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Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 7-10 Linear and Circular Motion Compared Slide 7-11 Linear and Circular Kinematics Compared Slide 7-12 Tangential Acceleration at = α r Slide 7-14 Torque Which force would be most effective in opening the door? Slide 7-15 Interpreting Torque Torque is due to the component of the force perpendicular to the radial line. τ = rF⊥ = rF sin φ Slide 7-16 A Second Interpretation of Torque τ = r⊥ F = rF sin φ Slide 7-17 Example (text problem 8.12): The pull cord of a lawnmower engine is wound around a drum of radius 6 6.00 00 cm 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? F 75 N F=75 R=6.00 cm τ = r⊥ F = rF = (0.06 m )(75.0 N ) = 4.5 Nm 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 17 m tall, and they pull with a force of 4200 N at an angle l off 65° to t the th h horizontal. i t l Wh Whatt iis th the ttorque they th exertt on the statue? If they are standing to the right of the statue, is the torque positive or negative? Slide 7-18 Newton’s Second Law for Rotation α =τ /I I = moment of inertia. Objects j with larger g moments of inertia are harder to get rotating. I = ∑ mi ri 2 Slide 7-24 Moments of Inertia of Common Shapes Slide 7-25 Example (text problem 8.2): What is the rotational inertia of a solid iron disk of mass 49 49.0 0 kg with a thickness of 5 5.00 00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it? From table 8.1: 1 1 2 2 I = MR = (49.0 kg )(0.2 m ) = 0.98 kg m 2 2 2 12 Example: (a) Find the moment of inertia of the system below. Th masses are m1 and The d m2 and d th they are separated t db by a distance r. Assume the rod connecting the masses is massless. ω m1 r1 r2 m2 r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) p y respectively. 13 Example continued: Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m. 2 I = ∑ mi ri 2 = m1r12 + m2 r22 i =1 = (2.00 kg k )(0.33 m ) + (1.00 kg k )(0.67 m ) 2 2 = 0.67 kg m 2 (b) What is the moment of inertia if the axis is moved so that is passes through m1? 2 I = ∑ mi ri 2 = m1r12 + m2 r22 i =1 = (2.00 kg )(0.00 m ) + (1.00 kg )(1.00 m ) 2 = 1.00 kg m 2 2 14 Example (text problem 8.2): What is the rotational inertia of a solid iron disk of mass 49 49.0 0 kg with a thickness of 5 5.00 00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it? 1 1 2 2 I = MR = (49.0 kg )(0.2 m ) = 0.98 kg m 2 2 2 15 Rotational and Linear Dynamics Compared Slide 7-26 Example The motor in a CD player exerts a torque of 7.0 x 10-4 N · m. What is the disk’s angular acceleration? (A CD has a diameter of 12.0 cm and d a mass off 16 g.)) Slide 7-28 Example A baseball bat has a mass of 0 0.82 82 kg and is 0 0.86 86 m long long. It’s It s held vertically and then allowed to fall. What is the bat’s angular acceleration when it has reached 20° from the vertical? ti l? (Model (M d l th the b batt as a uniform if cylinder). li d ) Slide 7-29 Constraints Due to Ropes and Pulleys Slide 7-30 Example How long does it take the small mass to fall 1 1.0 0 m when released from rest? Slide 7-31 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. 21 Example (text problem 8.25): A flywheel of mass 182 kg has a radius of 0.62 0 62 m (assume the flywheel is a hoop) hoop). (a) What is the torque required to bring the flywheel from rest to a speed of 120 rpm in an interval of 30 sec? ω f = 120 d ⎞⎛ 1 min i ⎞ rev ⎛ 2π rad ⎟ = 12.6 rad/sec ⎜ ⎟⎜ min ⎝ 1 rev ⎠⎝ 60 sec ⎠ ⎛ Δω ⎞ τ = rF = r (ma ) = rm(rα ) = mr ⎜ ⎟ ⎝ Δt ⎠ ω f − ωi ⎞ ωf ⎞ 2⎛ 2⎛ ⎟⎟ = mr ⎜⎜ ⎟⎟ = 29.4 Nm = mr ⎜⎜ ⎝ Δt ⎠ ⎝ Δt ⎠ 2 22 Example continued: (b) How much work is done in this 30 sec period? W = τΔθ = τ (ωav Δt ) ⎛ ωi + ω f = τ ⎜⎜ ⎝ 2 ⎞ ⎛ωf ⎟⎟Δt = τ ⎜⎜ ⎝ 2 ⎠ ⎞ ⎟⎟Δt = 5600 J ⎠ 23 Rotational KE and Inertia For a rotating solid body: K rot n 1 1 1 1 2 2 2 = m1v1 + m2 v2 + … + mn vn = ∑ mi vi2 2 2 2 i =1 2 For a rotating body vi = ωri where ri is the distance from the rotation axis to the mass mi. 1 1⎛ n 1 2 2 2⎞ 2 = ∑ mi (ωri ) = ⎜ ∑ mi ri ⎟ω = Iω 2 ⎝ i =1 2 i =1 2 ⎠ n K rot 24 Equilibrium The conditions for equilibrium