Angular acceleration α



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Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 7-0

Linear and Circular Motion Compared Slide 7-

Linear and Circular Kinematics Compared Slide 7-

Tangential Acceleration a αr t Slide 7-4

Torque Which force would be most effective in opening the door? Slide 7-5

Interpreting Torque Torque is due to the component of the force perpendicular to the radial line. τ rf rf sinφ Slide 7-6

A Second Interpretation of Torque τ r F rf sinφ Slide 7-7

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 τ r rf ( )( ) R6.00 cm 0.06 m 75.0 N 4.5 Nm F 8

Example Revolutionaries attempt to pull down a statue of the Great Leader by pulling on a rope tied to the top of his head. The statue is 7 m tall, and they pull with a force of 400 N at an angle of 65 to the horizontal. What is the torque they exert on the statue? If they are standing to the right of the statue, is the torque positive or negative? Slide 7-8

Newton s Second Law for Rotation α τ / I I moment of inertia. Objects with larger moments of inertia are harder to get rotating. I m r i i Slide 7-4

Moments of Inertia of Common Shapes Slide 7-5

Example (text problem 8.): 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

Example: (a) Find the moment of inertia of the system below. The masses are m and m and dthey are separated dby a distance r. Assume the rod connecting the masses is massless. ω r and r are the distances between mass and the rotation axis and mass m r r m and the rotation axis (the dashed, vertical line) respectively. 3

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 ( )( ) ( )( ).00 kg 0.33 m +.00 kg 0.67 m 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 + m r (.00 kg )( 0.0000 m ) + (.00 kg )(.00 m ).00 kg m 4

Example (text problem 8.): 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? I MR ( 49.0 kg)( 0. m) 0.98 kg m 5

Rotational and Linear Dynamics Compared Slide 7-6

Example The motor in a CD player exerts a torque of 7.0 x 0-4 N m. What is the disk s angular acceleration? (A CD has a diameter of.0 cm and a mass of 6 g.) Slide 7-8

Example A baseball bat has a mass of 0.8 kg and is 0.86 m long. It s held vertically and then allowed to fall. What is the bat s angular acceleration when it has reached 0 from the vertical? (Model the bat as a uniform cylinder). Slide 7-9

Constraints Due to Ropes and Pulleys Slide 7-30

Example How long does it take the small mass to fall.0 m when released from rest? Slide 7-3

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

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 r mr ( ma) rm( rα ) mr Δω Δ t ω f ωi ω f mr 9.4 Nm Δ t Δ t

Example continued: (b) How much work is done in this 30 sec period? W ( ω Δ ) τδθ τ t av ωi + ω f Δ ω f τ t τ Δ t 5600 J 3

Rotational KE and Inertia For a rotating solid body: K rot m v + m v + + m v n n n i m i v i For a rotating body v i ωr i where r i is the distance from the rotation axis to the mass m i. K rot n i n m ωri i ω ( ) m r Iω i i i 4

Equilibrium The conditions for equilibrium are: F 0 τ 0 5

Example (text problem 8.35): A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.00 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 0.00 N. What is the tension in the cable and what are the horizontal and vertical forces exerted on the boom by the hinge? 6

Example continued: y FBD for the bar: F y T X F x θ x w bar F sb () F F T cos θ 0 Apply the conditions for equilibrium to the bar: () Fy Fy wbar Fsb + T sinθ 0 x x L ( 3) τ wbar Fsb θ ( L) + ( T sin ) x 0 7

Example continued: Equation (3) can be solved for T: T L wbar + F xsin θ 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 8

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. 9

Example continued: y F b hinge (elbow joint) w F ca x τ F b F b x wx wx Fca x3 0 + Fcax3 30 N x 30

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 K + K tot T rot mv cm + Iω If the object rolls without slipping then v cm Rω. 3

3

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. 33

Example continued: Apply conservation of mechanical energy: + + K U K U E E f f i i f i + + + + 0 0 I R v I mv I mv mgh ω + mgh v R I m mgh + R I m mgh v Solving for v: 34

Example continued: The moments of inertia are: I disk I sphere mr 5 mr For the disk: For the sphere: v v disk sphere 4 gh 3 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 35

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 Στ 0. This means that the object cannot rotate when only these two forces are applied. x 36

Add friction: y τ F r Iα FBD: N Fx wsinθ Fs macm F s N wcos θ 0 F y s θ w x Also need a cm αr and 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. 37

Angular Momentum Δp ΔL Fnet lim τ net lim Δ t 0 Δt Δ t 0 Δt p mvv L I ω 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 ). 38

Example (text problem 8.69): A turntable of mass 5.00 kg has a radius of 0.00 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk. ω rev π rad 0.500 sec rev 3.4 rad/sec L Iω MR ω 0.079 kg m /s 39

Example (text problem 8.75): A skater is initially spinning at a rate of 0.0 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? The skater is on ice, so we can ignore external torques. i L i L f I ω I ω f i f I I ω ω i i f f.50 kg m.60 kg m ( 0.00 rad/sec ) 5.6 rad/sec 40

The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with a right-hand rule. 4

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. 4

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. 43

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. 44

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? 45

It is not. The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here. 46