceleration act in the downward direction during the motion.

4: When force is applied perpendicular to the direction of motion, then it causes change in the direction of the velocity.

A simple change of direction also constitutes change in velocity and, therefore, acceleration. Consider the case of a uniform circular motion in which a particle moves along a circular path at constant speed “v”. Let v 1 and v 2 be the velocities of the particle at two time instants, then

| v 1 | = | v 2 | = a constant

Uniform circular motion

A central force perpendicular to motion causes change in direction.

In the adjoining figure, the vector segments OC and OD represent v 1 and v 2 . Knowing that vector difference Δv is directed from initial to final position, it is represented by vector CD. Using the adjoining vector triangle,

OC + CD = OD v = CD = v 2 - v 1

The important thing to realize here is that direction of Δ v is along CD, which is directed towards the origin. This result is in complete agreement of what we know about uniform circular motion (The topic of uniform circular motion is covered in separate module). We need to apply a force (causing acceleration to the moving particle) across (i.e. perpendicular) to the motion to change direction. If the force (hence acceleration) is perpendicular to velocity, then magnitude of velocity i.e. speed remains same, whereas the direction of motion keeps changing.

Acceleration

Problem : An object moves along a quadrant AB of a circle of radius 10 m with a constant speed 5 m/s. Find the average velocity and average acceleration in this interval.

Solution : Here, we can determine time interval form the first statement of the question. The particle covers a distance of 2πr/4 with a speed v. Hence, time interval,

t = distance speed = 2 π r 4 v = 3.14 x 10 2 x 5 = 3.14 s

Magnitude of average velocity is given by the ratio of the magnitude of displacement and time :

Motion along circular path

| v avg | = = | AB | Time = ( 10 2 + 10 2 ) 3.14 14.14 3.14 = 4.5 m / s

Average velocity is directed along vector AB i.e. in the direction of displacement vector.

Magnitude of average acceleration is given by the ratio :

Motion along circular path

Direction of acceleration

a avg = = | Δ v | Δ t

Here,

| Δ v | = | v 2 - v 1 | | Δ v | = ( 5 2 + 5 2 ) = 5 2 m / s a avg = 5 2 3.14 = 3.1 m / s 2

Average acceleration is directed along the direction of vector v 2 - v 1 i.e. directed towards the center of the circle (as shown in the figure).

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5: When force is applied at a certain angle with the motion, then it causes change in both magnitude and direction of the velocity. The component of force in the direction of motion changes its magnitude (an increase or decrease), while the component of force perpendicular to the direction of motion changes its direction.

The motion of a small spherical mass “m”, tied to a fixed point with the help of a string illustrates the changes taking place in both the direction and magnitude of the velocity. Saving the details for discussion at a later point in the course, here tension in the string and tangential force provide for the change in the direction and magnitude of velocity respectively.

When the string makes at angle, θ, with the vertical, then :

Motion of a pendulum

Forces acting on the pendulum bob

(a) Tension in the string, T is :

T = mg cos θ

This force acts normal to the path of motion at all points and causes the body to continuously change its direction along the circular path. Note that this force acts perpendicular to the direction of velocity, which is along the tangent at any given point on the arc.

(b) Tangential force, F, is :

F = mg sin θ

This force acts tangentially to the path and is in the direction of velocity of the spherical mass, which is also along the tangent at that point. As the force and velocity are in the same direction, this force changes the magnitude of velocity. The magnitude of velocity increases when force acts in the direction of velocity and magnitude of velocity decreases when force acts in the opposite direction to that of velocity.

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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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