Torque and Angular Momentum

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1 Torque and Angular Momentum Rotational Kinetic Energy Rotational Inertia Torque Work Done by a Torque Equilibrium (revisited) Rotational Form of Newton s nd Law Rolling Objects Angular Momentum

2 Rotational KE and Inertia For a rotating solid body: n i i i n n v m m v m v m v K rot For a rotating body v i ωr i where r i is the distance from the rotation axis to the mass m i. ( ) rot ω ω ω I m r r m K n i i i n i i i

3 The quantity n I i m i r i is called rotational inertia or moment of inertia. Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in table 8. for extended bodies. 3

4 Example: (a) Find the moment of inertia of the system below. The masses are m and m and they are separated by a distance r. Assume the rod connecting the masses is massless. ω r r m m r and r are the distances between mass and the rotation axis and mass and the rotation axis (the dashed, vertical line) respectively. 4

5 Example continued: Take m.00 kg, m.00 kg, r 0.33 m, and r 0.67 m. I mir i i m r + m r 0.67 kg m (b) What is the moment of inertia if the axis is moved so that is passes through m? I mir i i m r.00 kg m 5

6 Example (text problem 8.4): What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 0.0 cm, about an axis through its center and perpendicular to it? From table 8.: I MR ( 49.0 kg)( 0. m) 0.98 kg m 6

7 Torque A rotating (spinning) body will continue to rotate unless it is acted upon by a torque. hinge P u s h Q: Where on a door do you push to open it? A: Far from the hinge. 7

8 Torque method : Top view of door F F Hinge end θ F τ rf r the distance from the rotation axis (hinge) to the point where the force F is applied. F is the component of the force F that is perpendicular to the door (here it is Fsinθ). 8

9 The units of torque are Nm (not joules!). By convention: When the applied force causes the object to rotate counterclockwise (CCW) then τ is positive. When the applied force causes the object to rotate clockwise (CW) then τ is negative. 9

10 Torque method : τ r F r is called the lever arm and F is the magnitude of the applied force. Lever arm is the perpendicular distance to the line of action of the force. 0

11 Top view of door F Hinge end r θ θ Lever arm Line of action of the force r sinθ r r r sinθ The torque is: τ r F Same as rf sinθ before

12 Example (text problem 8.): The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine. What magnitude torque does the cord apply to the drum? F75 N R6.00 cm τ r rf F ( 0.06 m)( 75.0 N) 4.5 Nm

13 Example: Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F acts in the middle of the bar. 45 X F 30 N 30 F 3 0 N 0 F 5 N 3

14 Example continued: Lever arm for F F 30 N 30 F 3 0 N 45 X 0 F 5 N Lever arm for F 3 The lever arms are: r ( m) r r 3 0 sin m ( 4m) sin m 4

15 Example continued: τ 0 The torques are: τ + (.73 m)( 30 N) τ Nm ( m)( 0 N) Nm The net torque is Nm and is the sum of the above results. 5

16 Work done by a Torque The work done by a torque τ is W τδθ. where Δθ is the angle (in radians) the object turns through. 6

17 Example (text problem 8.5): A flywheel of mass 8 kg has a radius of 0.6 m (assume the flywheel is a hoop). (a) What is the torque required to bring the flywheel from rest to a speed of 0 rpm in an interval of 30 sec? ω f rev π rad min 0.6 rad/sec min rev 60 sec τ rf mr r ( ma) rm( rα ) ω f ωi Δt mr ω f Δt mr Δω Δt 9.4 Nm 7

18 8 (b) How much work is done in this 30 sec period? ( ) 5600 J Δ Δ + Δ Δ t t t W f f i av ω τ ω ω τ ω τ θ τ Example continued:

19 Equilibrium The conditions for equilibrium are: F 0 τ 0 9

20 Example (text problem 8.36): A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at a 35 angle with the boom, is attached at a distance of.38 m from the hinge at the wall. The weight of the sign is 0.0 N. What is the tension in the cable and what are the horizontal and vertical forces exerted on the boom by the hinge? 0

21 Example continued: y FBD for the bar: F y T X F x θ x w bar F sb Apply the conditions for equilibrium to the bar: () () (3) F x F y F x F τ w y T bar w cosθ L bar F F sb sb 0 + T sinθ 0 ( L) + ( T sinθ ) x 0

