Dynamics powerpoints for first 1 or 2 chapters

Engineering Mechanics: Dynamics • Dynamics – Branch of mechanics that deals with the motion of bodies under the action of forces (Accelerated Motion)

• Two distinct parts: – Kinematics • study of motion without reference to the forces that cause motion or are generated as a result of motion

– Kinetics • relates the action of forces on bodies to their resulting motions ME101 - Division III

Kaustubh Dasgupta

1

Engineering Mechanics: Dynamics • Basis of rigid body dynamics – Newton’s 2nd law of motion • A particle of mass “m” acted upon by an unbalanced force “F” experiences an acceleration “a” that has the same direction as the force and a magnitude that is directly proportional to the force • a is the resulting acceleration measured in a nonaccelerating frame of reference

ME101 - Division III

Kaustubh Dasgupta

2

Engineering Mechanics: Dynamics • Space – Geometric region occupied by bodies • Reference system – Linear or angular measurements

• Primary reference system or astronomical frame of reference – Imaginary set of rectangular axes fixed in space – Validity for measurements for velocity < speed of light – Absolute measurements

• Reference frame attached to the earth??

• Time • Mass ME101 - Division III

Kaustubh Dasgupta

3

Engineering Mechanics: Dynamics • Newton’s law of gravitation m1m2 F G 2 r

– F :: mutual force of attraction between two particles – G :: universal gravitational constant • 6.673x10-11 m3/(kg.s2)

ME101 - Division III

Kaustubh Dasgupta

4

Engineering Mechanics: Dynamics • Weight – Only significant gravitational force between the earth and a particle located near the surface

mM e W G 2 r

W  mg

• g = GMe/r2 :: acceleration due to gravity (9.81m/s2) • Variation of g with altitude

g  g0

R

2

R  h 

2

g is the absolute acceleration due to gravity at altitude h g0 is the absolute acceleration due to gravity at sea level

R is the radius of the earth ME101 - Division III

Kaustubh Dasgupta

5

Engineering Mechanics: Dynamics Effect of Altitude on Gravitation • Force of gravitational attraction of the earth on a body depends on the position of the body relative to the earth • Assuming the earth to be a perfect homogeneous sphere, a mass of 1 kg would be attracted to the earth by a force of: • 9.825 N if the mass is on the surface of the earth • 9.822 N if the mass is at an altitude of 1 km • 9.523 N if the mass is at an altitude of 100 km • 7.340 N if the mass is at an altitude of 1000 km • 2.456 N if the mass is at an altitude of equal to the mean radius of the earth, 6371 km ME101 - Division III

Kaustubh Dasgupta

6

Engineering Mechanics: Dynamics • Effect of earth’s rotation – g from law of gravitation • Fixed set of axes at the centre of the earth – Absolute value of g

• Earth’s rotation – Actual acceleration of a freely falling body is less than absolute g – Measured from a position attached to the surface of the earth

ME101 - Division III

Kaustubh Dasgupta

7

Engineering Mechanics: Dynamics • Effect of earth’s rotation

sea-level conditions

– Engineering applications :: variation of g is ignored ME101 - Division III

Kaustubh Dasgupta

8

Kinematics of Particles • Motion – Constrained :: confined to a specific path – Unconstrained :: not confined to a specific path

• Choice of coordinates – Position of P at any time t • rectangular (i.e., Cartesian) coordinates x, y, z • cylindrical coordinates r, θ, z • spherical coordinates R, θ, Φ

– Path variables • Measurements along the tangent t and normal n to the curve ME101 - Division III

Kaustubh Dasgupta

9

Kinematics of Particles â&#x20AC;¢ Choice of coordinates

ME101 - Division III

Kaustubh Dasgupta

10

Kinematics of Particles Rectilinear Motion • Motion along a straight line

t t+Δt

ME101 - Division III

Kaustubh Dasgupta

11

Kinematics of Particles :: Rectilinear Motion t t+Δt •Motion along a straight line Position at any instance of time t :: specified by its distance s measured from some convenient reference point O fixed on the line :: (disp. is negative if the particle moves in the negative s-direction).

