1. Rotational Motion Honors Physics

2. Rotational Motion Objectives: Learn how to describe and measure rotational motion Learn how torque changes rotational velocity Explore factors that determine the stability of an object Learn the nature of centrifugal and Coriolis “forces”

3. Rotational Motion There are 360 degrees in one revolution There are 2 π radians in one revolution A radian is abbreviated “rad” and is technically unitless

4. Angular Displacement, ϴ The Greek letter theta, ϴ, is used to denote angular displacement. The angle of revolution is measured from the positive x axis, and goes counterclockwise, as shown on the graphic on the last page. The Earth makes one complete revolution, or 2 π radians in 24 hours. Through what angle, in radians, does the Earth rotate in 6 hours? π/2 radians

5. Angular Velocity

6. Angular Velocity, ω

7. As speed with a direction is called velocity, angular speed with a direction is called angular velocity. To assign a direction to a rotation we use a right hand rule as follows: 1. Rest the heel of your right hand on the rotating object. 2.Make sure your fingers are curled in the direction of rotation. 3.Your extended thumb points in the direction of the angular velocity. FYI  Angular velocity always points perpendicular to the plane of motion! Angular velocity – right hand rule  = 2  / T = 2  f =  / t relation between , T and f r v 

8. PRACTICE: Find the angular velocity (in rad s -1 ) of the wheel on the shaft. It is rotating at 30.0 rpm (revolutions per minute). SOLUTION:  The magnitude of  is given by  = (30.0 rev / 60 s)(2  rad / rev) = 3.14 rad s -1.  The direction of  is given by the right hand rule: “Place heel of right hand so fingers are curled in direction of rotation. Thumb gives the direction.” Angular velocity - example  = 2  / T = 2  f =  / t relation between , T and f

9. Angular Acceleration, α

10. Summary x = x o + v o t + a t t 2 1212 v = v o + a t t v 2 = v o 2 + 2a t  x θ = θ o + ω o t + αt 2 1212 ω = ω o + αt ω 2 = ω o 2 + 2α  θ The kinematic equations can be transformed into rotational variables :

11. Rotational Dynamics

12. How do you start something rotating? For a given applied force, the change in angular velocity depends on the lever arm. The lever arm is the perpendicular distance from the axis of rotation to the point where the force is exerted. For the door above, the lever arm is the distance from the hinges to the point where force was applied at the handle.

13. Lever Arms If the force is not exerted perpendicular to the axis of rotation, as shown at right, the lever arm is reduced. To find the lever arm, extend the line of force backwards until it forms a right angle with a line from the center of rotation. The distance from between the axis of rotation and that right angle is the lever arm and is equal to rsin as shown

14. Torque The tendency of a force to rotate an object about some axis is measured by a quantity called torque τ τ = Fd torque = Force x distance There are two ways to calculate torque. I’m going to show you both, but I prefer the 2 nd one.

15. On the left, the lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. Here, d = L sin Ф, so if τ = Fd then τ = FLsinФ L

16. On the right, you can see r = L, but instead we resolve the force into parallel and perpendicular components of the force. Here, if τ = Fd then τ = (FsinФ)(L) = FLsinФ L

17. Torque Both methods work out to the torque being equal to FLsinФ, but it seems much more logical to simply partition the force into the parallel and perpendicular components (relative to the wrench). The parallel component has zero torque associated with it. The perpendicular component is responsible for all of the torque that occurs. r = L

18. The Sign of Torque If the motion is turning counterclockwise, torque is positive. If the motion is turning clockwise, torque is negative. So the torque F 1 on the pivot point O is counterclockwise and positive. The torque for F 2 on O is clockwise and negative. Note that the units of torque τ = Fd are force times distance, or units like Newtons x meters.

19. Torque example The figure at the right shows a crate (with center of mass at the center of the box) that is being acted on by two forces. Each force (alone) produces a clockwise rotation, so the torques are both negative. τ 1 + τ 2 = F 1 d 1 + F 2 d 2 = -(500N)(0.50m) –(500N)(0.50m) = - 500 Nm Note how the lever arms d 1 and d 2 are constructed !!