22 Example continued: Equation (3) can be solved for T: T L wbar + F xsinθ 35 N sb ( L) Equation () can be solved for F x : F x T cos θ 88 N Equation () can be solved for F y : F y w + F T sinθ bar.00 N sb

23 8.5 Equilibrium in the Human Body Example (text problem 8.43): Find the force exerted by the biceps muscle in holding a one liter milk carton with the forearm parallel to the floor. Assume that the hand is 35.0 cm from the elbow and that the upper arm is 30.0 cm long. The elbow is bent at a right angle and one tendon of the biceps is attached at a position 5.00 cm from the elbow and the other is attached 30.0 cm from the elbow. The weight of the forearm and empty hand is 8.0 N and the center of gravity is at a distance of 6.5 cm from the elbow. 3

24 Example continued: F b hinge (elbow joint) w F ca τ F b F b x wx wx Fcax3 0 + Fcax3 30 N x 4

25 8.6 Rotational Form of Newton s nd Law τ Iα Compare to F ma 5

26 Example (text problem 8.57): A bicycle wheel (a hoop) of radius 0.3 m and mass kg is rotating at 4.00 rev/sec. After 50 sec the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? τ I α MR α ω ω i f rev sec π rad rev 5. rad/sec α Δω Δt ω f ω i Δt 0.50 rad/s τ av MR α 0.09 Nm 6

27 Rolling Objects An object that is rolling combines translational motion (its center of mass moves) and rotational motion (points in the body rotate around the center of mass). For a rolling object: K tot K T + K mv cm rot + Iω If the object rolls without slipping then v cm Rω. 7

28 Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first? h θ The object with the largest linear velocity (v) at the bottom of the ramp will win the race. 8

29 R I m mgh v v R I m mgh R v I mv I mv mgh K U K U E E f f i i f i ω Apply conservation of mechanical energy: Example continued: Solving for v:

30 Example continued: The moments of inertia are: I I disk sphere 5 mr mr For the disk: For the sphere: v v disk sphere 4 3 gh 0 7 gh Since V sphere > V disk the sphere wins the race. Compare these to a box sliding down the ramp. v box gh 30

31 How do objects in the previous example roll? y FBD: N w Both the normal force and the weight act through the center of mass so Στ0. This means that the object cannot rotate when only these two forces are applied. x 3

32 Add friction: FBD: F s y N τ F F x y F r s Iα wsinθ F N wcosθ s ma 0 cm θ w Also need a cm αr and x v v + aδx 0 The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.3.) It is static friction that makes an object roll. 3

33 Angular Momentum F net p lim Δ t 0 mv Δp Δt τ net lim Δ t 0 L Iω ΔL Δt Units of p are kg m/s Units of L are kg m /s When no net external forces act, the momentum of a system remains constant (p i p f ) When no net external torques act, the angular momentum of a system remains constant (L i L f ). 33

34 Example (text problem 8.70): A turntable of mass 5.00 kg has a radius of 0.00 m and spins with a frequency of rev/sec. What is the angular momentum? Assume a uniform disk. ω rev π rad sec rev 3.4 rad/sec L Iω MR ω kg m /s 34

35 Example (text problem 8.79): A skater is initially spinning at a rate of 0.0 rad/sec with I.50 kg m when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to.60 kg m? He/she is on ice, so we can ignore external torques. I ω i L i i L I f f ω f ω f I I i f ω i.50 kg m.60 kg m ( 0.0 rad/sec) 5.6 rad/sec 35

36 The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with a right-hand rule. 36

37 Curl the fingers of your right hand so that they curl in the direction a point on the object moves, and your thumb will point in the direction of the angular momentum. 37

38 Consider a person holding a spinning wheel. When viewed from the front, the wheel spins CCW. Holding the wheel horizontal, they step on to a platform that is free to rotate about a vertical axis. 38

39 Initially, nothing happens. They then move the wheel so that it is over their head. As a result, the platform turns CW (when viewed from above). This is a result of conserving angular momentum. 39

40 Initially there is no angular momentum about the vertical axis. When the wheel is moved so that it has angular momentum about this axis, the platform must spin in the opposite direction so that the net angular momentum stays zero. Is angular momentum conserved about the direction of the wheel s initial, horizontal axis? 40

41 It is not. The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here. 4

42 Summary Rotational Kinetic Energy Moment of Inertia Torque (two methods) Conditions for Equilibrium Newton s nd Law in Rotational Form Angular Momentum Conservation of Angular Momentum 4

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