Velocity of the particle:

ds v  s dt

Acceleration of the particle:

Both are vector quantities

ME101 - Division III

or

d 2s a  2  s dt

+ve or –ve depending on whether velocity is increasing or decreasing

+ve or –ve depending on +ve or –ve displacement

vdv  a ds

dv a  v dt

or Kaustubh Dasgupta

s ds  s ds 12

Kinematics of Particles Rectilinear Motion: Graphical Interpretations Using s-t curve, v-t & a-t curves can be plotted.

Area under v-t curve during time dt = vdt == ds • Net disp from t1 to t2 = corresponding area under v-t curve  s t

s

2

1

ds   vdt 2

t1

or s2 - s1 = (area under v-t curve) Area under a-t curve during time dt = adt == dv • Net change in vel from t1 to t2 = corresponding area under a-t curve  v t

v

2

1

dv   adt 2

t1

or v2 - v1 = (area under a-t curve)

ME101 - Division III

Kaustubh Dasgupta

13

Kinematics of Particles Rectilinear Motion: Graphical Interpretations Two additional graphical relations: Area under a-s curve during disp ds= ads == vdv • Net area under a-s curve betn position coordinates s1 and s2  v2 s2

v

1

vdv  ads s1

or ½ (v22 – v12) = (area under a-s curve) Slope of v-s curve at any point A = dv/ds • Construct a normal AB to the curve at A. From similar triangles:

CB dv  v ds •

dv  CB  v  a (accelerati on) ds

Vel and posn coordinate axes should have the same numerical scales so that the accln read on the x-axis in meters will represent the actual accln in m/s2 ME101 - Division III

Kaustubh Dasgupta

14

Kinematics of Particles :: Rectilinear Motion Analytical Integration to find the position coordinate Acceleration may be specified as a function of time, velocity, or position coordinate, or as a combined function of these.

(a) Constant Acceleration At the beginning of the interval  t = 0, s = s0, v = v0 For a time interval t: integrating the following two equations

a

dv dt

vdv  a ds Substituting in the following equation and integrating will give the position coordinate: ds v dt Equations applicable for Constant Acceleration and for time interval 0 to t ME101 - Division III

Kaustubh Dasgupta

15

Kinematics of Particles :: Rectilinear Motion Analytical Integration to find the position coordinate (b) Acceleration given as a function of time, a = f(t) At the beginning of the interval  t = 0, s = s0, v = v0 For a time interval t: integrating the following equation

a

dv  dv f (t )  dt dt

Substituting in the following equation and integrating will give the position coordinate:

ds v dt Alternatively, following second order differential equation may be solved to get the position coordinate:

d 2s a  2  s dt ME101 - Division III



s  f (t ) Kaustubh Dasgupta

16

Kinematics of Particles :: Rectilinear Motion Analytical Integration to find the position coordinate (c) Acceleration given as a function of velocity, a = f(v) At the beginning of the interval  t = 0, s = s0, v = v0 For a time interval t: Substituting a and integrating the following equation

a

dv  dt

f (v ) 

dv dt

Solve for v as a function of t and integrate the following equation to get the position coordinate: ds

v

dt

Alternatively, substitute a = f(v) in the following equation and integrate to get the position coordinate :

vdv  a ds ME101 - Division III

Kaustubh Dasgupta

17

Kinematics of Particles: Rectilinear Motion Analytical Integration to find the position coordinate (d) Acceleration given as a function of displacement, a = f(s) At the beginning of the interval  t = 0, s = s0, v = v0 For a time interval t: substituting a and integrating the following equation vdv  a ds

Solve for v as a function of s : v = g(s), substitute in the following equation and integrate to get the position coordinate: v

ds dt

It gives t as a function of s. Rearrange to obtain s as a function of t to get the position coordinate.

In all these cases, if integration is difficult, graphical, analytical, or computer methods can be utilized.

ME101 - Division III

Kaustubh Dasgupta

18

Kinematics of Particles: Rectilinear Motion Example Position coordinate of a particle confined to move along a straight line is given by s = 2t3 – 24t + 6, where s is measured in meters from a convenient origin and t is in seconds. Determine: (a) time reqd for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) acceleration of the particle when v = 30 m/s, and (c) net disp of the particle during the interval from t = 1 s to t = 4 s.

