Dynamics 8-1 Overview Dynamicsthe study of moving objects. - - PowerPoint PPT Presentation

Professional Publications, Inc.

FERC

8-1 Dynamics

Overview

Dynamics—the study of moving objects. Kinematics—the study of a body’s motion independent of the forces on the body. Kinetics—the study of motion and the forces that cause motion.

Professional Publications, Inc.

FERC

8-2 Dynamics

Kinematics—Rectangular Coordinates

Professional Publications, Inc.

FERC

8-3a1 Dynamics

Kinematics—Polar Coordinates

Professional Publications, Inc.

FERC

8-3a2 Dynamics

Kinematics—Polar Coordinates

Professional Publications, Inc.

FERC

8-4a Dynamics

Kinematics—Circular Motion

Professional Publications, Inc.

FERC

8-4b1 Dynamics

Kinematics—Circular Motion

Angular velocity = Angular acceleration = Tangential acceleration = Normal acceleration =

Professional Publications, Inc.

FERC

8-4b2 Dynamics

Kinematics—Circular Motion

Example (FEIM): A turntable starts from rest and accelerates uniformly at 1.5 rad/s2. How many revolutions does it take for the rotational frequency to reach 33.33 rpm? = 2f = 2 rad rev

• 33.33 rev

min

• 1 min

60 s

• = 3.49rad

s =

f t

dt =

f t

tdt = tf

2

2 t =

• =

2

2 = 3.49 rad

s

( )

2

2

( ) 1.5 rad

s2

( )

= 4.06 rad n = 4.06 rad 2 rad rev = 0.65 revolution

Professional Publications, Inc.

FERC

8-5a1 Dynamics

Kinematics—Projectile Motion

s = s0 + v0t + 1

2 at 2

v = v0 +a0t v

2 = v0 2 + 2a0 s s0

( )

Constant acceleration formulas:

Professional Publications, Inc.

FERC

8-5a2 Dynamics

Kinematics—Projectile Motion

Example 1 (FEIM): A projectile is launched at 180 m/s at a 30° incline. The launch point is 150 m above the impact plane. Find the maximum height, flight time, and range.

Professional Publications, Inc.

FERC

8-5a3 Dynamics

Kinematics—Projectile Motion

• h

max = v0 sin +at = 0

tmax = v0 sin a = 180m s

• 0.5

( )

9.8m s

• = 9.18 s

h = h0 + v0t sin + 1

2 at 2

hmax = 150 m+ 180 m s

2

• (9.18 s)(0.5)

1 2

( ) 9.8 m

s

2

• (9.18 s)

2 = 563 m

himpact = 0 = 150 m+90t 4.9t

2

By the quadratic equation, timpact = 19.9 s cos30° = 0.866 Rimpact = R0 + v0timpact cos = 0+ 180m s

• (19.9 s)(0.866) = 3100 m

Professional Publications, Inc.

FERC

8-5b1 Dynamics

Kinematics—Projectile Motion

Example 2 (FEIM): A bomber flies horizontally at 275 km/h and an altitude of 3000 m. At what viewing angle from the bomber to the target should the bomb be dropped?

Professional Publications, Inc.

FERC

8-5b2 Dynamics

Kinematics—Projectile Motion

himpact = 1

2 at 2 = 1 2

( ) 9.8 m

s

2

• t

2 = 3000 m

t = 2himpact g = 24.7 s 275km hr

• 1000 m

km

• 1 hr

3600 s

• = 76.39 m/s

Rimpact = R0 + v0t = 0+ 76.39m s

• 24.7 s

( ) = 1887 m

= arctanRimpact himpact = arctan1887 m 3000 m = 32.2°

Professional Publications, Inc.

FERC

8-6a Dynamics

Kinetics—Newton’s 2nd Law of Motion

For a constant mass, One-dimension motion

Professional Publications, Inc.

FERC

8-6b1 Dynamics

Kinetics—Newton’s 2nd Law of Motion

Example (FERM prob. 6, p. 15-5):

Professional Publications, Inc.

