Lecture L4 - Google Groups

Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Physics 201

Lecture 4 Motion in Two and Three Dimensions  Position and Displacement, Velocity, Acceleration  Projectile Motion  Uniform Circular Motion  Relative Motion in One and Two Dimensions

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions

 In the previous lecture, we consider motions along a straight

line, that are, motions in One Dimension only.

 This lecture extends the materials of the previous lecture to

Two and Three Dimensions.  The ideas of Position and Displacement, Velocity, and Acceleration are now more complex.  We are going to use Vectors in this lecture.

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Position and Displacement, Average Velocity  r  xiˆ  yˆj  zkˆ

 During a time interval Δt the particle moves from P1 to P2, the displacement during this interval is    r  r  r  ( x  x )iˆ  ( y  y ) ˆj  ( z  z )kˆ 2

1

2

1

 xiˆ  yˆj  zkˆ

2

1

2

1

 The average velocity during this interval is defined as

    r x ˆ y ˆ z ˆ  vav   i j k  vav x  vav y  vav z t t t t

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Instantaneous Velocity

The average velocity is defined as     r x ˆ y ˆ z ˆ  vav   i j k  vav x  vav y  vav z t t t t

The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time.

   r d r v  lim  t  0  t dt     dx dy ˆ dz ˆ v  v x  v y  v z  iˆ  j k dt dt dt

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Instantaneous Velocity

Example

A robotic vehicle, or rover, is exploring the surface of Mars. The landing craft is in the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time:

x  2.0m  (0.25m/s 2 )t 2 y  (1.0m/s)t  (0.025m/s 3 )t 3 (a) Find the rover's coordinates and its distance from the lander at t = 2.0 s. (a) • At time t = 2.0 s the rover's coordinates are x = 2.0 m - (0.25 m/s2)( 2.0 s)2 = 1.0 m y = (1.0m/s)(2.0s) + (0.025m/s 3 )(2.0s)3 = 2.2m • The rover's distance from the origin at this time is

r  x 2  y 2  (1.0m) 2  (2.2m) 2  2.4m

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Instantaneous Velocity

Example x  2.0m  (0.25m/s 2 )t 2 y  (1.0m/s)t  (0.025m/s 3 )t 3 (b) Find the rover's displacement and average velocity vectors during the interval from t = 0.0 s to t = 2.0 s. • To find the displacement and average velocity, we express the position vector as a function of time t  r  xiˆ  yˆj  [ 2.0m  (0.25m/s 2 )t 2 ]iˆ  [(1.0m/s )t  (0.025 m/s 3 )t 3 ] ˆj  •At time t = 0.0 s the position vector is r0  (2.0m)iˆ

• From part (a) the position vector at time t = 2.0 s is  r2  (1.0m )iˆ  ( 2.2m ) ˆj • Therefore the displacement from t = 0.0 s to t = 2.0 s is    r  r2  r0  (1.0m - 2.0m)iˆ  (2.2m) ˆj  (-1.0m)iˆ  (2.2m) ˆj • The average velocity during this interval is the displacement divided by the elapsed time:   r (-1.0m)iˆ  (2.2m) ˆj vavg   t 2.0s  (0.5m/s)iˆ  (1.1m/s) ˆj

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions

x  2.0m  (0.25m/s 2 )t 2 y  (1.0m/s)t  (0.025m/s 3 )t 3 (c) Derive a general expression for the rover's instantaneous velocity vector. Express the velocity at t = 2.0 s in component form and also in terms of magnitude and direction. • The displacement as a function of time t is of the form  r  xiˆ  yˆj  [ 2.0m  (0.25m/s 2 )t 2 ]iˆ  [(1.0m/s )t  (0.025 m/s 3 )t 3 ] ˆj

Instantaneous Velocity

• The instantaneous velocity can be calculated as   dr dx ˆ dy ˆ v  i j dt dt dt  2 ˆ v  [( 0.5m/s )t ]i  [(1.0m/s )  (0.075 m/s 3 )t 2 ] ˆj • At time t = 2.0 s, the components of instantaneous velocity are v2 x  [(0.5m/s 2 )(2.0s)]  1.0m/s v2 y  1.0m/s  (0.075m/s 3 )(2.0s)  1.3m/s • Please express the velocity at t = 2.0 s in terms of magnitude and direction.