Solution Differentiating s = 2t3 – 24t + 6

 v = 6t2 – 24 m/s  a = 12t m/s2

(a) v = 72 m/s  t = ± 4 s (- 4 s happened before initiation of motion  no physical interest.) t=4s (b) v = 30 m/s  t = 3 sec  a = 36 m/s2 (c) t = 1 s to 4 s. Using s = 2t3 – 24t + 6 Δs = s4 – s1 = [2(43) – 24(4) +6] – [2(13) – 24(1) + 6] Δs = 54 m ME101 - Division III

Kaustubh Dasgupta

19

Kinematics of Particles Plane Curvilinear Motion Motion of a particle along a curved path which lies in a single plane.

For a short time during take-off and landing, planes generally follow plane curvilinear motion

ME101 - Division III

Kaustubh Dasgupta

20

Kinematics of Particles Plane Curvilinear Motion: Between A and A’: Average velocity of the particle : vav = Δr/ Δt  A vector whose direction is that of Δr and whose magnitude is magnitude of Δr/ Δt Average speed of the particle = Δs/ Δt

Instantaneous velocity of the particle is defined as the limiting value of the average velocity as the time interval approaches zero   v is always a vector tangent to the path

Extending the definition of derivative of a scalar to include vector quantity: Derivative of a vector is a vector having a magnitude and a direction.

Magnitude of v is equal to speed (scalar) ME101 - Division III

Kaustubh Dasgupta

21

Kinematics of Particles Plane Curvilinear Motion Magnitude of the derivative:

dr / dt  r  s  v  v  Magnitude of the velocity or the speed Derivative of the magnitude:

d r / dt  dr / dt  r  Rate at which the length of the position vector is changing Velocity of the particle at A  tangent vector v Velocity of the particle at A’  tangent vector v’  v’ – v = Δv  Δv Depends on both the change in magnitude of v and on the change in direction of v.

ME101 - Division III

Kaustubh Dasgupta

22

Kinematics of Particles Plane Curvilinear Motion Between A and A’: Average acceleration of the particle : aav = Δv/ Δt  A vector whose direction is that of Δv and whose magnitude is the magnitude of Δv/ Δt Instantaneous accln of the particle is defined as the limiting value of the average accln as the time interval approaches zero 

By definition of the derivative:

 In general, direction of the acceleration of a particle in curvilinear motion neither tangent to the path nor normal to the path.  Acceleration component normal to the path points toward the center of curvature of the path.

ME101 - Division III

Kaustubh Dasgupta

23

Kinematics of Particles Plane Curvilinear Motion Motion of a particle along a curved path which lies in a single plane.

For a short time during take-off and landing, planes generally follow plane curvilinear motion

ME101 - Division III

Kaustubh Dasgupta

1

Kinematics of Particles Plane Curvilinear Motion: Between A and A’: Average velocity of the particle : vav = Δr/ Δt  A vector whose direction is that of Δr and whose magnitude is magnitude of Δr/ Δt Average speed of the particle = Δs/ Δt

Instantaneous velocity of the particle is defined as the limiting value of the average velocity as the time interval approaches zero   v is always a vector tangent to the path

Extending the definition of derivative of a scalar to include vector quantity: Derivative of a vector is a vector having a magnitude and a direction.

Magnitude of v is equal to speed (scalar) ME101 - Division III

Kaustubh Dasgupta

2

Kinematics of Particles Plane Curvilinear Motion Magnitude of the derivative:

dr / dt  r  s  v  v  Magnitude of the velocity or the speed Derivative of the magnitude:

d r / dt  dr / dt  r  Rate at which the length of the position vector is changing Velocity of the particle at A  tangent vector v Velocity of the particle at A’  tangent vector v’  v’ – v = Δv  Δv Depends on both the change in magnitude of v and on the change in direction of v.

ME101 - Division III

Kaustubh Dasgupta

3

Kinematics of Particles Plane Curvilinear Motion Between A and A’: Average acceleration of the particle : aav = Δv/ Δt  A vector whose direction is that of Δv and whose magnitude is the magnitude of Δv/ Δt Instantaneous accln of the particle is defined as the limiting value of the average accln as the time interval approaches zero 

By definition of the derivative:

 In general, direction of the acceleration of a particle in curvilinear motion neither tangent to the path nor normal to the path.  Acceleration component normal to the path points toward the center of curvature of the path.

ME101 - Division III

Kaustubh Dasgupta

4

Kinematics of Particles Plane Curvilinear Motion Visualization of motion: Hodograph

Acceleration has the same relation to velocity as the velocity has to the position vector.