FERC

8-6b2 Dynamics

Kinetics—Newton’s 2nd Law of Motion

Example (FERM prob. 6, p. 15-5):

Professional Publications, Inc.

FERC

8-7a Dynamics

Kinetics—Impulse and Momentum

Impulse (constant mass in one dimension) Momentum Impulse-Momentum Principle

Professional Publications, Inc.

FERC

8-7b1 Dynamics

Kinetics—Impulse and Momentum

Example 1 (FEIM): A 0.046 kg marble attains a velocity of 76 m/s in a slingshot. Contact with the slingshot is 1/25 of a second. What is the average force on the marble during the launch? F

ave = mv

t = (0.046 kg) 76m s

• 0.04 s

= 87.4 N

Professional Publications, Inc.

FERC

8-7b2 Dynamics

Kinetics—Impulse and Momentum

Example 2 (FEIM): A 2000 kg cannon fires a 10 kg projectile horizontally at 600 m/s. It takes 0.007 s for the projectile to pass through the barrel. What is the recoil velocity if the cannon is not restrained? What average force must be exerted on the cannon to keep it from moving? F = mv t = (10 kg)(600m s ) 0.007 s = 8.5710

5 N

mprojectilevprojectile = mcannonvcannon (10 kg)(600m s ) = (2000 kg)(vcannon) vcannon = 3m s = initial recoil velocity

Professional Publications, Inc.

FERC

8-8a Dynamics

Work & Energy

Work Kinetic Energy of a Mass Kinetic Energy of a Rotating Body Potential Energy

• Gravity
• Spring (linear)

Fs = kx where the spring is compressed a distance x W = KE2 KE1 = 1

2 m(v2 2 v1 2)

W = KE2 KE1 = 1

2 I(2 2 1 2)

W = PE2 PE1 = mg(h2 h1) W = PE2 PE1 = 1

2 k(x2 2 x1 2)

Professional Publications, Inc.

FERC

8-8b Dynamics

Work & Energy

Conservation of Energy

• For a closed system (no external work), the change in potential

energy equals the change in kinetic energy. PE1 PE2 = KE2 KE1 PE1 +KE1 = PE2 +KE2

• For a system with external work, W equals ΔPE + ΔKE.

W12 = (PE1 PE2)+(KE2 KE1) PE1 +KE1 +W12 = PE2 +KE2

Professional Publications, Inc.

FERC

8-8c Dynamics

Work & Energy

Impacts: Momentum is always conserved. Elastic Impacts: Kinetic energy is conserved. m1v1 + m2v2 = m1 v

1 + m2

v

2

m1 = m2sov1 + v2 = v

1 +

v

2 = 0.85 0.53 = 0.32 1 2 m v1 2 + 1 2 m v2 2 = 1 2 m

v

1 2 + 1 2 m

v

2 2

v1

2 + v2 2 =

v

1 2 +

v

2 2

Solving two equations and two unkowns: Therefore, (A) is correct.

• v

1 = 0.53 m/s

• v

2 = 0.85 m/s

Example 1 (FEIM): Two identical balls collide along their centerlines in an elastic

• collision. The initial velocity of

ball 1 is 0.85 m/s. The initial velocity of ball 2 is –0.53 m/s. What is the relative velocity of each ball after the collision? (A) –0.53 m/s and 0.85 m/s (B) –0.72 m/s and 1.2 m/s (C) –5.1 m/s and 1.2 m/s (D) 0.98 m/s and 1.8 m/s

Professional Publications, Inc.

FERC

8-8d Dynamics

Work & Energy

Example 2 (FEIM): Ball A of 200 kg is traveling at 16.7 m/s. It strikes stationary ball B of 200 kg along the centerline. What is the velocity of ball A after the collision? Assume the collision is elastic. (A) –16.7 m/s (B) –8.35 m/s (C) (D) 8.35 m/s There are two possible solutions for these equations. Since there must be a change in the collision, ball A’s velocity must be 0. Therefore, (C) is correct. mA = mBsovA + vB = v

A +

v

B = 16.7 m/s

vA

2 + vB 2 =

v

A 2 +

v

B 2

• v

A = 0,

v

B = 16.7 m/s

• r
• v

A = 16.7 m/s,

v

B = 0

Professional Publications, Inc.

