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Rotational Dynamics Why do objects spin? Objects can travel in different ways: Translation all points on the body travel in parallel paths Rotation all points on the body move around a fixed point An object can be both translating and rotating simultaneously. 2
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Rigid Body object or system of particles in which the distances between particles remains fixed and constant Examples Non Examples 4
What causes linear motion to change? answer: Net Force What causes rotational motion to change? answer: Torque 5
Torque Tendency of a force to rotate an object about an axis Torque is affected by: 1. the strength of the force, 2. where it is applied, and 3. how it is pointed. 6
Calculating Torque Torque = Lever Arm x Force τ =l x F = r F sin θ x = Cross Product recall: W = F Δr now: τ =l x F Cross Product Simulator Scalar : Answer Answer : Vector Dot Product the product of the components of the two vectors that point in the same direction Cross Product the product of the components of the two vectors that are perpendicular to each other 7
Recall: A B = A B cos θ A B = A x B x + A y B y Now: AxB = A B sin θ direction comes from the right hand rule example: A = 4 i + 5 j B = 2i + 6 j AxB = A x B y A y B x example: direction comes from the right hand rule U = 2i 3j k V = 1i + 4j 2k UxV = (U 2 V 3 U 3 V 2 )i (U 1 V 3 U 3 V 1 )j + (U 1 V 2 U 2 V 1 )k 8
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The direction of the torque To find the direction of the torque, use one of two right hand rules. Caution: for either of the above to work, the two tails of the vectors must be together! examples: A B Direction of A x B? Direction of B x A? D Direction of C x D? C Direction of D x C? 10
example: F = 20.0 N r =.75 m Line of Action 11
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Equilibrium of a Rigid Body question: A rigid body is in equilibrium if: answer: 1. Στ = 0 (zero angular acceleration) 2. ΣF = 0 (zero translational acceleration) What is the tension in the supporting bar, and what is the hinge force? Mass of Sign = 2.50 kg Length of Horiz Bar = 80.0 cm Mass of Bar = Negligible 15
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Center of Gravity recall: Center of mass one point that can be thought of as the average location for the total mass V cm = m 1 v 1 + m 2 v 2 m 1 + m 2 19
now: Center of Gravity Center of Gravity point on a rigid body at which the weight can be considered to act when the torque due to the weight is being calculated x cg = w 1 x 1 + w 2 x 2 w 1 + w 2 20
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example: This Olympic bar weighs 44.0 pounds (mass of 19.958 kilograms). If a 45 pound (mass of 20.412 kg) plate is put on each end of the bar, and an additional 5.0 pound (mass of 2.268 kg) plate is put on the right side, where would you need to lift it to raise it horizontally? x cg = w 1 x 1 + w 2 x 2 w 1 + w 2 22
Stable Equilibrium object's center of gravity will stay inside the area of support when a "small" displacement produces a force or torque that restores the object back to its original position. Unstable Equilibrium object's center of gravity will leave the area of support when a "small" displacement produces a force or torque that displaces the object farther from its original position. 23
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this is just for one particle Rotational Version of Newton's Second Law F = ma a tan = rα F = mrα ΣΤ = Fr = mr 2 α multiply both sides by r mr 2 =relationship between T and α ="rotational inertia" of the particle ="moment of inertia" of the particle extended object = many particles I= m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 +... ΣΤ = Iα 25
what's doing the accelerating what's doing the twisting Σ F = m a what's resisting the accelerating Σ T = I α what's resisting the twisting Newton's Second Law the effect of the accelerating Rotational Equivalent of Newton's Second Law the effect of the twisting 26
Calculating Moments of Inertia of Common Shapes 27
Calculating Moments of Inertia of Common Shapes 28
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Rotational Work and Kinetic Energy recall: We can go directly from work done by a force to work done by a torque: r θ F s W = F s W = F r θ s =r θ W = θ We can also derive an expression for rotational power: P = W = θ = t t ω 34
