Chapter 9. Rotational Dynamics

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Chapter 9. Rotational Dynamics"

Transcription

1 Chapter 9 Rotational Dynamics

2 In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of translation and rotation.

3 1) Torque Produces angular acceleration F1 F2 If F1 = - F2,! F 1 +! F 2 = m! a CM 0 = m! a CM Object rotates about CM (or fixed axis) ==> CM remains at rest

4 The amount of torque depends on where and in what direction the force is applied, as well as the location of the axis of rotation.

5 φ r DEFINITION OF TORQUE Magnitude of Torque = (Magnitude of the force) x (Lever arm) τ = Frsinφ Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis. SI Unit of Torque: newton x meter (N m)

6 Example In San Francisco a very simple technique is used to turn around a cable car when it reaches the end of its route. The car rolls onto a turntable, which can rotate about a vertical axis through its center. Then, two people push perpendicularly on the car, one at each end, as shown in the drawing. The turntable is rotated one-half of a revolution to turn the car around. If the length of the car is 9.20 m and each person pushes with a 185-N force, what is the magnitude of the net torque applied to the car?

7 Example You are installing a new spark plug in your car, and the manual specifies that it be tightened to a torque that has a magnitude of 45 N m. Using the data in the drawing, determine the magnitude, F, of the force that you must exert on the wrench.

8 2) Rigid objects in equilibrium A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero. ax = a y = 0 α = 0

9 Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all of the external forces acting on the object. 3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4. Apply the equations that specify the balance of forces at equilibrium. (Set the net force in the x and y directions equal to zero.) 5. Select a convenient axis of rotation. Set the sum of the torques about this axis equal to zero. 6. Solve the equations for the desired unknown quantities.

10 Example A person exerts a horizontal force of 190 N in the test apparatus shown in the drawing. Find the horizontal force M (magnitude and direction) that his flexor muscle exerts on his forearm.

11 Example 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.

12 3) Torque due to gravity; centre-of-gravity a) Torque due to discrete point masses m1 m2 axis W1 W2 r2 r1 τ 1 = W 1 r 1 τ 2 = W 2 r 2 τ = τ 1 + τ 2

13 (b) Continuous mass Wi ri τ = ΔW i r i

14 (c) Centre-of-gravity (definition) The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. Wr cg = ΔW i r i r cg = ΔW i r i W

15 (d) Locating the c.g. For 2 discrete masses (extends to any number) τ = Wr cg = W 1 r 1 + W 2 r 2 r cg = W 1 r 1 + W 2 r 2 W 1 + W 2 Uniform gravity W i = m i g r cg = m 1r 1 + m 2 r 2 m 1 + m 2 = r cm For symmetrical objects, cg is at the geometrical centre

16 Example Find the torque of a symmetrical meter stick with a mass of 1 kg about a fulcrum 15 cm from one end.

17 Balancing r cg = W 1 r 1 + W 2 r 2 W 1 + W 2

18 Finding the centre of gravity Image reprinted with permission of John Wiley and Sons, Inc.

19 Example The drawing shows a person (weight 584 N) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming the person holds this position.

20 Example A uniform board is leaning against a smooth vertical wall. The board is at an angle θ above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is Find the smallest value for the angle θ, such that the lower end of the board does not slide along the ground.

21 4) Moment of Inertia; 2nd law for rotation Translation F = ma m resists acceleration Rotation τ = Iα - I resists angular acceleration

22 Moment of inertia of a point object I = mr 2 FT m r F T = ma T rf T = ma T r τ = mr 2 α τ = Iα Moment of inertia of a rigid object I = 2 m i r i

23

24 5) Angular momentum linear momentum: p = mv angular momentum: τ = Iα = I Δω Δt τ = ΔL Δt L = Iω For a point particle on a circular path, L = mvr Units: kg m 2 /s

25 Conservation of angular momentum τ = ΔL Δt If external torque is zero, L is constant. L = Iω I = 2 m i r i If I decreases, angular speed increases.

26 Example Satellite in elliptical orbit L = Iω L A = L P I A ω A = I P ω P mv A r A = mv P r P v A = r P r A v P Image reprinted with permission of John Wiley and Sons, Inc.

27 Example A uniform board is leaning against a smooth vertical wall. The board is at an angle θ above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is Find the smallest value for the angle θ, such that the lower end of the board does not slide along the ground.

28 Example C&J In outer space two identical space modules are joined together by a massless cable. These modules are rotating about their center of mass, which is at the center of the cable because the modules are identical (see the drawing). In each module, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is v 0 = 17 m/s. Then they pull together until the distance between them is reduced by a factor of two. Determine the final tangential speed v f for each module. Answer: 34 m/s

29 Example A small kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m The angular speed is 6.28 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the centre of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

30 Example A thin rod has a length of 0.25 m and rotates in a circle on a frictionless table-top. The axis is perpendicular to the length of the rod at one of its ends. The rod has angular velocity of 0.32 rad/s and a moment of inertia of kg m 2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug ( mass = kg) gets where it s going, what is the angular velocity of the rod? Answer: 0.26 rad/s