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A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. A ...
Typology: Exercises
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Today’s Objectives: Students will be able to:
In-Class Activities:
(continued)
A roller coaster car travels down a fixed, helical path at a constant speed.
How can we determine its position or acceleration at any instant?
If you are designing the track, why is it important to be able to predict the acceleration of the car?
(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.
A particle moves along a curve defined by the path function, s.
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.
If the particle moves a distance s along the curve during time interval t, the displacement is determined by vector subtraction: r = r’ - r
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 t, the average acceleration during that increment is: aavg = v / t = ( v - v’ )/ t
The instantaneous acceleration is the time- derivative of velocity: a = d v /dt = d^2 r /dt^2
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.
(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 position of the particle can be defined at any instant by the position vector r = x i + y j + z k. The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t).
The magnitude of the position vector is: r = (x^2 + y^2 + z^2 )^0.^5 The direction of r is defined by the unit vector: ur = (1/r) r
The direction of a is usually not tangent to the path of the particle.
The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector):
a = d v /dt = d^2 r /dt^2 = ax i + ay j + az k
where ax = = = dvx /dt, ay = = = dvy /dt,
az = = = dvz /dt
The magnitude of the acceleration vector is
a = [(ax)^2 + (ay)^2 + (az)^2 ]^0.^5
Given: The motion of two particles (A and B) is described by the position vectors rA = [3t i + 9t(2 – t) j ] m rB = [3(t^2 – 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.
(continued)
Speed is the magnitude of the velocity vector. vA = (3^2 + 18^2 ) 0.^5 = 18.2 m/s vB = (6^2 + 3^2 ) 0.^5 = 6.71 m/s
vA = d rA /dt = = [3 i + (18 – 18t) j ] m/s At t = 2 s: vA = [3 i – 18 j ] m/s
x. A i yA j
vB = d rB /dt = xB i + yB j = [(6t – 6) i + 3 j ] m/s At t = 2 s: vB = [6 i + 3 j ] m/s
C) 1 m/s D) 2 m/s
(continued)
Solution:
x-components:
y-components: Position: y = 0. 5x^2 = 0. 5(2. 5t^2 )^2 = (3. 125t^4 ) ft Velocity: vy = dy/dt = d (3. 125t^4 ) /dt = (12. 5t^3 ) ft/s Acceleration: ay = vy = d (12. 5t^3 ) /dt = (37. 5t^2 ) ft/s^2
Velocity: vx = x = dx/dt = (5t) ft/s
Position: = => x = (5/2)t^2 = (2. 5t^2 ) ft
Acceleration: ax = x = v•• x = d/dt (5t) = 5 ft/s^2
x dx 0
t
0
5 t dt
(continued)
At t = 1 s, r = (2. 5 i + 3. 125 j ) ft
Distance: d = r = (2. 52 + 3. 1252 ) 0.^5 = 4. 0 ft
Acceleration vector: a = [5 i + 37. 5t^2 j ] ft/s^2
Magnitude: a = (5^2 + 37. 52 )^0.^5 = 37.8 ft/s^2
The magnitude of the acceleration vector is calculated as:
