What is linear velocity? (With exercises solved)

The linear velocity it is defined as one that is always tangential to the path followed by the particle, regardless of its shape. If the particle always moves in a rectilinear path, there is no problem in imagining how the velocity vector follows this straight line.

However, in general the movement is carried out on a curve of arbitrary shape. Each portion of the curve can be modeled as if it were part of a circle of radius to, which at every point is tangent to the path followed.

In this case, the linear velocity is accompanying the curve tangentially and at all times at each point of it..

Mathematically the instantaneous linear velocity is the derivative of the position with respect to time. Be r the position vector of the particle at an instant t, then the linear velocity is given by the expression:

v = r'(t) = dr / dt

This means that linear velocity or tangential velocity, as it is also often called, is nothing other than the change in position with respect to time..

Article index

• 1 Linear velocity in circular motion
  • 1.1 Linear velocity, angular velocity, and centripetal acceleration
  • 1.2 -Solved exercise 1 
  • 1.3 -Solved exercise 2
• 2 References

Linear speed in circular motion

When the movement is on a circumference, we can go next to the particle at each point and see what happens in two very special directions: one of them is the one that always points towards the center. This is the address radial.

The other important direction is the one that passes on the circumference, this is the direction tangential and linear velocity always has it.

In the case of uniform circular motion, it is important to realize that the velocity is not constant, since the vector changes its direction as the particle rotates, but its modulus (the size of the vector), which is the speed, yes remains unchanged.

For this movement the position as a function of time is given by s (t), where s is he arc run Y t It's the time. In that case the instantaneous speed is given by the expression v = ds / dt and it's constant.

If the magnitude of the speed also varies (we already know that the direction always does, otherwise the mobile could not turn), we are facing a varied circular movement, during which the mobile, in addition to turning, can brake or accelerate.

Linear velocity, angular velocity, and centripetal acceleration

The motion of the particle can also be seen from the point of view of the swept angle, instead of doing it from the archway. In such a case we speak of the angular velocity. For a movement on a circumference of radius R, there is a relationship between arc (in radians) and angle:

s = R θ

Deriving with respect to time on both sides:

ds / dt = R (dθ/ dt)

Calling the derivative of θ with respect to t What angular velocity and denoting it with the Greek letter ω "omega", we have this relationship:

v = ωR

Centripetal acceleration

All circular motion has centripetal acceleration, which is always directed towards the center of the circumference. She ensures that the speed changes to move with the particle as it rotates.

Centripetal acceleration to_c or to_R it always points to the center (see figure 2) and is related to linear velocity like this:

to_c = v^two / R

And with the angular velocity as:

to_c = (ωR)^two / R = ω^twoR

For uniform circular motion, the position s (t) is of the form:

s (t) = so + vt

Furthermore, the varied circular motion must have a component of acceleration called tangential acceleration to_T, which deals with changing the magnitude of linear velocity. Yes to_T  it's constant, the position is:

s (t) = s_or + v_ort + ½ a_Tt^two

With v_or as initial velocity.

Solved problems of linear velocity

The solved exercises help to clarify the proper use of the concepts and equations given above..

-Solved exercise 1 

An insect moves on a semicircle of radius R = 2 m, starting from rest at point A while increasing its linear speed, at a rate of p m / s^two. Find: a) After how long it reaches point B, b) The linear velocity vector at that instant, c) The acceleration vector at that instant.

Solution

a) The statement indicates that the tangential acceleration is constant and is equal to π m / s^two, then it is valid to use the equation for uniformly varied motion:

s (t) = s_or + v_ort + ½ a_T.t^two

With s_or = 0 and v_or = 0:

s (t) = ½ a_T.t^two

s = πR (Half the length of the circumference)

t = (2. πR /to_T) ^½ s = (2π.2 / π)^½s = 2 s

b) v (t) = v_or + to_T. t = 2π m / s

When at point B, the linear velocity vector points in the vertical direction down in the direction (-Y):

v (t) = 2π m / s(-Y)

c) We already have the tangential acceleration, the centripetal acceleration is missing to have the velocity vector to:

to_c = v^two / R = (2π)^two / 2 m / s^two = 2π^two m / s^two

to = a_c (-x) + a_T (-Y) = 2π^two(-x) + π (-Y) m / s^two

-Exercise solved 2

A particle rotates in a circle of radius 2.90 m. At a particular instant, its acceleration is 1.05 m / s^two in a direction such that it forms 32º with its direction of movement. Find its linear velocity at: a) This moment, b) 2 seconds later, assuming that the tangential acceleration is constant.

Solution

a) The direction of movement is precisely the tangential direction:

to_T = 1.05 m / s^two . cos 32º = 0.89 m / s^two ; to_C = 1.05 m / s^two . sin 32º = 0.56 m / s^two

Speed ​​clears from to_c = v^two / R What:

v = (R.a_c)^1/2  = 1.27 m / s

b) The following equation is valid for uniformly varied motion: v = v_or + to_Tt = 1.27 + 0.89 .2^two m / s = 4.83 m / s

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 84-88.
2. Figueroa, D. Physics Series for Sciences and Engineering. Volume 3rd. Edition. Kinematics. 199-232.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6^th… Ed Prentice Hall. 62-64.
4. Relative Motion. Recovered from: courses.lumenlearning.com
5. Wilson, J. 2011. Physics 10. Pearson Education. 166-168.