1. Physics - BRHS Rotational dynamics torque moment of inertia
angular momentum
conservation of angular momentum
rotational kinetic energy

2. Lightning Review Last lecture: Rotations
angular displacement, velocity (rate of change of angular displacement), acceleration (rate of change of angular velocity)
motion with constant angular acceleration
Gravity laws
Review Problem: A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

3. Sample Review Problem
An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. Find the angle at which the curve should be banked if a typical car rounds it at a 50.0-m radius and a speed of 13.4 m/s.

4. Rotational Equilibrium and Rotational Dynamics

5. Torque
Consider force required to open door. Is it easier to open the door by pushing/pulling away from hinge or close to hinge?
close to hinge
away from hinge
Farther from from hinge, larger rotational effect!
Physics concept: torque

6. Torque
Torque, , is the tendency of a force to rotate an object about some axis
is the torque
d is the lever arm (or moment arm)
F is the force
Door example:

7. Lever Arm
The lever arm, d, is the shortest (perpendicular) distance from the axis of rotation to a line drawn along the the direction of the force
d = L sin Φ
It is not necessarily the distance between the axis of rotation and point where the force is applied

8. Direction of Torque Torque is a vector quantity
The direction is perpendicular to the plane determined by the lever arm and the force
Direction and sign:
If the turning tendency of the force is counterclockwise, the torque will be positive
If the turning tendency is clockwise, the torque will be negative
Direction of torque:
out of page
Units SI
Newton meter (Nm)
US Customary
Foot pound (ft lb)

9. An Alternative Look at Torque
The force could also be resolved into its x- and y-components
The x-component, F cos Φ, produces 0 torque
The y-component, F sin Φ, produces a non-zero torque L
F is the force
L is the distance along the object
Φ is the angle between force and object

10. ConcepTest 1
You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?
1. No
2. Yes

11. Convince your neighbor!
ConcepTest 1
You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?
1. No
2. Yes
Convince your neighbor!

12. ConcepTest 2
You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

13. Convince your neighbor!
ConcepTest 2
You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:
Convince your neighbor!

14. What if two or more different forces act on lever arm?

15. Net Torque
The net torque is the sum of all the torques produced by all the forces
Remember to account for the direction of the tendency for rotation
Counterclockwise torques are positive
Clockwise torques are negative

16. Example 1: Determine the net torque: N 4 m 2 m Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d1=4 m
d2=2 m
Find:
St = ?
500 N
800 N
1. Draw all applicable forces
2. Consider CCW rotation to be positive
Rotation would be CCW

17. Where would the 500 N person have to be relative to fulcrum for zero torque?

18. Example 2: N’ y d2 m 2 m Given: weights: w1= 500 N w2 = 800 N
lever arms: d1=4 m
St = 0
Find:
d2 = ?
500 N
800 N
1. Draw all applicable forces and moment arms
According to our understanding of torque there would be no rotation and no motion!
What does it say about acceleration and force?
Thus, according to 2nd Newton’s law SF=0 and a=0!

19. Torque and Equilibrium
First Condition of Equilibrium
The net external force must be zero
This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium
This is a statement of translational equilibrium
Second Condition of Equilibrium
The net external torque must be zero
This is a statement of rotational equilibrium

20. Axis of Rotation
So far we have chosen obvious axis of rotation
If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque
The location of the axis of rotation is completely arbitrary
Often the nature of the problem will suggest a convenient location for the axis
When solving a problem, you must specify an axis of rotation
Once you have chosen an axis, you must maintain that choice consistently throughout the problem

21. Center of Gravity
The force of gravity acting on an object must be considered
In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point

22. Calculating the Center of Gravity
The object is divided up into a large number of very small particles of weight (mg)
Each particle will have a set of coordinates indicating its location (x,y)
The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm
We wish to locate the point of application of the single force , whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles.
This point is called the center of gravity of the object

23. Coordinates of the Center of Gravity
The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object
The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry.
Often, the center of gravity of such an object is the geometric center of the object.

24. Example: Find center of gravity of the following system: Given:
masses: m1= 5.00 kg
m2 = 2.00 kg
m3 = 4.00 kg
lever arms: d1=0.500 m
d2=1.00 m
Find:
Center of gravity

25. Experimentally Determining the Center of Gravity
The wrench is hung freely from two different pivots
The intersection of the lines indicates the center of gravity
A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object’s center of gravity

26. Equilibrium, once again
A zero net torque does not mean the absence of rotational motion
An object that rotates at uniform angular velocity can be under the influence of a zero net torque
This is analogous to the translational situation where a zero net force does not mean the object is not in motion

27. Example of a Free Body Diagram
Isolate the object to be analyzed
Draw the free body diagram for that object
Include all the external forces acting on the object

28. Example
Suppose that you placed a 10 m ladder (which weights 100 N) against the wall at the angle of 30°. What are the forces acting on it and when would it be in equilibrium?

29. Example: Given: weights: w1= 100 N length: l=10 m angle: a=30° St = 0
Find:
f = ?
n=?
P=? mg a
1. Draw all applicable forces
2. Choose axis of rotation at bottom corner (t of f and n are 0!)
Torques: Forces:
Note: f = ms n, so

30. So far: net torque was zero.
What if it is not?

31. Torque and Angular Acceleration
When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration
The angular acceleration is directly proportional to the net torque
The relationship is analogous to ∑F = ma
Newton’s Second Law

32. Torque and Angular Acceleration
torque t
dependent upon object and axis of rotation. Called moment of inertia I. Units: kg m2
The angular acceleration is inversely proportional
to the analogy of the mass in a rotating system

33. Example: Moment of Inertia of a Uniform Ring
Image the hoop is divided into a number of small segments, m1 …
These segments are equidistant from the axis

34. Other Moments of Inertia

35. Newton’s Second Law for a Rotating Object
The angular acceleration is directly proportional to the net torque
The angular acceleration is inversely proportional to the moment of inertia of the object
There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object.
The moment of inertia also depends upon the location of the axis of rotation

36. Let’s watch the movie!

37. Example:
Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m and weight of 9.8 N hanging from rope wrapped around flywheel.
What are forces acting on flywheel and weight? Find acceleration of the weight. mg

38. Example: Given: masses: M = 5 kg weight: w = 9.8 N radius: R=0.2 mMg T
Given:
masses: M = 5 kg
weight: w = 9.8 N
radius: R=0.2 m
Find:
Forces=? mg
1. Draw all applicable forces
Forces: Torques:
Tangential acceleration at the edge of flywheel (a=at):

39. ConcepTest 3
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.

40. Convince your neighbor!
ConcepTest 3
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?
Convince your neighbor!
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.

41. Return to our example:
Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg
If flywheel initially at rest and then begins to rotate, a torque must be present:
Define physical quantity:

42. Angular Momentum
Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum
Angular momentum is defined as L = I ω
If the net torque is zero, the angular momentum remains constant
Conservation of Linear Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero.
That is, when
(compare to )

43. Return to our example once again:
Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg
Each small part of the flywheel is moving with some velocity. Therefore, each part and the flywheel as a whole have kinetic energy!
Thus, total KE of the system:

44. Total Energy of Rotating System
An object rotating about some axis with an angular speed, ω, has rotational kinetic energy ½Iω2
Energy concepts can be useful for simplifying the analysis of rotational motion
Conservation of Mechanical Energy
Remember, this is for conservative forces, no dissipative forces such as friction can be present

45. ConcepTest 4
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.

46. Convince your neighbor!
ConcepTest 4
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?
Convince your neighbor!
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.