ME101 - Division III

Kaustubh Dasgupta

5

Kinematics of Particles Plane Curvilinear Motion Derivatives and Integration of Vectors: same rules as for scalars

V is a function of x, y, and z, and an element of volume is Integral of V over the volume is equal to the vector sum of the three integrals of its components.

ME101 - Division III

Kaustubh Dasgupta

6

Kinematics of Particles Plane Curvilinear Motion Three coordinate systems are commonly used for describing the vector relationships (for plane curvilinear motion of a particle): 1. Rectangular Coordinates  x-y 2. Normal and tangential coordinates  n-t 3. Polar coordinates  r-θ (special case of 3-D motion in which cylindrical coordinates r, θ, z are used) Choice of coordinate systems depends on  the manner in which the motion is generated  or the form in which the data is specified.

ME101 - Division III

Kaustubh Dasgupta

7

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) If all motion components are directly expressible in terms of horizontal and vertical coordinates

Time derivatives of the unit vectors are zero because their magnitude and direction remains constant.

Also, dy/dx = tan θ = vy /vx ME101 - Division III

Kaustubh Dasgupta

8

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Projectile Motion ď&#x192; An important application Assumptions: neglecting aerodynamic drag, Neglecting curvature and rotation of the earth, and altitude change is small enough such that g can be considered to be constant ď&#x192; Rectangular coordinates are useful for the trajectory analysis For the axes shown in the figure, the acceleration components are: ax = 0, ay = - g Integrating these eqns for the condition of constant accln (slide 11) will give us equations necessary to solve the problem.

ME101 - Division III

Kaustubh Dasgupta

9

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Projectile Motion Horizontal Motion: ax = 0 Integrating this eqn for constant accln condition

v  v0  at

 v x  v0 x

1 x  x0  v0 t  at 2  x  x0  v0 x t 2 v 2  v02  2ax  x0   v x  v0 x

Subscript zero denotes initial conditions: x0 = y0 = 0

+

For the conditions under discussion:  x- and y- motions are independent  Path is parabolic

Vertical Motion: ay = - g Integrating this eqn for constant accln condition

v  v0  at

 v y  v0  y  gt

1 2 1 at  y  y0  v0  y t  gt 2 2 2 2 v 2  v02  2a y  y0   v 2y  v0  y  2 g  y  y0  y  y 0  v0 t 

ME101 - Division III

Kaustubh Dasgupta

+ 10

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Common descriptions of curvilinear motion uses Path Variables: measurements made along the tangent and normal to the path of the particle.

• Positive n direction: towards the center of curvature of the path Velocity and Acceleration en = unit vector in the n-direction at point A et = unit vector in the t-direction at point A During differential increment of time dt, the particle moves a differential distance ds from A to A’.

ρ = radius of curvature of the path at A’  ds = ρ dβ Magnitude of the velocity: v = ds/dt = ρ dβ/dt  In vector form

Differentiating: Unit vector et has non-zero derivative because its direction changes.

ME101 - Division III

Kaustubh Dasgupta

11

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Determination of ėt:  change in et during motion from A to A’  The unit vector changes to e’t The vector difference det is shown in the bottom figure.

• In the limit det has magnitude equal to length of the arc │et│ dβ = dβ • Direction of det is given by en  We can write: det = en dβ



Dividing by dt: det /dt = en (dβ/dt) en  Substituting this and v = ρ dβ/dt =

in equation for acceleration:



Here:

an 

v2



  2  v

at  v  s ME101 - Division III

a  an2  at2

Kaustubh Dasgupta

12

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Important Equations •

•

•

v  

In n-t coordinate system, there is no component of velocity in the normal direction because of constant ρ for any section of curve (normal velocity would be rate of change of ρ).

an 

v2



  2  v

at  v  s a  an2  at2

Normal component of the acceleration an is always directed towards the center of the curvature  sometimes referred as centripetal acceleration.  If the particle moves with constant speed, at = 0, and a = an = v2/ρ  an represents the time rate of change in the dirn of vel.

Tangential component at will be in the +ve t-dirn of motion if the speed v is increasing, and in the - ve t-direction if the speed is decreasing.  If the particle moves in a straight line, ρ = ∞ an = 0, and a =  at represents the time rate of change in the magnitude of velocity.

