1. 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 θ. 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). ***** τ = F r sin θ. *****

3. INTRODUCTION/ Lever Arm Do Example Problem 1 p. 202 L = r sin θ L =.25(sin 60°) L =.217 m τ = F r sin θ 35 = F (.25)(sin 60°) 35 =.217 F 161.29 N = F Do Practice Problems p. 203 # 11-15

4. FINDING NET TORQUE τ = F g r τ 1 – τ 2 = 0 F g1 r 1 – F g2 r 2 = 0

5. FINDING NET TORQUE Do Example Problem 2 p. 204 F gK r K = F gA r A m K gr K = m A gr A 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 object’s distance from the axis of rotation. It is denoted by “I”. It is measured in kg*m 2. I = mr 2 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 I Single Mass = mr 2 I =.3(.325) 2 I =.3(.105625) I =.032 kg*m 2 (moment of Inertia if rotated about midpoint) Entire Baton I = 2I Single Mass I = 2(.032) I =.064 kg*m 2 (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 I Single Mass = mr 2 I =.3(.65) 2 I =.3(.4225) I =.127 kg*m 2 (moment of Inertia if rotated about one end) Entire Baton I = I Single Mass I =.127 kg*m 2 (moment of Inertia if rotated about one end)

9. NEWTON’S SECOND LAW FOR ROTATIONAL MOTION Newton’s 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. *****α = τ Net / I *****

10. NEWTON’S SECOND LAW FOR ROTATIONAL MOTION Do Example Problem 4 p. 209 First Second α = Δω / ΔtI = ½ mr 2 α = 2Π(8 rev/s) – 0 / 15I = ½ (15)(.22) 2 α = 16(3.14) / 15 I = 7.5(.0484) α = 50.24 / 15I =.363 kg*m 2 α = 3.35 rad/s 2

11. NEWTON’S SECOND LAW FOR ROTATIONAL MOTION Finally α = τ Net / I 3.35 = τ Net /.363 1.216 N*m = τ Net Solve for Force τ = Fr 1.216 = F(.22) 5.53 N = F Do 8.2 Section Review p. 210 # 30-35