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Constrained motion

Problem : Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A moving down is 10 m/s. What is the velocity of B when angle θ = 60° ?

Motion of a leaning rod

One end of the ros is moving with a speed 10 m/s in vertically downward direction.

Solution : The velocity of B is not an independent velocity. It is tied to the velocity of the particle “A” as two particles are connected through a rigid rod. The relationship between two velocities is governed by the inter-particles separation, which is equal to the length of rod.

The length of the rod, in turn, is linked to the positions of particles “A” and “B” . From figure,

x = L 2 y 2

Differentiatiting, with respect to time :

v x = d x d t = 2 y 2 L 2 y 2 X d y d t = y v y L 2 y 2 = v y tan θ

Considering magnitude only,

v x = v y tan θ = 10 tan 60 0 = 10 3 m s

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Nature of velocity

Problem : The position vector of a particle is :

r = a cos ω t i + a sin ω t j

where “a” is a constant. Show that velocity vector is perpendicular to position vector.

Solution : In order to prove as required, we shall use the fact that scalar (dot) product of two perpendicular vectors is zero. Now, we need to find the expression of velocity to evaluate the dot product as intended. We can obtain the same by differentiating the expression of position vector with respect to time as :

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

To check whether velocity is perpendicular to the position vector, we evalaute the scalar product of r and v , which should be equal to zero.

r . v = 0

In this case,

r . v = ( a cos ω t i + a sin ω t j ) . ( - a ω sin ω t i + a ω cos ω t j ) - a 2 ω sin ω t cos ω t + a 2 ω 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. It is pertinent to mention here that this result can also be inferred from the plot of motion. An inspection of position vector reveals that it represents uniform circular motion as shown in the figure here.

Circular motion

The particle describes a circular path.

The position vector is always directed radially, whereas velocity vector is always tangential to the circular path. These two vectors are, therefore, perpendicular to each other.

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Problem : Two particles are moving with the same constant speed, but in opposite direction. Under what circumstance will the separation between two remains constant?

Solution : The condition of motion as stated in the question is possible, if particles are at diametrically opposite positions on a circular path. Two particles are always separated by the diameter of the circular path. See the figure below to evaluate the motion and separation between the particles.

Motion along a circular path

Two particles are always separated by the diameter of the circle transversed by the particles.

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Comparing velocities

Problem : A car of width 2 m is approaching a crossing at a velocity of 8 m/s. A pedestrian at a distance of 4 m wishes to cross the road safely. What should be the minimum speed of pedestrian so that he/she crosses the road safely?

Solution : We draw the figure to illustrate the situation. Here, car travels the linear distance (AB + CD) along the direction in which it moves, by which time the pedestrian travels the linear distance BD. Let pedestrian travels at a speed “v” along BD, which makes an angle “θ” with the direction of car.

Motion of a car and a pedestrian

The pedestrian crosses the road at angle with direction of car.

We must understand here that there may be number of combination of angle and speed for which pedestrian will be able to safely cross before car reaches. However, we are required to find the minimum speed. This speed should, then, correspond to a particular value of θ.

We can also observe that pedestrian should move obliquely. In doing so he/she gains extra time to cross the road.

From triangle BCD,

tan ( 90 - θ ) = cot θ = CD BC = CD 2 CD = 2 cot θ

Also,

cos ( 90 - θ ) = sin θ = BC BD = 2 BD BD = 2 sin θ

According to the condition given in the question, the time taken by car and pedestrian should be equal for the situation outlined above :

t = 4 + 2 cot θ 8 = 2 sin θ v

v = 8 2 sin θ + cos θ

For minimum value of speed, v θ = 0 ,

v θ = - 8 x ( 2 cos θ - sin θ ) ( 2 sin θ + cos θ ) 2 = 0 ( 2 cos θ - sin θ ) = 0 tan θ = 2

In order to evaluate the expression of velocity with trigonometric ratios, we take the help of right angle triangle as shown in the figure, which is consistent with the above result.

Trigonometric ratio

The tangent of angle is equal to 2.

From the triangle, defining angle “θ”, we have :

sin θ = 2 5

and

cos θ = 1 5

The minimum velocity is :

v = 8 2 x 2 5 + 1 5 = 8 5 = 3.57 m / s

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Questions & Answers

water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
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motion
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can anyone tell who founded equations of motion !?
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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
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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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