are: ∑F = 0 ∑τ = 0 25 Example (text problem 8.35): A sign is supported by a uniform horizontal boom of length 3 3.00 00 m and weight 80 80.0 0N N. A cable cable, inclined at a 35° angle with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of th sign the i iis 120 120.0 0 N. N What Wh t iis th the ttension i iin th the cable bl and d what h t are the horizontal and vertical forces exerted on the boom by the hinge? g 26 Example continued: y FBD for the bar: Fy T X θ Fx x wbar Apply the conditions for equilibrium to the bar: Fsb (1) ∑ Fx = Fx − T cos θ = 0 (2) ∑ Fy = Fy − wbar − Fsb + T sin θ = 0 ⎛L⎞ (3) ∑τ = − wbar ⎜ ⎟ − Fsb (L ) + (T sin θ )x = 0 ⎝2⎠ 27 Example continued: Equation (3) can be solved for T: ⎛L⎞ wbar ⎜ ⎟ + Fsb (L ) ⎝2⎠ T= x sin i θ = 352 N Equation (1) can be solved for Fx: Fx = T cos θ = 288 N Equation (2) can be solved for Fy: Fy = wbar + Fsb − T sin θ = −2.00 N 28 Equilibrium in the Human Body Example (text problem 8.43): Find the force exerted by the biceps p muscle in holding g 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 h fforearm and d empty h hand d iis 18 18.0 0 N and d the h center off gravity is at a distance of 16.5 cm from the elbow. 29 Example continued: y Fb “hinge”” “hi (elbow joint) x w ∑τ = F x b 1 Fca − wx2 − Fca x3 = 0 wx2 + Fca x3 = 130 N Fb = x1 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 tot = K T + K rot = 1 2 1 2 mvcm + Iω 2 2 If the object rolls without slipping then vcm = Rω. 31 32 Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp 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: Ei = E f U i + Ki = U f + K f 1 2 1 2 1 2 1 ⎛v⎞ mgh + 0 = 0 + mv + Iω = mv + I ⎜ ⎟ 2 2 2 2 ⎝R⎠ 2 1⎛ I ⎞ mgh = ⎜ m + 2 ⎟v 2 2⎝ R ⎠ Solving for v: 2mgh v= I ⎞ ⎛ m + ⎜ 2 ⎟ R ⎠ ⎝ 34 Example continued: I disk The moments of inertia are: I sphere For the disk: vdisk 4 = gh 3 For the sphere: vsphere 10 = gh 7 1 = mR 2 2 2 = mR R2 5 Since Vsphere> Vdisk the sphere wins the race race. Compare these to a box sliding down the ramp. vbox = 2 gh 35 How do objects in the previous example roll? y N FBD: w x Both the normal force and the weight act through the center of mass so Στ = 0. 0 This means that the object cannot rotate when only these two forces are applied. 36 Add friction: y FBD: s N Fs ∑τ = F r = Iα ∑ F = w sin θ − F = ma ∑ F = N − w cosθ = 0 x s cm y θ w Also need acm = αR and x v 2 = v02 + 2aΔx 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 8.13.) It is static friction that makes an object roll. 37 Angular Momentum Δp Fnet = lim Δt →0 Δt p = mv Units of p are kg m/s When no net external forces act, the momentum of a system remains constant (pi = pf) ΔL τ net = lim Δt →0 Δt L = Iω Units of L are kg m2/s When no net external torques act, the angular momentum of a system remains constant (Li = Lf). 38 Example (text problem 8.69): A turntable of mass 5.00 kg has a radius of 0 0.100 100 m and spins with a frequency of 0 0.500 500 rev/sec. What is the angular momentum? Assume a uniform disk. rev ⎛ 2π rad ⎞ ω = 0.500 ⎜ ⎟ = 3.14 rad/sec sec ⎝ 1 rev ⎠ ⎛1 2⎞ L = Iω = ⎜ MR ⎟ω = 0.079 kg g m 2 /s ⎝2 ⎠ 39 Example (text problem 8.75): A skater is initially spinning at a rate of 10.0 10 0 rad/sec with II=2.50 2 50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2? The skater is on ice, so we can ignore external torques. Li = L f I iωi = I f ω f ⎛ Ii ω f = ⎜⎜ ⎝If ⎞ ⎛ 2.50 kg m 2 ⎞ ⎟ωi = ⎜ ⎜ 1.60 kg m 2 ⎟⎟(10.0 rad/sec) = 15.6 rad/sec ⎟ ⎝ ⎠ ⎠ 40 The Vector Nature of Angular Momentum Angular momentum is a vector. Its direction is defined with g rule. a right-hand 41 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. 42 Consider C id a person holding a spinning wheel. When viewed from the front, the wheel spins CCW. Holding g the wheel horizontal, they y 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 head. As a result 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 g 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 i not. t The Th floor fl exerts t a torque t on the th system t (platform ( l tf + person), thus angular momentum is not conserved here. 46