1. Section 10.6: Torque

2. Translation-Rotation Analogues & Connections
Displacement x θ
Velocity v ω
Acceleration a α
Mass (moment of inertia) m I
Kinetic Energy (K) (½)mv (½)Iω2
Force F ?
CONNECTIONS
s = rθ, v = rω, at= rα, ac = (v2/r) = ω2r

3. Newton’s 1st Law (rotational language version): “A rotating body will continue to rotate at a constant angular velocity unless an external TORQUE acts.”
Clearly, to understand this, we need to define the concept of TORQUE.
Newton’s 2nd Law (rotational language version): Also needs torque.

4.  Introduce the torque concept. Angular acceleration α  F.
To cause an object to rotate requires a FORCE, F. (Cause of angular acceleration α).
BUT: The location of the force on the object & the direction it acts are also important!
 Introduce the torque concept.
Angular acceleration α  F.
But also α  (distance from the point of application of F to the hinge = Moment Arm or Lever Arm, d)
 From experiment! 

5. α  (force) (moment arm) = Fd
Moment Arm  d = distance of the axis of rotation from the “line of action” of force F
d = Distance which is  to both the axis of rotation and to an imaginary line drawn along the direction of the force (“Line of Action”).
Line of Action  Imaginary line extending out both ends of the force vector. Experiment finds that angular acceleration
α  (force) (moment arm) = Fd
 Define: TORQUE
τ  Fd τ causes α
(Just as in the linear motion case, F causes a)
Lower case
Greek “tau” 

6. Newton’s laws & rotational motion
We want to find a rotational analogue to force
The figure is a top view of a door that is hinged on the left: The 4 pushing forces are of equal strength. Which of these will be the most effective at opening the door?
F1 will open the door.
F2 will not.
F3 will open the door, but not as easily as F1.
F4 will open the door – it has same magnitude as F1, but we know it is not as effective as pushing at the outer edge of the door.

7. 1. The magnitude F of the force.
Ability of force F to cause a rotation or twisting motion depends on 3 factors:
1. The magnitude F of the force.
2. The distance r from point of
application to pivot.
3. The angle at which F is applied.
Make these idea quantitative.
Figure is a force F applied at 
one point on a rigid body.
τ depends on the 3 properties, & is the rotational analogy to force.

8. Let’s see this again

9. Torque Units:
Newton-meter = N m
Sign convention: A torque that tends to rotate an object in a counterclockwise direction is positive. torque that tends to rotate an object in clockwise direction is positive. See Figure.

10. Section 10.7: Rigid Object Under a Net Torque
Consider the object shown.
The force F1 will tend to cause
a counterclockwise rotation about O.
The force F2 will tend to cause
a clockwise rotation about O
The net torque is:
τnet = ∑τ = τ1+ τ2
= F1d1 – F2d2

11. We’ve seen that toque is the rotational analogue to force.
Now we need to learn what toque does.
Figure is a model rocket engine attached to one end of lightweight, rigid rod.
Tangential acceleration at & the angular acceleration α are related.
Tangential component of force.
Multiply both sides by r:
Therefore, torque τ causes angular acceleration α. This relation is analogous to Newton’s 2nd Law F = ma

12. Newton’s 2nd Law for Rotations
Figure: A rigid body undergoes pure rotational motion about a fixed, frictionless, & unmoving axis. Net torque on the object is the sum of the torques on all the individual particles.
Definition: Moment of Inertia
Newton’s 2nd law for rotational motion

13. Ex. 10.7: Net Torque on a Cylinder
Double cylinder, shaped as shown. Attached to axle through center. Large part, radius R1 has rope around it & is pulled, with tension T1 to right. Small part, radius R2 has a rope around it & is pulled down with tension T2.
(A) Net torque?
∑τ = τ1+ τ2 = T2R2 – T1R1
(B) T1 = 5 N, to R1 = 1 m, T2 = 15 N, R2 = 0.5 m.
Net torque? Which direction will it rotate?
∑τ = (15)(0.5) – (1)(5) = 2.5 N m
Direction is counterclockwise (positive torque)

14. Ex. 10.8: Rotating Rod

15. Ex. 10.10: Angular Acceleration of a Wheel
Wheel, radius R, mass M, moment of inertia I. A cord is wrapped around it & attached to mass m. System is released & m falls & wheel rotates. Find the tension T in the cord, acceleration a of falling m, angular acceleration α of wheel.
Newton’s 2nd Law for wheel: ∑τ = Iα (1)
Tension T is tangential force on wheel & is the only force producing a torque. So: ∑τ = TR (2)
m moves in a straight line
Newton’s 2nd Law for m: SFy = ma = mg – T (3) No slipping of cord.  a = αR (4)
From (1) & (2): α = (∑τ/I) = (TR)/I (5)
From (4) & (5): a = (mg – T)/m = (TR2)/I (6)
Solve for T from (6): T = (mg)/[1 + (mR2/I)]
Put into (6) to get a: a = g/[1 + (I/mR2)]
Put into (4) to get α: α = g/[R + (I/mR)]
 a