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Physics Chapter 8: Rotational Equilibrium and Dynamics Rotational Equilibrium and Dynamics Rotational Dynamics What is Involved in Rotating an Object Force Distance from the Point of Rotation Rotational Equilibrium and Dynamics Rotational Dynamics To swing open a door, you exert a force. The doorknob is near the outer edge of the door. You exert the force on the doorknob at right angles to the door, away from the hinges. Rotational Equilibrium and Dynamics Rotational Dynamics What is Involved in Rotating an Object Force Distance from the Point of Rotation Torque (t) t Fr Rotational Equilibrium and Dynamics Torque (t) Product of the Rotational Force and the Length of a Lever Arm t Fr Rotational Equilibrium and Dynamics Torque Magnitude Fr Direction + When Rotation is Counterclockwise - When Rotation is Clockwise t Fr Rotational Equilibrium and Dynamics Torque Units Force = Newton Length = meter Torque = Newton*Meter (N*m) t Fr Rotational Equilibrium and Dynamics Torque Direction is Not Always Perpendicular to the Radius t Fr sin Rotational Equilibrium and Dynamics Problem The gardening tool shown is used to pull weeds. If a 1.23N*m torque is required to pull a given weed, what force did the weed exert on the tool? Rotational Equilibrium and Dynamics Solution t = 1.23 N*m r = 0.040m F = ? t Fw (0.040 m) 1.23 N m 0 1.23 N m Fw 31 N 0.040 m Rotational Equilibrium and Dynamics Problem • What is the torque on a bolt produced by a 15N force exerted perpendicular to a wrench that is 25cm long. Rotational Equilibrium and Dynamics Solution F = 15N r = 25cm t Fr sin t (15N )(0.25m)(sin 90.0 ) 3.8N m o Rotational Equilibrium and Dynamics Problem A person slowly lowers a 3.3kg crab trap over the side of a dock. What torque does the trap exert about the person’s shoulder? Rotational Equilibrium and Dynamics Solution t = ? r = 0.70m m = 3.3kg m t r mg (0.70 m)(3.3 kg) 9.81 2 23 N m s Rotational Equilibrium and Dynamics Problem At the local playground, a 16kg child sits on the end of a horizontal teeter-totter, 1.5 m from the pivot point. On the other side of the pivot an adult pushes straight down on the teetertotter with a force of 95N. In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot? Rotational Equilibrium and Dynamics Solution rchild = 1.5m mchild = 16kg radult = 3.0m Fadult = 95N t 0 rc Fc rA FA rc (mc g ) (1.5m)(16kg)(9.81m / s 2 ) rA 2.48m FA 95 N 3.0 m > 2.48 m; the adult has enough momentum to push down and push the child up. Rotational Equilibrium and Dynamics Problem At the local playground, a 16kg child sits on the end of a horizontal teeter-totter, 1.5 m from the pivot point. On the other side of the pivot an adult pushes straight down on the teetertotter with a force of 95N. In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5m from the pivot? Rotational Equilibrium and Dynamics Solution rchild = 1.5m mchild = 16kg radult = 2.5m Fadult = 95N t 0 rc Fc rA FA rc (mc g ) (1.5m)(16kg)(9.81m / s 2 ) rA 2.48m FA 95 N 2.5 m > 2.48 m; the adult has just barely enough momentum to push the child up. Rotational Equilibrium and Dynamics Problem At the local playground, a 16kg child sits on the end of a horizontal teeter-totter, 1.5 m from the pivot point. On the other side of the pivot an adult pushes straight down on the teetertotter with a force of 95N. In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0m from the pivot? Rotational Equilibrium and Dynamics Solution rchild = 1.5m mchild = 16kg radult = 2.0m Fadult = 95N t 0 rc Fc rA FA rc (mc g ) (1.5m)(16kg)(9.81m / s 2 ) rA 2.48m FA 95 N 2.0 m < 2.48 m; the adult does not have enough momentum to hold the child up, and the child goes down. Rotational Equilibrium and Dynamics Center of Mass The center of mass of an object is the point on the object that moves in the same way that a point particle would move. The path of center of mass of the object is a straight line. Rotational Equilibrium and Dynamics Center of Mass To locate the center of mass of an object, suspend the