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

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Angular Mechanics - Rotational Ke

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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?

Whiteboards: Rotational KE

1 | 2 | 3

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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

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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

Whiteboards: Rolling with Energy

1 | 2 | 3

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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 s-1

= 18 s-1

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

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Angular Mechanics – Pulleys and such

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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

Angular Mechanics – yo yo ma

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Find acceleration of system (assume it is a cylinder)r1 = 6.720 cmr2 = 0.210 cmm = 273 g

r1

r2

h = 1.0 m