1. Rotational Dynamics

2. The analogies between linear & rotational motion continue.
Rotational Dynamics
The Causes of rotational motion!
The analogies between linear & rotational motion continue.
Newton’s 3 Laws are still valid!
But, here we write them using rotational language and notation.

3. Translational-Rotational Analogues Continue!
Translation Rotation
Displacement x θ
Velocity v ω
Acceleration a α
Force F τ (torque)

4. Torque
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.

5.  Introduce the torque concept. Angular acceleration α  F.
To cause a body to rotate about an axis requires a FORCE, F. (Cause of angular acceleration α).
BUT: The location of the force on the body and the direction it acts are also important!
 Introduce the torque concept.
Angular acceleration α  F.
But also α  (the distance from the point of application of F to the hinge = Lever Arm, r)
 From experiment! 

6. newton · meter (N · m) or (N · m)
Definition of Torque
τ  Fr
r  Lever Arm
Direction
The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation.
SI Unit forTorque:
newton · meter (N · m) or (N · m)

7. RA, RB ≡ “Lever Arms” for FA & FB.
Angular acceleration α  force F, but also  distance
from the point of application of F to the hinge (“Lever Arm”)
FA = FB, but which
gives a greater α ?
Hinge
RA, RB ≡ “Lever Arms” for FA & FB.
α  “Lever Arm”

8. The line of action is an extended line drawn colinear with the force
The line of action is an extended line drawn colinear with the force. The lever arm is the distance between the line of action and the axis of rotation, measured on a line that is perpendicular to both. The torque is represented by the symbol t (Greek letter tau), and its magnitude is defined as the magnitude of the force times the lever arm.

9. Lever Arm
Lever Arm  r = distance of the axis of rotation from the “line of action” of force F
r = Distance which is  to both the axis of rotation and to an imaginary line drawn along the direction of the force (“line of action”).
Find: Angular acceleration
α  (force) (lever arm) = Fr
 Define: TORQUE
τ  Fr τ causes α
(Just as in the linear motion case, F causes a)
Lower case
Greek “tau” 

10. Door Hinge Torques: Forces at angles are less effective   rA  
Due to FA: τA = rAFA
Due to FC : τC = rCFC
Due to FD: τD = 0
(Since the lever arm is 0)
τC < τA (For FC = FA)
  rA  
rC is the Lever
Arm for FC rC
The lever arm for FA is the distance from the knob to the hinge. The lever arm for FD is zero. The lever arm for FC is as shown.

11. F= F sinθ r= r sinθ F = F cosθ τ = rF sinθ τ = rF sinθ Units of τ:
OR, resolve F into
components F & F
τ = rF 
 These are the
same, of course!  
In general, write τ = rF 
F= F sinθ
F = F cosθ
τ = rF sinθ
r= r sinθ
τ = rF sinθ
Units of τ:
N m = m N

12. Torque In general, write τ = rF
Or, resolving F into components F|| and F:
τ = rF
Even more generally:
τ = rF sinθ
Units of torque: Newton-meters (N m)

13. Example: Biceps Torque
τ = rF = 35 m N
τ = rF = 30 m N

14. Exercise B

15. More than one torque? α  τnet = ∑τ = sum of torques
If there is more than one torque:
α  τnet = ∑τ = sum of torques
Always use the following sign convention!
Counterclockwise rotation  + torque
Clockwise rotation  - torque

16. Example τ = τA + τB = - 6.7 m N --------------> τA= rAFA
r= rBsin60º
>
τB = -rBFBsin60º
2 thin disk-shaped wheels, radii rA = 30 cm & rB = 50 cm, are attached to each other on an axle through the center of each. Calculate the net torque on this compound wheel due to the 2 forces shown, each of magnitude 50 N.
τ = τA + τB
= m N
τA= rAFA

17. Net torque: ∑τ = τA + τB + τC + τfr = -1.4 m N
Problem
τA = - (0.24 m)(18 N)
= m N
τB = + (0.24 m)(28 N)
= m N
τC = - (0.12 m)(35 N)
= m N
τfr = m N
Net torque: ∑τ = τA + τB + τC + τfr = -1.4 m N
35 N
28 N
12 cm
24 cm
18 N

18. Translational-Rotational Analogues & Connections Continue!
Translation Rotation
Displacement x θ
Velocity v ω
Acceleration a α
Force (Torque) F τ
Mass m ?
CONNECTIONS
v = rω
atan= rα
aR = (v2/r) = ω2r
τ = rF