1. Torque Looking back Looking ahead Newton’s second law.
The mathematics of circular motion.
Looking ahead
Apply rigid-body model to extended object.
Understand the rotation of a rigid body around a fixed axis. 1

2. Torque Definition ¿ = +- r Ftangential = r F sin Á
Rotational effect of force. Tells how effective force is at twisting or rotating an object.
¿ = +- r Ftangential = r F sin Á
Units N m
Sign, CCW rotation is positive
Depends on 3 properties:
1) Magnitude of force F.
2) Its distance r from pivot point.
3) Angle Á.

3. Concept Quiz
The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.
In which of the cases is the torque on the nut the biggest?
A. Case 1 B. Case 2 C. Case 3
CORRECT 10

4. Concept Quiz
The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.
In which of the cases is the torque on the nut the smallest?
A. Case 1 B. Case 2 C. Case 3
CORRECT 12

5. Torque Example and Quiz
A person raises one leg to an angle of 30 degrees. An ankle weight (89 N) attached a distance of 0.84 m from her hip. What is to torque due to this weight?
1) Draw Diagram
2) t = F r sin q
= F r sin(90 – 30)
If she raises her leg higher, the torque due to the weight will
A) Increase
B) Same
C) Decrease 30
F=89 N
r = 0.84 m
= 65 N m 17

6. Equilibrium Quiz
A rod is lying on a table and has two equal but opposite forces acting on it. What is the net force on the rod?
A) Up B) Down C) Zero
Will the rod move? A) Yes B) No
Y direction: S Fy = may
+F – F = 0 y x F
Yes, it rotates! F 21

7. Equilibrium Conditions for Equilibrium S F = 0 Translational EQ
S t = Rotational EQ
Can choose any axis of rotation…. Choose Wisely!
Steel from podium 25

8. Equilibrium Example
A meter stick is suspended at the center. If a 1 kg weight is placed at x=0. Where do you need to place a 2 kg weight to balance it?
A) x = 25 B) x=50 C) x=75 D) x=100
E) 1 kg can’t balance a 2 kg weight.
Steel from podium
9.8 N
17.6 N
50 cm d
S t = 0
9.8 (0.5) – (17.6)d = 0
d = 25
Balance Demo 25

9. Static Equilibrium and Center of Mass
pivot d
Gravitational Force Weight = mg
Acts as force at center of mass
Torque about pivot due to gravity:
¿ = WrsinÁ =mgd
Object not in static equilibrium
W=mg

10. Static Equilibrium Not in equilibrium Equilibrium
Center of mass
pivot d
W=mg
Torque about pivot  0
Center of mass
pivot
Torque about pivot = 0
Not in equilibrium
Equilibrium
A method to find center of mass of an irregular object

11. Equilibrium Quiz and Example
The picture below shows two people lifting a heavy log. Which of the two people is supporting the greatest weight?
1. The person on the left is supporting the greatest weight
2. The person on the right is supporting the greatest weight
3. They are supporting the same weight
CORRECT
Look at torque about center:
FR L – FL L/2 = 0
FR= ½ FL mg FL FR
L/2 L 34

12. Example
A 75 kg painter stands at the center of a 50 kg 3 meter plank. The supports are 1 meter in from each edge. Calculate the force on support A.
1 meter
1 meter A B FA FB Mg mg
1) Draw FBD
2) SF = 0
3) Choose pivot
4) St = 0
1 meter
0.5meter
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle
FA + FB – mg – Mg = 0
-FA (1) sin(90)+ FB (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0
FA = 0.5 mg Mg = Newtons 40

13. Concept Quiz
If the painter moves to the right, the force exerted by support A
A) Increases B) Unchanged C) Decreases
1 meter
1 meter A B FA FB Mg mg
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle
¿A also decreases
¿M decreases 43

14. Example
How far to the right of support B can the painter stand before the plank tips?
1 meter
1 meter A B
Just before board tips, force from A becomes zero FB Mg mg
1) Draw FBD
2) SF = 0
3) Choose pivot
4) St = 0
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle
FB – mg – Mg = 0
0.5meter x
FB (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = 0
0.5 m = x M 47

15. Summary Torque = Force that causes rotation Equilibrium t = F r sin q
S F = 0
S t = 0
Can choose any axis.

16. Torque and rotational motion
Torque is the quantity that changes rotational motion.
For rotations, it is the object that replaces force as the cause of angular acceleration.

17. Torque and Rotational Motion
Linear motion: F = ma (Newton’s 2nd law linear motion)
If force is tangential to a circular path:
F=Ft and a = at=r®
Rotational motion:
¿ = Ft r = m atr=(mr2)® =I®
(Newton’s 2nd law rotational motion)
Extended object:
I =  mr2 F t r

18. Moment of inertia I = i miri2
Net torque leads to angular acceleration, but we need the moment of inertia to tell us how large an acceleration results.
Moment of inertia in rotational problems takes the place that mass has in translational problems.
Depends upon the mass and location of the mass from the axis of rotation.
Large moment of inertia requires large torque to accelerate object (just like large mass requires large force)

19. Concept Quiz
Consider a rod of uniform density with an axis of rotation through its center and an identical rod with the axis of rotation through one end. Which has the larger moment of inertia?
A IC > IE
B IC < IE
C IC = IE

20. Concept Quiz A: IA > IB B: IA < IB C: IA = IB
Consider two masses, each of size 2m at the ends of a light rod of length L with the axis of rotation through the center of the rod. The rod is doubled in length and the masses are halved. What happens to I?                                               
A: IA > IB
B: IA < IB
C: IA = IB

21. Concept Test
Consider a solid disk with an axis of rotation through the center (perpendicular to the diagram).
Two holes are cut out near the center and the material is placed near the rim. Which has the larger I?                                
A IA > IB
B IA < IB
C IA = IB

22. Concept Test
A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates.
               
The magnitude of the torque on the pulley is..
A: greater than mgR
B: equal to mgR
C: less than mgR
  (Hint: Is the tension in the string = mg?)

23. Acceleration of the mass m?
¿ = Ts R= Icyl ®
® = T_sR/ (0.5MR^2)= 2Ts/MR
Tension of the string from linear motion of block:
Ts-mg=may
ay = at = - ® R
® = -ay/R = -(Ts-mg)/(mR)
Ts = -mR ®+mg
® = 2Ts/MR = 2(-mR® +mg)R/MR = 2(mR®+mg)/M