object from any point. When the object stops swinging, the center of mass is along the vertical line drawn from the suspension point. Draw the line, and then suspend the object from another point. Again, the center of mass must be below this point. Rotational Equilibrium and Dynamics Center of Mass Draw a second vertical line. The center of mass is at the point where the two lines cross. Rotational Equilibrium and Dynamics Center of Mass and Stability An object is said to be stable if an external force is required to tip it. The object is stable as long as the direction of the torque due to its weight, τw tends to keep it upright. This occurs as long as the object’s center of mass lies above its base. To tip the object over, you must rotate its center of mass around the axis of rotation until it is no longer above the base of the object. To rotate the object, you must lift its center of mass. The broader the base, the more stable the object is. Rotational Equilibrium and Dynamics Center of Mass and Stability If the center of mass is outside the base of an object, it is unstable and will roll over without additional torque. If the center of mass is above the base of the object, it is stable. If the base of the object is very narrow and the center of mass is high, then the object is stable, but the slightest force will cause it to tip over. Rotational Equilibrium and Dynamics Center of Mass and Stability Rotational Equilibrium and Dynamics Moment of Inertia (I) The Inertia of a Rotating Mass For a Point Mass… I mr 2 Rotational Equilibrium and Dynamics Moment of Inertia (I) However, Smi changes with different objects Rotational Equilibrium and Dynamics Moment of Inertia (I) Observe how the moment of inertia depends on the location of the rotational axis. Hold a book in the upright position and put your hands at the bottom of the book. Feel the torque needed to rock the book towards and away from you. Repeat with your hands at the middle of the book. Less torque is needed as the average distance of the mass from the axis is less. Rotational Equilibrium and Dynamics Zero Torque and Static Equilibrium No Change in Linear Motion No Acceleration No Change in Rotational Motion In Both X and Y Components Rotational Equilibrium and Dynamics Zero Torque and Static Equilibrium F1 + F2 = mg Rotational Equilibrium and Dynamics Static Equilibrium SFx 0 SFy 0 Rotational Equilibrium and Dynamics Static Equilibrium SFx 0 and SFy 0 St 0 Rotational Equilibrium and Dynamics Conditions for Equilibrium An object is said to be in static equilibrium if both its velocity and angular velocity are zero or constant. First, it must be in translational equilibrium The net force exerted on the object must be zero. Second, it must be in rotational equilibrium The net torque exerted on the object must be zero. Rotational Equilibrium and Dynamics Conditions for Equilibrium Rotational Equilibrium and Dynamics Problem • A 12.5kg board, 4.00m long, is being held up on one end by Jessica. She calls for help, and Bobbi responds. What is the least force that Bobbi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? Rotational Equilibrium and Dynamics Solution m = 12.5kg r = 4.00m F mg If Bobbi lifts the opposite end she will only lift half of the mass of the board. 1 2 F (12.5kg)(9.8m / s ) 61.2 N 2 Rotational Equilibrium and Dynamics Problem • A 12.5kg board, 4.00m long, is being held up on one end by Jessica. She calls for help, and Bobbi responds. What is the greatest force that Bobbi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? Rotational Equilibrium and Dynamics Solution m = 12.5kg r = 4.00m F mg If Bobbi lifts the center she will lift the mass of the board. F (12.5kg)(9.8m / s ) 122 N 2 Rotational Equilibrium and Dynamics Problem • A car’s specifications state that its weight distribution is 53 percent on the front tires and 47 percent on the rear tires. The wheel base is 2.46m. Where is the car’s center of mass? Rotational Equilibrium and Dynamics t front t rear F frontrfront Frear rrear (0.53Fg )rfront (0.47 