Directions of tangential components of acceleration are shown in the figure.

ME101 - Division III

Kaustubh Dasgupta

13

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Circular Motion: Important special case of plane curvilinear motion • Radius of curvature becomes constant (radius r of the circle).

• Angle β is replaced by the angle θ measured from any radial reference to OP Velocity and acceleration components for the circular motion of the particle:

v   an 

v2



  2  v

at  v  s



a  an2  at2 general motion

ME101 - Division III

circular motion

Kaustubh Dasgupta

14

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example The curvilinear motion of a particle is defined by vx = 50 â&#x20AC;&#x201C; 16t and y = 100 â&#x20AC;&#x201C; 4t2.

At t = 0, x = 0. vx is in m/s2, x and y are in m, and t is in s. Plot the path of the particle and determine its velocity and acceleration at y = 0.

Solution:

Calculate x and y for various t values and plot ME101 - Division III

Kaustubh Dasgupta

15

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example Solution: When y = 0  0 = 100 – 4t2  t = 5 s

ME101 - Division III

Kaustubh Dasgupta

16

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example: The rider jumps off the slope at 300 from a height of 1 m, and remained in air for 1.5 s. Neglect the size of the bike and of the rider. Determine: (a) the speed at which he was travelling off the slope, (b) the horizontal distance he travelled before striking the ground, and (c) the maximum height he attains.

Solution: Let the origin of the coordinates be at A.

ME101 - Division III

Kaustubh Dasgupta

17

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example: Solution: For projectile motion: ax = 0, ay = -g = -9.81 m/s2  Constant Acceleration (a) speed at which he was travelling off the slope?

Let v0 be the initial velocity of the bike at A. For vertical Motion: ay = -g; subsequent integrations will give following equations

v  v0  at

 v y  v0  y  gt

1 2 1 at  y  y0  v0  y t  gt 2 2 2 2 v 2  v02  2a y  y0   v 2y  v0  y  2 g  y  y0  y  y 0  v0 t 

+

Using second eqn: -1 = 0 + (v0)y(1.5) – 0.5(9.81)(1.5)2 Initial velocity along y-direction (v0)y = v0 sin30 = 0.5v0  -1 = 0 + 0.5v0(1.5) – 0.5(9.81)(1.5)2  Initial Velocity of the bike: v0 = 13.38 m/s (velocity at A) ME101 - Division III

Kaustubh Dasgupta

18

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example: Solution: For projectile motion: ax = 0, ay = -g = -9.81 m/s2  Constant Acceleration (b) horizontal distance he travelled before striking the ground?

Let R be the horizontal distance between A and B. For horizontal Motion: ax = 0; subsequent integrations will give following equations

v  v0  at

 v x  v0 x

1 2 at  x  x0  v0 x t 2 v 2  v02  2ax  x0   v x  v0 x x  x0  v0 t 

+

Using second eqn: R = 0 + (v0)x(1.5) = 13.38cos30(1.5)  Horz distance: R = 17.4 m

ME101 - Division III

Kaustubh Dasgupta

19

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) Example: Solution: For projectile motion: ax = 0, ay = -g = -9.81 m/s2  Constant Acceleration (c) Maximum height attained by the bike?

Let (h - 1) m be the maximum height attained from x-axis at point C. For Vertical Motion: ay = -g,

v  v0  at

 v y  v0  y  gt

1 2 1 at  y  y0  v0  y t  gt 2 2 2 2 v 2  v02  2a y  y0   v 2y  v0  y  2 g  y  y0  y  y 0  v0 t 

+

Using the third eqn between A and C: All the quantities are known except the height of point C (y = h-1) and the velocity at point C  vy = 0 at C  0 = (0.5x13.38)2 – 2(9.81)(h – 1 – 0)  h = 3.28 m (total height attained above ground level) ME101 - Division III

Kaustubh Dasgupta

20

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Example: At the position shown, the driver applies brakes to produce a uniform deceleration. Speed of the car is 100 km/h at A (bottom of the dip), and 50 km/h at C (top of the hump). Distance between A and C is 120 m along the road.

Passengers experience a total acceleration of 3 m/s2 at A. Radius of curvature of the hump at C is 150 m. Calculate: (a) radius of curvature at A (b) total acceleration at inflection point B, and (c) total acceleration at C.