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Average Acceleration

 During the time interval from tl to t2, the vector change in velocity is    v  v  v  (v  v )iˆ  (v  v ) ˆj  (v  v )kˆ 2

1

2x

1x

2y

1y

2z

1z

 The average acceleration of the car during this time interval is defined as the velocity change divided by the time interval Δt = t2- t1     v  v v aav  2 1  t 2  t1 t

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Instantaneous Acceleration The instantaneous acceleration is equal to the instantaneous rate of change of velocity with time    v dv a  lim  t 0 t dt

     dv dv x dv y dv z a    dt dt dt dt

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Instantaneous Acceleration

Example

 v  [(0.5m/s 2 )t ]iˆ  [(1.0m/s)  (0.075m/s 3 )t 2 ] ˆj (d) Derive a general expression for the rover's instantaneous acceleration vector. Express the acceleration at t = 2.0 s in component form and also in terms of magnitude and direction. • We can derive the instantaneous acceleration vector a as   dv dv x ˆ dv y ˆ a  i j  a x iˆ  a y ˆj dt dt dt  a  [0.5m/s 2 ]iˆ  [0.15m/s 3 )t ] ˆj • At time t = 2.0 s, the components of instantaneous acceleration are a x  0.5m/s 2 a y  [0.15m/s 3 )(2.0s)]  0.30m/s 2 10

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Components of Acceleration The instantaneous acceleration is equal to the instantaneous rate of change of velocity with time    v dv a  lim  t 0 t dt

  dv     a  atangent  anormal  at  an dt

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Uniform Circular Motion

 A particle is in Uniform Circular Motion (UCM) if it travels in a circle or circular arc at constant speed.  Although the speed does not vary, the particle is accelerating.

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Components of Acceleration

A skier moves along a ski-jump ramp as shown in figure. The ramp is straight from point A to point C and curved from point C onward. The skier picks up speed as she moves downhill from point A to point E, where her speed is maximum. She slows down after passing point E. Draw the direction of the acceleration vector at points B, D, E, and F.

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance.

 We'll neglect the effects of air resistance and the curvature and rotation of the earth.  Projectile motion is always confined to a vertical plane determined by the direction of the initial velocity. Thus projectile motion is two-dimensional.  The key to analyzing projectile motion is that we can treat the x- and y-coordinates separately. 14

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion We can analyze projectile motion as a combination of • horizontal motion with constant velocity • and vertical motion with constant acceleration.  Let’s assume that the projectile is launched with some initial velocity

   v0  v0 x  v0 y  v0 x iˆ  v0 y ˆj

 The components v0x and v0y can be written in the form

v0 x  v0 cos  0 v0 y  v0 sin  0

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

The horizontal motion

 Let’s assume that the projectile is point (x0,y0) at t = 0. Since there is no acceleration on the horizontal direction

v x  v0 x  v0 cos 0  constant

 The displacement

x  x0  v x t  v xo t  (v0 cos  0 )t

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

The vertical motion

The vertical motion is exactly the free-fall motion we discussed in the previous lecture vy  vy0  gt  v0 sin0  gt

1 1 y  y0  vy0t  gt2  (v0 sin0)t  gt2 2 2

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion  Two independent motions Along x-direction

Along y-direction

ax = 0

ay = -g = -9.8 m/s2

vx = v0cosα0

vy = v0sinα0 - gt

x = x0 + (v0cosα0)t

1 y  y0  (v0 sin0 )t  gt2  0 2 v2y  (v0 sin0)2  2g(y  y0)

 Only time t appears in both sets of equations → coupling through time t. 18

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

The equation of the path

 Along the x-axis

x  x0  v x t  v xot  (v0 cos 0 )t  Along the y-axis

1 1 y  y0  vy0t  gt2  (v0 sin0)t  gt2 2 2  For simplicity, let’s assume that x0 = y0 = 0, then the equation of the path (trajectory) is of the form