20. Torque example Here if you use the 2 nd method described, the lever arm is the full 2.0m. But then you have to find the component of the force that is perpendicular to the lever arm. The y-component of the 300N force is 300sin60, or 260N. The rotation is counterclockwise (+). τ = Fd = (260N)(2.0m) = +520 Nm

21. Equilibrium There are two conditions for mechanical equilibrium: ΣF = 0 Στ = 0 The second condition asserts that if an object is in rotational equilibrium, the net torque acting on it about any axis is zero. If the object is in equilibrium, it does not matter what is chosen as the axis of rotation for calculating net torque – it is completely arbitrary.

22. Torque example A bolt needs to be tightened with a torque of 35 N∙m. What force should be applied to the end of wrench? Using first method: Length of lever arm = L sin ϴ Lever arm = (0.25)sin60 = 0.22 m Solve for the force τ = Frsin 35 N∙m = F (0.22) F = 35 N∙m / (0.22) F = 160 N

23. Torque example A bolt needs to be tightened with a torque of 35 N∙m. What force should be applied to the end of wrench? Using second method: Length of lever arm = 25 cm =0.25m Perpendicular component of force: F sin 60 Solve for the force τ = (Fsin)(r) 35 N∙m = F sin60(0.25) F = 35 N∙m / (sin60)(0.25) F = 160 N Easier to understand? F perp

24. Example - equilibrium Kariann (left) 56kg Aysha (right) 43 kg Total length of see saw 1.75 m. To be balanced, where should they place the pivot point? Note 1.75 m = r K + r A Kariann rotates counterclockwise (+) Aysha rotates clockwise (-)

25. Example - equilibrium Kariann (left) 56kg Aysha (right) 43 kg F = mg = (56)(9.8) = 550N F= mg = (43)(9.8) = 420N When no rotation, sum of torques = 0 so τ K = τ A F gK r K = F gA r A and r K + r A = 1.75 m (550N)(1.75-r A ) = (420N)(r A ) (550N)(1.75) = (550+420)r A r A = 0.99m so r k = 0.76m

26. Moment of Inertia How does an object rotate when a torque is exerted on it? Recall for F = ma, the amount of mass creates the inertia (resistance to movement) of the object. Similar situation with rotation….. The resistance to rotation is called the moment of inertia, which is represented by the symbol I. I = mr 2 Units are mass x distance 2. Distance is the object’s distance from the axis of rotation.

27. Moments of inertia for various objects

28. Moment of Inertia The moment of inertia of a book depends on the axis of rotation. The moment of inertia in the top book is larger than the moment of inertia in the bottom book because the average distance of the book’s mass from the rotational axis is larger.

29. Example Compare the moments of inertia for the two identical rods, rotating about different axes. Top rod: r = ½ L = ½ (0.65m) = 0.33 m I single mass = mr 2 = (0.30kg)(0.33m) 2 = 0.033kg∙m 2 Two masses I total = 2∙I single mass = 0.066 kg∙m 2 Bottom rod: r= L, I = mr 2 =(0.30kg)(0.65m) 2 I total = 1.13 kg∙m 2

30. Example Rank the objects shown in the diagram according to their moments of inertia about the indicated axes. All spheres have equal masses, and all spacing is the same. B > C > D > A

31. Newton’s 2 nd law- rotational motion

32. Example

33. Rigid Objects in Equilibrium - Example Example 3 A Diving Board A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board.

34. Rigid Objects in Equilibrium - Example Remember counterclockwise (+) clockwise (-) F 1 no torque, goes through the axis

35. Rigid Objects in Equilibrium - Example Now still need to find F 1 Can use equilibrium of forces to find it Can’t use torque because F 1 contributed no torque.

36. 8.3 Equilibrium and Center of Mass You are responsible for reading this section and having a general understanding of the concept, but due to time constraints, we will not be working problems in this section. Not to worry, you will have lots of fun with this topic in AP Physics