ME101 - Division III

Kaustubh Dasgupta

21

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Example Solution: Converting the units of Velocity: vA = 100 km/h [1000/(60x60)] = 27.8 m/s vC = 50 km/h [1000/(60x60)] = 13.89 m/s For Constant Deceleration, we can use the following formulae: Using the third equation between A and C to find the v  v0  at constant deceleration of the car: 1 (13.89)2 = (27.8)2 + 2a(120 – 0) x  x0  v0 t  at 2 2  a = - 2.41 m/s2 This acceleration is the tangential component of the v 2  v02  2ax  x0  total acceleration  at = - 2.41 m/s2 (a) radius of curvature at A? Total accln at A is given as: a = 3 m/s2 Using the third eqn: (3)2 = (an)2 + (-2.41)2  an = 1.785 m/s2 Using the first eqn: ρA = (27.8)2/1.785  ρA = 432 m ME101 - Division III

Kaustubh Dasgupta

an 

v2



  2  v

at  v  s a  an2  at2 22

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Example Solution: vA = 100 km/h [1000/(60x60)] = 27.8 m/s vC = 50 km/h [1000/(60x60)] = 13.89 m/s (b) total acceleration at inflection point B? Tangential component of acceleration at B, at = -2.41 m/s2 At inflection point radius of curvature is infinity, Therefore, normal component of acceleration, an = 0  Total acceleration at B: a = at = -2.41 m/s2

an 

v2



  2  v

at  v  s a  an2  at2

(c) total acceleration at C? Tangential component of acceleration at C, at = -2.41 m/s2 Normal component can be found from first eqn: an = (13.89)2/150 = 1.286 m/s2 Total acceleration at C: a2 = (1.286)2 + (-2.41)2  Total acceleration at C: a = 2.73 m/s2 ME101 - Division III

Kaustubh Dasgupta

23

Kinematics of Particles: Plane Curvilinear Motion Rectangular Coordinates (x-y) If all motion components are directly expressible in terms of horizontal and vertical coordinates

Time derivatives of the unit vectors are zero because their magnitude and direction remains constant.

Also, dy/dx = tan θ = vy /vx ME101 - Division III

Kaustubh Dasgupta

1

Kinematics of Particles: Plane Curvilinear Motion Normal and Tangential Coordinates (n-t) Determination of ėt:  change in et during motion from A to A’  The unit vector changes to e’t The vector difference det is shown in the bottom figure.

• In the limit det has magnitude equal to length of the arc │et│ dβ = dβ • Direction of det is given by en  We can write: det = en dβ



Dividing by dt: det /dt = en (dβ/dt) en  Substituting this and v = ρ dβ/dt =

in equation for acceleration:



Here:

an 

v2



  2  v

at  v  s ME101 - Division III

a  an2  at2

Kaustubh Dasgupta

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Example (1) on normal and tangential coordinates

ME101 - Division III

Kaustubh Dasgupta

3

Example (1) on normal and tangential coordinates

ME101 - Division III

Kaustubh Dasgupta

4

Example (2) on normal and tangential coordinates

ME101 - Division III

Kaustubh Dasgupta

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Example (2) on normal and tangential coordinates

ME101 - Division III

Kaustubh Dasgupta

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Example (2) on normal and tangential coordinates

ME101 - Division III

Kaustubh Dasgupta

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Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) The particle is located by the radial distance r from a fixed point and by an angular measurement θ to the radial line.

• θ is measured from an arbitrary reference axis • er and eθ are unit vectors along +r & +θ dirns.

Location of particle at A: r = r er By definition: v = dr/dt and a = d2r/dt2 Therefore we need: ėr and ėθ During time dt, the coordinate directions rotate through an angle dθ: er  e’r and eθ  e’θ Vector change der is in the +ve θ direction Vector change deθ is in the -ve r direction As already seen in the previous section: magnitudes of der and deθ in the limit are equal to the unit vector (radius) times dθ  der = eθ dθ and deθ = - er dθ ME101 - Division III

Kaustubh Dasgupta

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Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) der = eθ dθ and deθ = - er dθ de r  e • Dividing by dθ  d de r d  e • Dividing by dt  dt dt

 r   e  e

de  e r d de d  e r dt dt

e    e r

Relations for Velocity: Differentiating r = r er wrt time Vector expression for velocity 

v  r  re r  re r



v  re r  r e

vr  r

Magnitudes can be calculated as: r-component of v is the rate at which the vector r stretches. θ component of v is due to the rotation of r along the circumference of a circle having radius r.