  2 g x y  (tan0)x  2  2(v0 cos0) 

Parabola 19

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

The horizontal range R

The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial (launch) height.  Along the x-axis

x  x0  (v0 cos 0 )t  R

 Along the y-axis

1 y  y0  (v0 sin0 )t  gt2  0 2

 Eliminating t between these two equations

R

2v02 v2 sin0 cos0  0 sin20 g g

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

The horizontal range R

x  x0  (v0 cos 0 )t  R 1 y  y0  (v0 sin0 )t  gt2  0 2

R

2v02 v2 sin0 cos0  0 sin20 g g

The horizontal range R has its maximum value when

sin20 10  45o 21

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Example

Projectile Motion

This is a typical “projectile motion”. For simplicity, we'll ignore air resistance for this example and use the projectile-motion equations to describe the motion. Let’s assume that the initial position is at the origin x0 = y0 = 0. (a) Position of the ball at t = 2.00 s  Along the x-axis

x  (v0 cos  0 )t  (37.0 m/s)(cos53.10 )2s  44.4 m  Along the y-axis

1 1 22 y  (v0 sin0)t  gt2  (37.0 m/s)(sin53.10)(2s)- (9.8m/s2)(2s)2  39.6 m 2 2

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

Example

(a)Velocity of the ball at t = 2.00 s  The x-component of velocity

v x  v0 cos 0  (37.0 m/s)cos53.10  22.2 m/s  The y-component of velocity

vy  v0 sin0  gt  (37.0 m/s)(sin53.10 ) - (9.8 m/s2 )(2s) 10.0 m/s  The magnitude and direction of the velocity are

v  vx2  v2y  (22.2 m/s)2  (10.0 m/s)2  24.3m/s

v

10.0 m/s  24.20 22.2 m/s

  arctany  arctan vx

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Projectile Motion

Example

(b) Find the time when the ball reaches the highest point of its flight and find its height h at this point.  At the highest point, the vertical velocity vy is zero. When does this happen? Call the time tl; then

vy  v0 sin0  gt1  0 v sin0 (37.0 m/s)(sin53.10) t1  0   3.02s g 9.8 m/s2

 The height h = ymax can be found as

1 1 h  ymax  (v0 sin0)t1  gt12  (37.0 m/s)(sin53.10)(3.02s)- (9.8m/s2)(3.02s)2  44.7 m 2 2  There is another way to find h, please try it and also part (c) on your own. 24

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Components of Acceleration The instantaneous acceleration is equal to the instantaneous rate of change of velocity with time    v dv a  lim  t 0 t dt

  dv     a  aparallel  aperpendicular  at  an dt

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Uniform Circular Motion

 A particle is in Uniform Circular Motion (UCM) if it travels in a circle or circular arc at constant speed.  Although the speed does not vary, the particle is accelerating.

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Uniform Circular Motion Let’s consider a particle in Uniform Circular Motion with speed v in a circle of radius r.  Two points p and q are symmetric w.r.t. the y axis.  The x and y components of the velocity vectors at those points are

   vp  vpx  vpy vpx  vcos vpy  vsin

   vq  vqx  vqy vqx  vcos vqy  vsin

t 

arc( pq) r2 

 The time required for the particle to move from p to q is v v  The x and y components of the average acceleration as the particle moves from p to q are

ax 

vqx vpx t

0

ay 

vqy vpy t



2vsin v2  sin     r(2) / v r  

vertically downward

 If the angle θ approaches zero, then p and q approach point P, the instantaneous acceleration at point P has a magnitude of 2

a

v  constant r

Centripetal acceleration

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Uniform Circular Motion Let’s consider a particle p moving at constant speed v around a circle of radius r.  The velocity vector can be written as    v  v x  v y  v x iˆ  v y ˆj  ( v sin  )iˆ  (v cos  ) ˆj

dy p dt dx p dt

vy p vx p  v  (v sin  )iˆ  (v cos ) ˆj  ( )iˆ  ( ) ˆj r r  The acceleration vector of the particle can then be calculated as follow   dv v dy p ˆ v dx p ˆ a  ( )i  ( )j dt r dt r dt

