Slide 1 : CURVILINEAR MOTION:
GENERAL & RECTANGULAR COMPONENTS Today’s Objectives:
Students will be able to:
Describe the motion of a particle traveling along a curved path.
Relate kinematic quantities in terms of the rectangular components of the vectors. In-Class Activities:
• Check Homework
• Reading Quiz
• Applications
• General Curvilinear Motion
• Rectangular Components of Kinematic Vectors
• Concept Quiz
• Group Problem Solving
• Attention Quiz
Slide 2 : READING QUIZ 1. In curvilinear motion, the direction of the instantaneous velocity is always
A) tangent to the hodograph.
B) perpendicular to the hodograph.
C) tangent to the path.
D) perpendicular to the path. 2. In curvilinear motion, the direction of the instantaneous acceleration is always
A) tangent to the hodograph.
B) perpendicular to the hodograph.
C) tangent to the path.
D) perpendicular to the path.
Slide 3 : APPLICATIONS The path of motion of a plane can be tracked with radar and its x, y, and z coordinates (relative to a point on earth) recorded as a function of time. How can we determine the velocity or acceleration of the plane at any instant?
Slide 4 : APPLICATIONS (continued) A roller coaster car travels down a fixed, helical path at a constant speed.
Slide 5 : GENERAL CURVILINEAR MOTION
(Section 12.4) A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. The position of the particle at any instant is designated by the vector
r = r(t). Both the magnitude and direction of r may vary with time.
Slide 6 : VELOCITY Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment Dt is
vavg = Dr/Dt .
The instantaneous velocity is the time-derivative of position
v = dr/dt .
The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length Ds approaches the magnitude of Dr as t?0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!
Slide 7 : ACCELERATION Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’ over a time increment Dt, the average acceleration during that increment is:
aavg = Dv/Dt = (v - v’)/Dt
The instantaneous acceleration is the time-derivative of velocity:
a = dv/dt = d2r/dt2 A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.
Slide 8 : CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Section 12.5) It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference. The magnitude of the position vector is: r = (x2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
Slide 9 : RECTANGULAR COMPONENTS: VELOCITY The magnitude of the velocity vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent to the path of motion.
Slide 10 : RECTANGULAR COMPONENTS: ACCELERATION The direction of a is usually not tangent to the path of the particle. The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]0.5
Slide 11 : EXAMPLE Given: The motion of two particles (A and B) is described by the position vectors
rA = [3t i + 9t(2 – t) j] m and
rB = [3(t2 –2t +2) i + 3(t – 2) j] m. Find: The point at which the particles collide and their speeds just before the collision. Plan: 1) The particles will collide when their position vectors are equal, or rA = rB .
2) Their speeds can be determined by differentiating the position vectors.
Slide 12 : EXAMPLE (continued) 1) The point of collision requires that rA = rB, so xA = xB and yA = yB . Solution: Set the x-components equal: 3t = 3(t2 – 2t + 2)
Simplifying: t2 – 3t + 2 = 0
Solving: t = {3 ? [32 – 4(1)(2)]0.5}/2(1)
=> t = 2 or 1 s Set the y-components equal: 9t(2 – t) = 3(t – 2)
Simplifying: 3t2 – 5t – 2 = 0
Solving: t = {5 ? [52 – 4(3)(–2)]0.5}/2(3)
=> t = 2 or – 1/3 s So, the particles collide when t = 2 s (only common time). Substituting this value into rA or rB yields
xA = xB = 6 m and yA = yB = 0
Slide 13 : EXAMPLE (continued) 2) Differentiate rA and rB to get the velocity vectors. Speed is the magnitude of the velocity vector.
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s vB = drB/dt = xB i + yB j = [ (6t – 6) i + 3 j ] m/s
At t = 2 s: vB = [ 6 i + 3 j ] m/s • •
Slide 14 : CHECK YOUR UNDERSTANDING QUIZ 1. If the position of a particle is defined by
r = [(1.5t2 + 1) i + (4t – 1) j ] (m), its speed at t = 1 s is
A) 2 m/s B) 3 m/s
C) 5 m/s D) 7 m/s 2. The path of a particle is defined by y = 0.5x2. If the component of its velocity along the x-axis at x = 2 m is
vx = 1 m/s, its velocity component along the y-axis at this position is
A) 0.25 m/s B) 0.5 m/s
C) 1 m/s D) 2 m/s
Slide 15 : GROUP PROBLEM SOLVING Given: The velocity of the particle is
v = [ 16 t2 i + 4 t3 j + (5 t + 2) k] m/s.
When t = 0, x = y = z = 0. Find: The particle’s coordinate position and the magnitude of its acceleration when t = 2 s.
Plan: Note that velocity vector is given as a function of time.
1) Determine the position and acceleration by integrating and differentiating v, respectively, using the initial conditions.
2) Determine the magnitude of the acceleration vector using t = 2 s.
Slide 16 : GROUP PROBLEM SOLVING (continued) Solution:
1) x-components: 2) y-components:
Slide 17 : GROUP PROBLEM SOLVING (continued) 3) z-components: Position vector: r = [ 42.7 i + 16 j + 14 k] m.
Acceleration vector: a = [ 64 i + 48 j + 5 k] m/s2
Magnitude: a = (642 + 482 +52)0.5 = 80.2 m/s2 4) The position vector and magnitude of the acceleration vector are written using the component information found above.
Slide 18 : ATTENTION QUIZ 1. If a particle has moved from A to B along the circular path in 4s, what is the average velocity of the particle ?
A) 2.5 i m/s
B) 2.5 i +1.25j m/s
C) 1.25 ? i m/s
D) 1.25 ? j m/s 2. The position of a particle is given as r = (4t2 i - 2x j) m. Determine the particle’s acceleration.
A) (4 i +8 j ) m/s2 B) (8 i -16 j ) m/s2
C) (8 i) m/s2 D) (8 j ) m/s2
Slide 19 : End of the Lecture Let Learning Continue