ME101 - Division III

v  v

Kaustubh Dasgupta

The term dθ/dt is called Angular Velocity (rad/s) since it represents time r rate of change of angle 2 2 θ vr  v 9

Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) Relations for Acceleration: Differentiating the expression v  re r  r e wrt time The derivative of the second term will produce three terms since all three factors are variable.



a  v  re r  re r   r e  re  r e  We know: e r   e



e    e r

Vector expression for acceleration 

a  re r  r e  r e  re  r 2 e r  a  r  r 2 e r  r  2r e









Magnitudes can be calculated as:

ar  r  r 2 a  r  2r 

a  a r2  a2 ME101 - Division III

The term d2θ/dt2 is called Angular Accln since it represents change made in angular vel during an instant of time (rad/s2)

  r dt

θ-component can be 1 d 2 a  r alternatively written as: 

Kaustubh Dasgupta

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Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) Geometric Interpretations of the equations Top figure shows velocity vectors and their r- and θ- components at positions A and A’ after an infinitesimal movement.

Changes in magnitudes and directions of these components are shown in the bottom figure. Following are the changes: (a) Magnitude change of vr : = increase in length of vr or dvr  dr  Accn term (in the + r-dirn): dr / dt  r (b) Direction change of vr : Magnitude of this change = vr d  rd  Accn term (in the + θ-dirn): rd / dt  r (c) Magnitude change of vθ : = change in length of vθ or d (r)  Accn term (in the + θ-dirn): d (r) / dt  r  r (d) Direction change of vθ : Magnitude of this change = v d  r d  Accn term (in the - r-dirn): r (d / dt )  r 2 ME101 - Division III

Kaustubh Dasgupta

11

Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) Geometric Interpretations of the equations Collecting terms gives same relations as obtained previously

ar  r  r 2 a  r  2r 

ME101 - Division III

Kaustubh Dasgupta

12

Kinematics of Particles: Plane Curvilinear Motion Polar Coordinates (r - θ) Circular Motion: For motion in a circular path, r is constant  The components of velocity and acceleration become:

vr  r v  r ar  r  r 2 a  r  2r 





vr  0 v  r ar  r 2 a  r 

 Same as that obtained with n- and t-components, where the θ and t-directions coincide but the +ve r-direction is along the –ve n-direction  ar = -an for circular motion centered at the origin of the polar coordinates.

Further the expressions for ar and aθ can also be obtained using rectangular coordinates x = rcosθ and y = rsinθ  and ay  y   ax  x these rectangular components can be resolved into r- and θ-components to get the same expressions as obtained above.

ME101 - Division III

Kaustubh Dasgupta

13

Example (1) on polar coordinates Rotation of the radially slotted arm is governed by θ = 0.2t + 0.02t3.

Simultaneously, the power screw in the arm engages the slider B and controls its distance from O according to r = 0.2 + 0.04t2. Calculate the magnitudes of the velocity and acceleration of the slider for the instance when t = 3 s. θ is in radians, r is in meters, and t is in seconds.

ME101 - Division III

Kaustubh Dasgupta

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Example (1) on polar coordinates Solution: Using the Polar Coordinates. v  r r Available Equations: v  r

ar  r  r 2 a  r  2r 

v  vr2  v2 a  a r2  a2

Obtaining the derivatives of r and θ at t = 3 s.

ME101 - Division III

Kaustubh Dasgupta

15

Example (1) on polar coordinates Solution: Substituting in these eqns:

vr  r v  r

ar  r  r 2 a  r  2r 

v  vr2  v2 a  a r2  a2

ME101 - Division III

Kaustubh Dasgupta

16

Example (2) on polar coordinates

ME101 - Division III

Kaustubh Dasgupta

17

Example (2) on polar coordinates

ME101 - Division III

Kaustubh Dasgupta

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Example (3) on polar coordinates

ME101 - Division III

Kaustubh Dasgupta

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Example (3) on polar coordinates

ME101 - Division III

Kaustubh Dasgupta

20

Example (3) on polar coordinates

ME101 - Division III

Kaustubh Dasgupta

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