 v yp  v cos

 v2 v2 a  (  cos  )iˆ  (  sin  ) ˆj r r

 v xp  v sin 

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Uniform Circular Motion  v2 v2 a  ( cos )iˆ  ( sin  ) ˆj r r  The magnitude of the acceleration vector is of the form a  a x2  a 2y 

v2 r

 To determine the direction of the acceleration vector, we find the angle ϕ ay   tan    tan  ax The acceleration vector is always directed along the radius toward the center of the circle  The period T of the motion can be found as

T

2r v

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Relative Motion in One Dimensions

 In general, when two observers measure the velocity of a moving body, they get different results if one observer is moving relative to the other.  The velocity of an object depends on the reference frame of the observer  relative velocity.  A reference frame is a coordinate system plus a time scale. We only consider Inertial reference frames that move at constant velocity with respect to each other.  Suppose that reference frame A is stationary, while reference frame B is moving at a constant velocity vBA w.r.t frame A.  At a given moment, the position xPA of particle P measured by frame A is

xPA  xPB  xBA

 The velocity vPA of P as measured by frame A is

vPA 

dxPA dxPB dxBA   dt dt dt

vPA  vPB vBA

 The acceleration aPA of P as measured by frame A is aPA  dvPA  dvPB  dvBA dt dt dt

aPA30 aPB

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Relative Motion in One Dimensions Example Alex (frame A), parked by the side of an east-west road, is watching car P, which is moving in westerly direction. Barbara (frame B) driving east at a speed 52 km/h, watches the same car. Take the easterly direction as positive. a) If Alex measures a speed of 78 km/h for car P, what velocity will Barbara measure?

vPA  vPB  vBA

vPB  vPA vBA  (78km/h) (52km/h)  130 km/h

b) If Alex sees car P brake to a halt in 10 s, what acceleration (assumed constant) will he measure for it?

aPA 

v  v0  at

0 vPA 0 (78km/h)   2.2 m/s2 t 10s

aPA  aPB c) What acceleration will Barbara measure for the braking car? In Barbara frame, the initial velocity v0 of the car is -130 km/h and the final velocity v is -52 km/h aPB 

(52km/h) (130km/h) 78km/h   2.2 m/s2 10s 10s

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Relative Motion in Two Dimensions

Sample Problem 4.6  The person on water sees the trajectory of the package as a parabola.  An observer on the airplane, however, sees the motion of the package as a straight line toward the Earth.  If, once it drops the package, the airplane continues to move horizontally with the same velocity, then the package hits the ground directly beneath the airplane.

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Relative Motion in Two Dimensions  Let’s assume that two inertial reference frames A and B are separating at a constant velocity vBA and, for simplicity, we further assume that their x and y axes remain parallel to each other as they do so.  At a given moment, the position rPA of particle    P measured by frame A is

rPA  rPB  rBA

 The velocity vPA of P as measured by frame A is

    dr dr dr vPA  PA  PB  BA dt dt dt

   vPA  vPB vBA

 The acceleration aPA of P as measured by frame A is       dv dv dv aPA  aPB aPA  PA  PB  BA dt dt dt 33

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Ngac An Bang, Physics Department, HUS

Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions Relative Motion in Two Dimensions A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The water in the river has a uniform speed of 5.00 km/h due east relative to the Earth. Determine the velocity of the boat relative to an observer standing on either bank.  The velocity vector of the boat relative to the Earth

   vPA  vPB  vBA

   vbE  vbr  vrE

 Using the Pythagorean theorem, the vbE can be found as 2 2 vbE  vbr  vrE 11.2 km/h

 The direction

v vbr

  arctanrE  26.60

The boat is moving at a speed of 11.2 km/h in the direction 26.6° east of north relative to the Earth.

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Ngac An Bang, Faculty of Physics, HUS

Motion in Two and Three Dimensions That’s enough for today  Problems will be given to you.  Please try the Sample Problems given in the textbook.  Feel free to contact me via email.

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