FERC

8-8e Dynamics

Work & Energy

Inelastic Impacts: Kinetic energy does not have to be conserved if some energy is converted to another form. Example 1 (FEIM): A ball is dropped from an initial height

• ho. If the coefficient of restitution is

0.90, how high will the ball rebound? (A) 0.45ho (B) 0.81ho (C) 0.85ho (D) 0.9ho Since there must be a change in the collision, ball A’s velocity must be 0. Therefore, (B) is correct.

• v

1

v

2 = e(v1 v2)

where e = coefficient of restitution

• v

1 = m2v2(1+e)+(m1 em2)v1

m1 + m2

• v

2 = m1v1(1+e)+(em1 m2)v2

m1 + m2 mgh = 1

2 mv 2

v = 2gho

• v = ev = e 2gho =

2g h

• h = e

2ho = (0.9) 2ho = 0.81ho

Professional Publications, Inc.

FERC

8-8f Dynamics

Work & Energy

Example 2 (FEIM): Two masses collide in a perfectly inelastic collision. What is the velocity of the combined mass after collision? (A) 0 (B) 4 m/s (C) –5m/s (D) 10 m/s Therefore, (B) is correct. m1 = 4m2 v1 = 10 m/s v2 = 20 m/s m1v1 + m2v2 = m3v3 m3 = m1 + m2 = 5m2 4m2 10m s

• + m2 20m

s

• = 5m2v3

5m2v3 = 40m s

• m2 20m

s

• m2

v3 = 4 m/s “Perfectly inelastic” means the masses collide and stick together.

Professional Publications, Inc.

FERC

8-9a Dynamics

Kinetics

Friction Example (FEIM): A snowmobile tows a sled with a weight of 3000 N. It accelerates up a 15° slope at 0.9 m/s2. The coefficient of friction between the sled and the snow is 0.1. What is the tension in the tow rope? F

slope = F rope (F friction +F gravity) = maslope

F

rope = F friction +F gravity + maslope

= mg sin15°+ mgµ cos15°+ maslope = (3000)(0.2588)+(3000 N)(0.1 )(0.9659) + 3000 N 9.8 m s

2

• 0.9 m

s

2

• = 1342 N

Professional Publications, Inc.

FERC

8-9b1 Dynamics

Kinetics

Plane Motion of a Rigid Body Similar equations can be written for the y-direction or any other coordinate direction. Example (FEIM): A 2500 kg truck skids with a deceleration of 5 m/s2. What is the coefficient of sliding friction? What are the frictional forces and normal reactions (per axle) at the tires?

Professional Publications, Inc.

FERC

8-9b2 Dynamics

Kinetics

The force of deceleration is equal to the friction force. The moment about the center of gravity, MA, must be equal to zero. F

deceleration = (2500 kg) 5 m

s

2

• = F

friction = µmg

MA =

• (4.5 m)NB (2 m)mg (1 m)(F

deceleration) = 0

= (4.5 m)NB (2 m)(2500 kg) 9.8 m s

2

• (2500 kg) 5 m

s

2

• = 0

NB = 13,667 N NA = mg NB = (2500 kg) 9.8 m s

2

• 13667 N = 10833 N

µ = 5 m s

2

• 9.8 m

s

2

• = 0.51

= µ(2500 kg) 9.8 m s

2

Professional Publications, Inc.

FERC

8-10a1 Dynamics

Rotation

Rotation about Fixed Axis

Professional Publications, Inc.

FERC

8-10a2 Dynamics

Rotation

There are three simultaneous equations for the movement of the mass, the cylinder, and the relationship between the two: Example (FEIM): A mass m is attached to a rope wound around a cylinder of mass mC and radius r. What is the acceleration of the falling mass? What is the rope tension? mg F = ma Fr = I r = a

Professional Publications, Inc.