Fg )( 2.46 rfront ) (0.53Fg )rfront (1.16 Fg ) (0.47 Fg )rfront (0.47 Fg )rfront (0.53Fg )rfront 1.16 Fg r front 1.16m Rotational Equilibrium and Dynamics Homework Pages 305-307 Problems 9 (26Nm) 11 (a, 6300Nm b, 550N) 21 (a, 392N b,Rx=339N, Ry=0N) 23 (11N, 1.6N & 7.1 N ) Rotational Equilibrium and Dynamics Torque and Angular Acceleration a r F ma F a m F mr rF r F 2 r mr mr rF t 2 mr I Rotational Equilibrium and Dynamics Torque and Angular Acceleration Remember that “I” Changes with the Object F mr rF r F 2 r mr mr rF t 2 mr I Rotational Equilibrium and Dynamics Newton’s Second Law of Rotational Motion The Torque Produced by this Force is… t I t I Rotational Equilibrium and Dynamics Problem • A bicycle wheel with a radius of 38cm is given an angular acceleration of 2.67rad/s2 by applying a force of 0.35N on the edge of the wheel. What is the wheel’s moment of inertia? Rotational Equilibrium and Dynamics t I Solution = 2.67rad/s2 r = 38cm F = 0.35N t Fr sin I o (0.35 N )(0.38m)(sin 90.0 ) 2 I 0.050kg m 2 2.67rad / s Rotational Equilibrium and Dynamics Angular Momentum (L) Mass ~ Moment of Inertia Velocity ~ Angular Velocity L I Rotational Equilibrium and Dynamics Angular Momentum (L) v L I (mr ) rmv rp r or L rp sin if the Momentum is Not 2 Perpendicular to the Tangent Rotational Equilibrium and Dynamics Angular Momentum (L) L I t I I t t L t t t Rotational Equilibrium and Dynamics Conservation of Angular Momentum L t t L t t Angular Impulse-Angular Momentum Theorem L L f Li tt Rotational Equilibrium and Dynamics Conservation of Angular Momentum L L f Li tt L f Li tt L f Li (if t 0) Rotational Equilibrium and Dynamics Rotational Kinetic Energy With Rotation mass ~ Moment of Inertia velocity ~ Angular Velocity 1 2 KE mv 2 KErot 1 2 I 2 Rotational Equilibrium and Dynamics Problem When a ceiling fan rotating with an angular speed of 2.55rad/s is turned off, a frictional torque of 0.220N*m slows it to a stop in 5.75s. What is the moment of inertia of the fan? Rotational Equilibrium and Dynamics Solution = 2.55rad/s t = 0.220N*m t = 5.75s t and t I , so t t I tt I (0.220 N m)(5.75 s) 2.55 rad s 0.496 kg m 2 Rotational Equilibrium and Dynamics Problem A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.25m and mass 8.40kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.322rad/s2. Rotational Equilibrium and Dynamics Solution = 0.322rad/s2 r = 3.25m m = 8.40kg 1 2 I mL . 12 1 1 rad 2 2 t I mL (8.40 kg)(3.25 m) 0.322 2 2.38 N m 12 12 s Rotational Equilibrium and Dynamics Problem Calculate the angular momentum of the Earth about its own axis, due to its daily rotation. Assume that the Earth is a uniform sphere. L Rotational Equilibrium and Dynamics Solution I = 2/5 mr2 m = 5.97x1024kg r = 6.38x106m t = 8.64x104s L I 2 2 L mr 5 2rad 5 7 . 27 x 10 rad / s 4 t 8.64 x10 s 2 (5.97 x10 24 kg)(6.38 x106 m) 2 (7.27 x10 5 rad / s) 7 x1033 kg * m 2 rad / s 5 Rotational Equilibrium and Dynamics Problem A torque of 0.12N*m is applied to an egg beater. If the egg beater starts at rest, what is its angular momentum after 0.50s? Rotational Equilibrium and Dynamics Solution t = 0.12N*m t = 0.50s L tt L (0.12 N * m)(0.50s) 0.06kg * m / s 2 Rotational Equilibrium and Dynamics Problem A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the studentstool system is 4.1kg*m2. A second student tosses a 1.5kg mass with a speed of 2.7m/s to the student on the stool, who catches it at a distance of 0.40m from the axis of rotation. What is the resulting angular speed of the student and the stool? Rotational Equilibrium and Dynamics Solution I = 4.1kg*m2 mp = 1.5kg vp = 2.7m/s r = 0.40m mvr I mvr ( I system mr ) 2 mvr I system mr 2 (1.5kg)( 2.7m / s)(0.40m) 0.37rad / s 2 2 4.1kg * m (1.5kg)(0.40m) Rotational Equilibrium and Dynamics Problem A person exerts a tangential force of 36.1N on the rim of a disk-shaped merry-go-round of radius 2.74m and mass 167kg. If the merry-goround starts at rest, what is its angular speed after the person has rotated it through an angle of 60.0°? Rotational Equilibrium and Dynamics