Physics - BRHS

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 perpendicularto 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

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. 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. 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: N Determine the net torque: 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 aFree 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 torquet 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: N T Mg 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. 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. 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.