Section 8'2 Rotational Dynamics

1
Section 8.2 Rotational Dynamics

•  Objectives
• Describe torque and the factors that determine
  it.
• Calculate net torque.
• Calculate the moment of inertia.

2
INTRODUCTION/ Lever Arm

• Go over Figure 8-3 p. 201.
• Where do you want to apply the force to open a
  door? The best place would be the door knob or
  at that end.
•  
• Lever Arm the perpendicular distance from the
  axis of rotation to the point where the force is
  exerted. With the door example, it is the
  distance from the hinges to the point where you
  exert the force. When the force is not
  perpendicular the perpendicular component must be
  found using L r sin ?.
•  
• Torque is a measure of how effectively a force
  causes rotation. It is the product of the force
  and the lever arm. It is denoted by Greek letter
  Tau. It is measured in Newton Meters (Nm).
• t F r sin ?.

3
INTRODUCTION/ Lever Arm

• Do Example Problem 1 p. 202
• L r sin ?
• L .25(sin 60)
• L .217 m
•  
• t F r sin ?
• 35 F (.25)(sin 60)
• 35 .217 F
• 161.29 N F
•  
• Do Practice Problems p. 203 11-15

4
FINDING NET TORQUE

• t Fgr
•  
• t1 t2 0
•  
• Fg1r1 Fg2r2 0

5
FINDING NET TORQUE

• Do Example Problem 2 p. 204
• FgKrK FgArA
• mKgrK mAgrA
• 56(9.8)(x) 43(9.8)(1.75 x)
• 548.8x 421.4(1.75 x)
• 548.8x 737.45 - 421.4x
• 970.2x 737.45
• x .76 m From Kariann
• or .99 m from Aysha
•  
• Do Practice Problems p. 205 16-20

6
THE MOMENT OF INERTIA

•  Moment of Inertia - is equal to the mass of the
  object times the square of the objects distance
  from the axis of rotation. It is denoted by I.
  It is measured in kgm2.
• I mr2
•  
• Table 8.2 p. 206

7
THE MOMENT OF INERTIA

• Do Example Problem 3 p. 207
• r ½ of L (length of rod)
• r ½ (.65)
• r .325 m
•  
• Rotating about the Midpoint
• Single Mass
• ISingle Mass mr2
• I .3(.325)2
• I .3(.105625)
• I .032 kgm2 (moment of Inertia if rotated
  about midpoint)
• Entire Baton
• I 2I Single Mass
• I 2(.032)
• I .064 kgm2 (Moment of Inertia of the entire
  Baton)

8
THE MOMENT OF INERTIA

• Example 3 p. 207 continued
• Rotating around one end of the baton
• Single Mass
• ISingle Mass mr2
• I .3(.65)2
• I .3(.4225)
• I .127 kgm2 (moment of Inertia if rotated
  about one end)
• Entire Baton
• I ISingle Mass
• I .127 kgm2 (moment of Inertia if rotated
  about one end)

9
NEWTONS SECOND LAW FOR ROTATIONAL MOTION

• Newtons Second Law for Rotational Motion
  states that the angular acceleration of an object
  is equal to the net torque on the object divided
  by the moment of inertia.
• a tNet / I

10
NEWTONS SECOND LAW FOR ROTATIONAL MOTION

• Do Example Problem 4 p. 209
• First Second
• a ?? / ?t I ½ mr2
• a 2?(8 rev/s) 0 / 15 I ½ (15)(.22)2
• a 16(3.14) / 15 I 7.5(.0484)
• a 50.24 / 15 I .363 kgm2
• a 3.35 rad/s2

11
NEWTONS SECOND LAW FOR ROTATIONAL MOTION

• Finally
• a tNet / I
• 3.35 tNet / .363
• 1.216 Nm tNet  
• Solve for Force
• t Fr
• 1.216 F(.22)
• 5.53 N F
•  
• Do 8.2 Section Review p. 210 30-35