Solution FT = 36.1N = 60o m = 167kg r = 2.74m 1 2 o Fr t , I mr , and 60.0 rad 2 3 1 2 W t If 2 2t 2 Fr F f 2 1 2 I mr mr 2 (36.1N )( / 3rad ) f 2 0.575rad / s (167kg)( 2.74m) Rotational Equilibrium and Dynamics Homework Pages 307-310 Problems 27 (1.12x10-2Nm) 35 (1.2rad/s) 37 (7.0m/s) 69 (a, 3.1m/s2 b, 27N, 9.3N) Rotational Equilibrium and Dynamics • Machines – Perform Work – Makes Tasks Easier – Changes Either the Magnitude or the Direction of an Applied Force as it Transmits Energy to the Task Rotational Equilibrium and Dynamics • Machines – Work • Input Work (Wi) – The Work that You Perform on the Machine • Output Work (Wo) – The Work that the Machine Performs Rotational Equilibrium and Dynamics • Machines – Work • According to the Law of Conservation of Energy, the Output Work Can Never be Greater than the Input Work Wo Wi Rotational Equilibrium and Dynamics • Machines – Mechanical Advantage • Input Force (Fin) – The Force You Exert on the Machine • Output Force (Fout) – The Force Exerted by the Machine • Distance from Input (din) – The Distance Achieved from Effort Force • Distance from Output (dout) – The Distance Achieved from Resistance Force Rotational Equilibrium and Dynamics • Machines – Mechanical Advantage (MA) • The Ratio of Forces • The Ratio of the Resistance Force to the Effort Force is the Mechanical Advantage Fout MA Fin Rotational Equilibrium and Dynamics • Machines – Mechanical Advantage (MA) • MA > 1.0 – The Machine Increases the Applied Force • MA < 1.0 – More Force is Applied than the Machine Exerts • MA = 1.0 – Machine Exerts the Same Magnitude of Force as what was Applied (but may change direction) Rotational Equilibrium and Dynamics • Machines – Ideal Mechanical Advantage (IMA) • Ratio of Distances Moved • The Ratio of the Distance from Effort to the Distance from Resistance is the Ideal Mechanical Advantage d in IMA d out Rotational Equilibrium and Dynamics • Machines – Efficiency • Some Energy May Be Lost by the Machine – Friction and Heat Wout % Efficiency x100 Win Rotational Equilibrium and Dynamics • Machines – Ideal Machine • Efficiency of 100% Wo Efficiency x100 Wi Rotational Equilibrium and Dynamics • Machines – Simple Machines • • • • • • Lever Pulley Wheel & Axle Inclined Plane Wedge Screw Rotational Equilibrium and Dynamics • Machines – Compound Machines • Two or More Simple Machines Linked so the Resistance Force of the First Machine Becomes the Effort Force of the Second Machine Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. How much effort force would Tyler apply if this were an ideal machine? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m Findin Fout d out Fout d out Fin d in (1200 N )(5.0m) Fin 300 N 20.0m Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What force is used to balance the friction force if the actual effort is 340 N? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N Fin F f Fin ideal F f Fin Fin ideal F f 340 N 300 N 40 N Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What is the work output? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N Wo Fout d out Wo (1200 N )(5.0m) 6 x10 J 3 Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What is the input work? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N Wi Fin d in Wi (340 N )( 20.0m) 6.8 x10 J 3 Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What is the mechanical advantage? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N Fout MA Fin 1200 N MA 3.53 340 N Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What is the ideal mechanical advantage? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N d in IMA d out 20m IMA 4 5m Rotational Equilibrium and Dynamics • Problem – Tyler raises a 1200N piano a distance of 5.00 m using a set of pulleys. Tyler pulls in 20.0 m of rope. What is the efficiency of the set of pulleys? Rotational Equilibrium and Dynamics • Solution – Fr = 1200N – dr = 5m – de = 20m – Fe = 340N – Fe ideal = 300N MA Efficiency x100 IMA 3.53 Efficiency x100 88% 4 Rotational Equilibrium and Dynamics • Homework