Angular Mechanics  Contents: Review Linear and angular Qtys Tangential Relationships Angular...

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Transcript of Angular Mechanics  Contents: Review Linear and angular Qtys Tangential Relationships Angular...
Angular Mechanics 
Contents:• Review
• Linear and angular Qtys• Tangential Relationships• Angular Kinematics
• Rotational KE• Example  Whiteboard
• Rolling Problems• Example  Whiteboard
Angular Mechanics  Angular Quantities
Linear:(m) s
(m/s) u(m/s) v
(m/s/s) a(s) t
(N) F (kg) m
Angular:  Angle (Radians)
o  Initial angular velocity (Rad/s)
 Final angular velocity (Rad/s)
 Angular acceleration (Rad/s/s)
t  Uh, time (s)
 Torque
I  Moment of inertiaTOC
Angular Mechanics  Angular kinematics
Linear:u + at = v
ut + 1/2at2 = su2 + 2as = v2
(u + v)t/2 = sma = F
1/2mv2 = Ekin
Fs = W
Angular: = o + t = ot + 1/2t2
2 = o2 + 2
= (o + )t/2* = I
*Not in data packet
Ek rot = 1/2I2
W = *
Angular Mechanics  Useful Substitutions
= I = rFso F = /r = I/rs = r, so = s/rv = r, so = v/ra = r, so = a/r
TOC
Angular Mechanics  Rotational Ke
TOC
Two types of kinetic energy:
Of course a rolling object has bothDemo – ring and cylinder
Translational: Ekin = 1/2mv2
Rotational: Ek rot = 1/2I2
Example: A 23.7 kg 45 cm radius cylinder is rolling at 13.5 m/s at the bottom of a hill.What is its translational kinetic energy?
What is its rotational kinetic energy?
What is the total kinetic energy? What was the height of the hill?
What is the rotational kinetic energy of an object with an angular velocity of 12 rad/s, and a moment of inertia of 56 kg m2?
Ek rot = 1/2I2
Ek rot = 1/2(56 kgm2)(12 rad/s)2
Ek rot = 4032 J = 4.0 x 103 J
4.0 x 103 J W
What must be the angular velocity of a flywheel that is a 22.4 kg, 54 cm radius cylinder to store 10,000. J of energy? (hint)
Ek rot = 1/2I2, I = 1/2mr2
Ek rot = 1/2(1/2mr2)2 = 1/4mr22
2 = 4(Ek rot)/mr2
=(4(Ek rot)/mr2)1/2=(4(10000J)/(22.4kg)(.54m)2)1/2
= 78.25 rad/s = 78 rad/s
78 rad/s W
What is the total kinetic energy of a 2.5 cm diameter 405 g sphere rolling at 3.5 m/s? (hint)
I=2/5mr2, = v/r, Ek rot=1/2I2 , Ekin=1/2mv2
Ek total= 1/2mv2 +1/2I2
Ek total= 1/2mv2 +1/2(2/5mr2)(v/r)2
Ek total= 1/2mv2 +2/10mv2 = 7/10mv2
Ek total= 7/10mv2 = 7/10(.405 kg)(3.5m/s)2
Ek total= 3.473 J = 3.5 J
3.5 J W
Angular Mechanics – Rolling with energy
TOC
mr  cylinder I = 1/2mr2
= v/rh
mgh = 1/2mv2 + 1/2I2
mgh = 1/2mv2 + 1/2(1/2mr2)(v/r)2
mgh = 1/2mv2 + 1/4mv2 = 3/4mv2
4/3gh = v2
v = (4/3gh)1/2
A 4.5 kg ball with a radius of 0.12 m rolls down a 2.78 m long ramp that loses 0.345 m of elevation. Set up the energy equation without plugging any of the knowns into it. Make substitutions for I and , but don’t simplify.
I = 2/5mr2, = v/rmgh = 1/2mv2 + 1/2I2
mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2
mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2 W
Solve this equation for v:mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2
mgh = 1/2mv2 + 2/10mr2v2/r2
mgh = 1/2mv2 + 2/10mv2 = 7/10mv2
10/7gh = v2
v = (10/7gh)1/2
v = (10/7gh)1/2 W
A 4.5 kg ball with a radius of 0.12 m rolls down a 2.78 m long ramp that loses 0.345 m of elevation.What is the ball’s velocity at the bottom? (v = (10/7gh)1/2)
v = (10/7(9.8m/s/s)(.345 m))1/2 = 2.1977 m/sv = 2.20 m/s
2.20 m/s W
A 4.5 kg ball with a radius of 0.12 m rolls down a 2.78 m long ramp that loses 0.345 m of elevation.What was the rotational velocity of the ball at the bottom? (v = 2.1977 m/s)
18 rad/s W
= v/r = (2.1977 m/s)/(.12 m) = 18.3 s1
= 18 s1
A 4.5 kg ball with a radius of 0.12 m rolls down a 2.78 m long ramp that loses 0.345 m of elevation.What was the linear acceleration of the ball down the ramp? (v = 2.1977 m/s)
0.869 m/s/s W
v2 = u2 + 2asv2/(2s) = a(2.1977 m/s)2/(2(2.78 m)) = 0.869 m/s/s
In General: I tend to solve all rotational dynamics problems using energy.
1. Set up the energy equation2. (Make up a height)3. Substitute linear for angular:
• = v/r• I = ?mr2
4. Solve for v5. Go back and solve for accelerations
TOC
Angular Mechanics – Pulleys and such
TOC
r
m1
m2
h
Find velocity of impact, and acceleration of systemr = 12.5 cmm1 = 15.7 kgm2 = 0.543 kgh = 0.195 m
Angular Mechanics – Pulleys and such
r
m1
m2
h = (made up)m3
Find acceleration of systemr = 46 cmm1 = 55 kgm2 = 15 kgm3 = 12 kgh = 1.0 m