Work Energy Theorem and Kinetic Energy W net = θ = I α θ ωf 2 = ωo 2 + 2αθ W net = I ( ) = 1 I ωf 2 1 I ωo 2 2 2 2 ωf 2 ωo 2 Since we know W = ΔKE, these must be the rotational KEs. K rot = 1 I ωf 2 2 35
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Angular Momentum recall: p = mv now: Angular Momentum = (moment of inertia) (angular velocity) L = I ω 41
Also recall: Impulse= J = ΣF Δt = Δp = mδv Now: Σ = ΔL Δt Rotational version of impulse momentum theorem Law of Conservation of Angular Momentum The total angular momentum of a rotating body remains constant if the net torque acting on it is zero. 42
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Vibrations and Waves Vibration or oscillation moving over the same distance again and again This motion is said to be periodic (occurring in regular intervals of time). 50
Simplest example of periodic motion: Mass on a spring on a frictionless surface Amplitude a scalar quantity that expresses the distance of both extreme displacements from the equilibrium position Restorative Force Force that attempts to return (restore) object to its equilibrium position. Frequency number of cycles per time interval (usually one second) Period (T) time it takes to complete one complete cycle T = 1 f Simple Harmonic Motion Restorative force is directly proportional to the distance from the equilibrium position (technically, proportional to the negative of the displacement). simple: only one frequency of motion harmonic: motion can be described with sines and cosines This system we have been describing is known as a Simple Harmonic Oscillator. 51
Energy in a Simple Harmonic Oscillator Recall: U s = k x 2 1 2 At end point: 1 E = k A 2 2 At middle point: E = mv o 2 1 2 1 2 At points between middle and endpoint: 1 E = m v 2 1 + k x 2 2 2 52
Graph of the Energy of a SHO 53
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Period of SHM V o = Circumference Period T = 2 π A Vo = 2π A T E = 1 2 mv o 2 1 E = k A 2 2 1 2 E = mv 2 o = mv o 2 = ka 2 1 E = k A 2 2 m k = A V o T = 2π 55
Position of Object in SHM x = A cos θ x = A cos ωt x = A cos 2πft ω = θ t ω = 2πf x = A cos [ 2πt ] T 56
Velocity of Object in SHM Where will you see v o on the wall? = v o v = v o sin θ v = v o sin ωt v = v o sin 2πft v = v o sin [ 2πt ] T ω = θ t ω = 2πf 57
A cos Acceleration of Object in SHM = v o F a = = m k kx m a = A cos [ 2πt] m T a = a o cos [ 2πt] T where a o = the max acceleration (since it's at the very end where the force is the strongest) x = A cos [ 2πt ] T where: a o = 58
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Simple Pendulum small object (bob) suspended from a lightweight cord Is this SHM? (It will be if the restoring force α displacement.) But, restoring force = mg sinθ So, NO, this is not SHM. 60
However, for very small angles (in radians), sin θ θ 61
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Vibrations and Waves Vibration or oscillation moving over the same distance again and again This motion is said to be periodic (occurring in regular intervals of time). 63
Simplest example of periodic motion: Mass on a spring on a frictionless surface Restorative Force Force that attempts to return (restore) object to its equilibrium position. Amplitude a scalar quantity that expresses the distance of both extreme displacements from the equilibrium position Frequency number of cycles per time interval (usually one second) Period (T) time it takes to complete one complete cycle T = 1 f Simple Harmonic Motion Restorative force is directly proportional to the distance from the equilibrium position (technically, proportional to the negative of the displacement). simple: only one frequency of motion harmonic: motion can be described with sines and cosines This system we have been describing is known as a Simple Harmonic Oscillator. 64
Energy in a Simple Harmonic Oscillator recall: U s = 1 k x 2 2 E = 1m v 2 + 1 k x 2 2 2 At end point: 65
F = mg Sin θ -mgθ (for small angles) This is approximately SHM. s = r θ x = L θ L F = -mgθ F = -mg x L x F = - mg L x F = k x In the form of Hooke's Law T = 2π T = 2π mg L T = 2π L g 66
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