Chapter 8: Rotational Equilibrium and Dynamics

Name: Chapter 8: Rotational Equilibrium and Dynamics
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Description: Chapter 8: Rotational Equilibrium and Dynamics

Chapter 8: Rotational Equilibrium and Dynamics
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 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 Dynamics What is Involved in Rotating an Object Force Distance from the Point of Rotation Torque (t)

Torque (t) Product of the Rotational Force and the Length of a Lever Arm

Torque Magnitude Fr Direction + When Rotation is Counterclockwise - When Rotation is Clockwise

Torque Units Force = Newton Length = meter Torque = Newton*Meter (N*m)

Torque Direction is Not Always Perpendicular to the Radius

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?

Solution t = 1.23 N*m r = 0.040m F = ?

Problem What is the torque on a bolt produced by a 15N force exerted perpendicular to a wrench that is 25cm long.

Solution F = 15N r = 25cm

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?

Solution t = ? r = 0.70m m = 3.3kg

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 teeter-totter 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?

Solution rchild = 1.5m mchild = 16kg radult = 3.0m Fadult = 95N 3.0 m > 2.48 m; the adult has enough momentum to push down and push the child up.

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 teeter-totter 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?

Solution rchild = 1.5m mchild = 16kg radult = 2.5m Fadult = 95N 2.5 m > 2.48 m; the adult has just barely enough momentum to push the child up.

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 teeter-totter 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?

Solution rchild = 1.5m mchild = 16kg radult = 2.0m Fadult = 95N 2.0 m < 2.48 m; the adult does not have enough momentum to hold the child up, and the child goes down.

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.

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.

Center of Mass Draw a second vertical line. The center of mass is at the point where the two lines cross.

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.

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.

Center of Mass and Stability

Moment of Inertia (I) The Inertia of a Rotating Mass For a Point Mass…

Moment of Inertia (I) However, Smi changes with different objects

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.

Zero Torque and Static Equilibrium No Change in Linear Motion No Acceleration No Change in Rotational Motion In Both X and Y Components

Zero Torque and Static Equilibrium F1 + F2 = mg

Static Equilibrium

Static Equilibrium and

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.

Conditions for Equilibrium

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?

Solution m = 12.5kg r = 4.00m If Bobbi lifts the opposite end she will only lift half of the mass of the board.

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?

Solution m = 12.5kg r = 4.00m If Bobbi lifts the center she will lift the mass of the board.

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?

Homework Pages Problems 9 (26Nm) 11 (a, 6300Nm b, 550N) 21 (a, 392N b,Rx=339N, Ry=0N) 23 (11N, 1.6N & 7.1 N )

Torque and Angular Acceleration

Torque and Angular Acceleration Remember that “I” Changes with the Object

Newton’s Second Law of Rotational Motion The Torque Produced by this Force is…

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?

Solution a = 2.67rad/s2 r = 38cm F = 0.35N

Angular Momentum (L) Mass ~ Moment of Inertia Velocity ~ Angular Velocity

Angular Momentum (L) or if the Momentum is Not Perpendicular to the Tangent

Angular Momentum (L)

Conservation of Angular Momentum Angular Impulse-Angular Momentum Theorem

Conservation of Angular Momentum

Rotational Kinetic Energy With Rotation mass ~ Moment of Inertia velocity ~ Angular Velocity

Problem When a ceiling fan rotating with an angular speed of 2.55rad/s is turned off, a frictional torque of N*m slows it to a stop in 5.75s. What is the moment of inertia of the fan?

Solution w = 2.55rad/s t = 0.220N*m t = 5.75s

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.

Solution a = 0.322rad/s2 r = 3.25m m = 8.40kg

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.

Solution I = 2/5 mr2 m = 5.97x1024kg r = 6.38x106m t = 8.64x104s

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?

Solution t = 0.12N*m t = 0.50s

Problem A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool 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?

Solution I = 4.1kg*m2 mp = 1.5kg vp = 2.7m/s r = 0.40m

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-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 60.0°?

Solution FT = 36.1N q = 60o m = 167kg r = 2.74m

Homework Pages Problems 27 (1.12x10-2Nm) 35 (1.2rad/s) 37 (7.0m/s) 69 (a, 3.1m/s2 b, 27N, 9.3N)

Machines Perform Work Makes Tasks Easier Changes Either the Magnitude or the Direction of an Applied Force as it Transmits Energy to the Task

Machines Work Input Work (Wi) The Work that You Perform on the Machine Output Work (Wo) The Work that the Machine Performs

Machines Work According to the Law of Conservation of Energy, the Output Work Can Never be Greater than the Input Work

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

Machines Mechanical Advantage (MA) The Ratio of Forces The Ratio of the Resistance Force to the Effort Force is the Mechanical Advantage

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)

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

Machines Efficiency Some Energy May Be Lost by the Machine Friction and Heat

Machines Ideal Machine Efficiency of 100%

Machines Simple Machines Lever Pulley Wheel & Axle Inclined Plane Wedge Screw

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

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?

Solution Fr = 1200N dr = 5m de = 20m

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?

Solution Fr = 1200N dr = 5m de = 20m Fe = 340N Fe ideal = 300N

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?

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?

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?

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?

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?

Homework

Chapter 8: Rotational Equilibrium and Dynamics

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