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UMass Amherst KIN 100 - angular-kinetics-2

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1Newton’s Laws of Motion: Angular AnaloguesNewton’s First Law: A rotating body will maintain a state of constant rotational motion unless acted on by an external torqueResistance to Acceleration• Resistance to linear acceleration:Mass• Resistance to angular acceleration:Moment of InertiaMoment of inertia is determined by mass andmass distribution relative to axis of rotationDetermining Moment of InertiaMoment of inertia is the sum of the products of all the mass elements of an object and the square of the distances of the mass elements from the axis of rotation:Iaxis= miri2Iaxis= m1r12+ m2r22+.... + mnrn2axisr1r2m2m1Determining Moment of InertiaA more practical approach:Iaxis= (mbody)(2)where “” is an experimentally determined length known as the radius of gyration, that applies to the whole object– is not the same as the distance to segment CM– magnitude of is different for different axes of rotationUnit for moment of inertia consists of unit of mass multiplied by unit of length squared (kgm2)2Human Body Moment of Inertia• Can be computed for:– individual body segments (forearm, thigh, etc)– the whole body• Typically expressed for an axis:– through the center of mass, or– through the proximal or distal end~13.5 kgm2~11.5 kgm2~4.5 kgm2~1.5 kgm2~2.5 kgm2moment of inertia of the whole body about different axes of rotationApplications:- Tuck vs. layout position of a diver or gymnast- Position of a runner's leg during the swing phasetucklayouthip- Chocking up on a baseball/softball batAngular MomentumMomentum:• For linear motion: M = m v• For angular motion: H = I Or: H = (m 2) • Factors that affect angular momentum:– mass of the object (m)– distribution of mass relative to axis of rotation ()– angular velocity of the object ()Units for angular momentum: kgm2sAngular MomentumA 57 g tennis ball is struck by a racket, giving it an angular momentum of 6.810-4kgm2/s. If the radius of gyration of the ball is 2.4 cm, how fast will it be spinning?topspinforehand shotH = I  = m 2 0.00068 kgm2/s = (0.057 kg)(0.024 m)20.00068 kgm2/s / 0.000033 kgm2=  = 20.6 rad/s = 1180 deg/s = 3.28 rev/s3Conservation of Angular MomentumThe total angular momentum of a given system remains constant in the absence of external torques (H is constant)However, I and  can change, so long as their product remains constant(recall that H = I )Conservation of Angular MomentumA 60 kg diver is in a layout position with radius of gyration of 0.5 m as he leaves the board with an angular velocity of 4 rad/s. What is the diver’s angular velocity when he assumes a tuck position, reducing his radius of gyration to 0.25 m? = 0.5 m  = 0.25 mConservation of Angular MomentumFirst, find H when diver leaves the board:H = m 2 H = (60 kg)(0.5 m)2(4 rad/s) = 60 kgm2/sH is constant, so now find  when is reduced to 0.25 m:60 kgm2/s = (60 kg)(0.25 m)2 = 16 rad/sTotal body angular momentum is constant while the body is airborne (no external torques) However, even with total angular momentum constant it is possible to:• Transfer angular velocity from one part of the body to another• Change the body axis of rotationTransfer of Angular Momentum4Transfer of Angular MomentumEarly in the dive, H is concentrated in upper body, while later in the dive H is concentrated in the lower bodyHTOTAL(constant)HLOWER(changes)HUPPER(changes)The Falling Cat PhenomenonH is constant during the fall, but the cat always rights itselfHow does it do this?Newton’s Second Law: A net torque produces angular acceleration of a body that is directly proportional to the magnitude of the torque, in the same direction as the torque, and inversely proportional to the body’s moment of inertiaT = I  (compare with F = m a)(and more properly: TP= IP)Newton’s Laws of Motion: Angular AnaloguesTorque - Angular AccelerationHow much torque does a board reaction force of 850 N create about the CM of a diver (r = 0.67 m)? What is the angular acceleration if the diver’s moment of inertia about his CM is 10.5 kgm2T = F  r = (850 N)(0.67 m) = 569.5 NmTcm= Icm = T / I = 569.5 Nm / 10.5 kgm2 = 54.2 rad/s2(>3000 deg/s2)Fr5Angular Impulse-Momentum• Depends on the magnitude and direction of the applied torque, and the duration (time) of torque application:Linear Impulse = force  time = FtAngular Impulse = torque  time = Tt• Impulse-momentum relationshipLinear: Ft = M Ft = (mv)f–(mv)iAngular: Tt = H Tt = (I )f–(I )iAngular Impulse-MomentumA person is swinging their leg forward at 200 /s (3.5 rad/s). If the leg has a moment of inertia about the hip of 0.7 kgm2, how much torque must be generated by the hip extensor muscles to stop the leg in 0.4 s?T  t = (I )f-(I )iT  t = (0.7 kgm2)(0) - (0.7 kgm2)(3.5 rad/s)T  t = -2.45 kgm2/sT = (-2.45 kgm2/s) / 0.4 s = -6.13 NmTAngular Work, Power, & Energy• When a torque (T) acts over an angular displacement (), mechanical work is doneW = T   ( must be in rad)• Mechanical power is the rate at which a torque does work on a systemP = W / t, or P = T  ( must be in rad/s)• Note that the units for angular work and power are J (Nm) and W (J/s), respectively, the same as for linear work and powerAngular Work, Power, & EnergyWhile doing dumbbell curls, a person generated an elbow joint flexion moment of 25.6 Nm, while flexing their elbow over a 120 range, how much mechanical work was done?It took the person 0.78 s to lift the weight (from A to B in the diagram); what was the mechanical power?W = T W = (25.6 Nm)(2.09 rad) = 53.5 JP = W / tP = 53.5 J / 0.78 s = 68.6 W6Angular Work, Power, & Energy• Total mechanical energy (TE) is the sum of kinetic energy (KE) and potential (PE) energyTE = KE + PE• Kinetic energy can be due to linear (TKE) and/or angular (RKE) motionTKE = ½ m v2RKE = ½ I 2• Therefore, the total mechanical energy of a system is given by:TE = ½ m v2+ ½ I 2+ m g hNewton’s Third Law: For every torque exerted by one body on another, there is an equal and opposite torque exerted by the second body on the firstNewton’s Laws of Motion: Angular AnaloguesAction-Reaction Torques• In running, the arms swing in opposition to the legs– This counters the torques applied to the feet during


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UMass Amherst KIN 100 - angular-kinetics-2

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