FERC

8-10a3 Dynamics

Rotation

Rearranging, Fr = I a r mg mC 2 a = ma F = mCr

2a

2r

2

= mC 2 a a = g m m + mC 2 Solving for F, F = mC 2 a = g mCm 2 m + mC 2

• =

mmC 2m + mC g

Professional Publications, Inc.

FERC

8-10b Dynamics

Rotation

Centripetal Force

• The force required to keep a body rotating about an axis.

(r is the distance from the center of mass to the center of rotation.) Centrifugal Force

• The “reaction” to centripetal force.
• The centrifugal force, like any inertia force, should not be used in

free-body diagrams. Example (FEIM): A 2000 kg car travels 65 km/hr around a curve of radius 60 m. What is the centripetal force? 65 km/hr = 18.06 m/s F

c = m v 2

r = (2000 kg) 18.06m s

• 2

60 m = 10900 N

Professional Publications, Inc.

FERC

8-10c Dynamics

Rotation

Banking Curves 64km hr

• 1000 m

km

• 1hr

3600 s

• = 17.8m

s = arctan v

2

gr

• = arctan

17.8m s

• 9.8m

s

• (150 m)
• = 12.1°

Example (FEIM): A 2000 kg car travels at 64 km/hr around a banked curve with a radius

• f 150 m. What should the angle between the roadway and the

horizontal be so tire friction is not needed to prevent sliding?

Professional Publications, Inc.

FERC

8-10d Dynamics

Rotation

Free Vibration Spring and Mass: Natural Frequency: Solution: For initial conditions x(0) = x0 and x'(0) = v0, For initial conditions x(0) = x0 and x'(0) = 0, Torsional Free Vibration: Natural frequency: Solution: n = kt I

Professional Publications, Inc.

FERC

8-10e1 Dynamics

Rotation

Example (FEIM): A 54 kg mass is supported by three springs, as shown. The starting position is 5.0 cm down from the equilibrium position. No external forces act on the mass after it is released. What are the maximum velocity and acceleration?

Professional Publications, Inc.

FERC

8-10e2 Dynamics

Rotation

From the solution to the differential equation, x(t) = x0 cosnt + v0 n

• sinnt

v(t) = x (t) = x0n sinnt + v0 cosnt But, v0 = 0; so v(t) = x0n sinnt x0 = 0.05 m k = k1 + k2 + k3 = 1750 N m +1750 N m + 4375 N m = 7875 N m = 0.604 m/s sin12.08t n = k m = 7875 N m 54 kg = 12.08 rad/s v = x = (0.05 m) 12.08 rad s

• sin12.08t

Professional Publications, Inc.

FERC

8-10e3 Dynamics

Rotation

The maximum velocity is when sin 12.08t = 1. Because the motion is

• scillatory, the maximum velocity occurs in both directions.

vmax = ±0.604 m/s a = v = (0.05 m) 12.08 rad s

• 2

cos12.08t amax = ±7.30 m/s

2

Professional Publications, Inc.

FERC

8-11a1 Dynamics

Review

A yoyo (mass m, inertia I about center of gravity) is placed on a horizontal surface with a coefficient of friction µ. The string, wrapped underneath, is pulled with a force F. Determine if the yoyo will slip on the floor and which direction it will rotate as a function of F. The two equations for the movement of the center of gravity and the rotation are F +R = ma Fri +Rro = I Example:

Professional Publications, Inc.

FERC

8-11a2 Dynamics

Review

No slip: R < µmg a = ro There are two possibilities for the third equation: slip or no slip. Solving these 3 equations with 3 unknowns (α, a, and R) leads to = F ro ri I + mro

2 < 0

a = Fro ro ri

( )

I + mro

2

a > 0 R = F I + mri

2

I + mro

2

The yoyo is moving to the right while wrapping itself on the string. The solution for R gives the condition for no slip. F < µmg I + mro

2

I + mri

2

Professional Publications, Inc.

FERC

8-11a3 Dynamics

Review

The yoyo always moves to the right, but can rotate forward or backard depending on the value for F. Slip: R = –µmg Solving these 3 equations lead to a = F µmg m = Fri µmgro I