Rotational Dynamics

1
Rotational Dynamics
2

• Consider a bicycle wheel to be a ring of radius
  30 cm and mass 1.5 kg. Neglect the mass of the
  axle and sprocket. If a force of 20 N is applied
  tangentially to a sprocket of radius 4.0 cm for
  4.0 s, what linear speed does the wheel achieve,
  assuming it rolls without slipping?
• a. 3.0 m/s
• b. 5.9 m/s
• c. 7.1 m/s
• d. 24 m/s

3

• M 1.5 kg, R 0.30 m, F 20 N, r 0.04 m
• t 4 s
• Thin hoop, I MR2 0.135 kg m2
• t Ia
• t/I
• 20 N 0.04 m/0.135 kg m2
• 5.9 rad/s2
• at 5.9 4 23.6 rad/s
• v wr 23.6 rad/s 0.3 m 7.1 m/s

4

• A wheel of moment of inertia of 5.00 kgm2 starts
  from rest and accelerates under a constant torque
  of 3.00 Nm for 8.00 s. What is the wheel's
  rotational kinetic energy at the end of 8.00 s?
• a. 57.6 J
• b. 64.0 J
• c. 78.8 J
• d. 122 J

5

• I 5 kgm2, t 3.00 Nm , t 8.00 s
• Krot ???
• Krot ½ I w2 1/2(5)w2
• To find omega, w, use w wo at.
• How do we find alpha, a?
• a t/I 3/5 0.6 rad/s2.
• So, w wo at 0 0.6 8 4.8 rad/s
• So,
• Krot ½ I w2 1/2(5)w2 57.6 J

6

• A solid sphere of mass 1.0 kg and radius 0.010 m
  is released from the top of a 1.0-m high 370
  inclined plane. What is the speed of the sphere
  when it reaches the bottom of the inclined plane?
• a. 3.7 m/s
• b. 4.4 m/s
• c. 5.6 m/s
• d. 6.3 m/s

7

• M 1.0 kg, R0.010 m, y 1.0 m
• V ???
• First, lets set up the linear and rotational
  dynamics equations for this sphere.
• F Ma
• Mgsinq f Ma
• Mgsinq Ma f
• Ia
• rf I a/r
• f Ia/r2 2/5MR2 (a/R2) 2/5 Ma
• Combining, Mgsinq Ma f 2/5 Ma
• Gives a 5/7 g sinq 4.2 m/s2

8

• v vo at 0 4.2 m/s2 t.
• Find t from, x ½ a t2 ,
• t (2x/a)1/2 (21.66 m/ 4.2 m/s2)1/2 0.89 s
• So, v 3.7 m/s.

9
Rotational Dynamics

• Torque, t I a
• Work, W tT
• Kinetic Energy of Rotation, K ½ I?2
• Conservation of Energy
• Angular Momentum, L I?
• Conservation of Angular Momentum

10
Newtons 2nd Law

• Linear motion
• Fnet ma

• Linear motion
• tnet Ia

11

• tnet Ia
• tnet R(T2 T1)
• I ½ MR2
• R(T2 T1) ½MR2a

Fnet m1 a Fnet T1 - µN N m1g T1 -
µm1gm1a T1 m1a -µm1g

N
Accel. a
Ang. Accel a
f µN
T1
M,R
T2
Fnet m2 a Fnet m2g T2 m2g T2 m2a T2
-m2a m2g
Accel a
m1g
m2g
12

T1 m1a -µm1gT2 T1 ½MRaT2 -m2a m2g -m2a
m2g m1a µm1g ½MRa -(m2 m1)a g(m2 µm1)
½MRa/R (m2 m1 ½M)a g(µm1 m2) a g(µm1
m2)/(m2 m1 ½M)

N
Accel. a
Ang. Accel a
f µN
T1
M,R
T2
Accel a
m1g
m2g
13

a g(µm1 m2)/(m2 m1 ½M) Let M 0.2 kg, m11
kg, m20.5 kg, µ0.1 a 3.7 m/s2, What if we
ignore pulley? a 3.92 m/s2, 6.7 higher!

N
Accel. a
Ang. Accel a
f µN
T1
M,R
T2
Accel a
m1g
m2g
14
It took work to rotate the pulley

• The pulley goes from zero angular speed to
  non-zero angular speed, ? at.
• This means that the pulleys final kinetic
  energy, Kf , is ½ I?2
• Kf ½(½ MR2)?2
• The work-energy theorem, W ?K
• W tT¼MR2?2

15

• Torque and Angular Momentum
• Recall, F ma m?v/?t ?p/?t, so making the
  correspondence to rotational motion,
• t Ia I??/?t ?(I?)/?t ?L/?t, where
• L I? angular momentum.
• t ?L/?t
• When no external forces are acting,
• torque 0, so
• ?L/?t 0 which means, Lf Lo or
• If?f Io?o