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UB PHY 101 - Chapter 8: Rotational Equilibrium and Dynamics

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PHY 101 1nd Edition Lecture 21 Outline of Last Lecture I. Summary: Angular & Linear QuantitiesII. Forces Causing Centripetal AccelerationIII. Problem Solving StrategiesIV. 7.5 Newtonian Gravitation RevisitedOutline of Current Lecture V. Chapter 8: Rotational Equilibrium and DynamicsVI. 8.1 Torquea. Force vs. TorqueVII. 8.2 Torque and Rotational EquilibriumCurrent LectureChapter 8: Rotational Equilibrium and Dynamics- Physics behind opening a door- Three factors that determine the “effectiveness” of the force in opening the door:o The magnitude of the forceo The position of the application of the forceo The angle at which the force is applied- Combining the 3 factors = torque8.1 Torque- Torque (τ): measures the effectiveness of a force to rotate an object about an axiso τ = rFsinΘ  SI Unit: N·mo R: magnitude of the position vector of the point where force is applied, measured from the rotational axis- If the force is perpendicular to the position vector, sinΘ = 1  τ = rFLever Arm- It is sometimes convenient to introduce the concept of lever armo d = rsinΘ- so thatThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.o τ = rFsinΘ = dFForce vs. Torque- forces cause (linear) acceleration- torques cause angular accelerations- like force, torque is a vector quantityo if the turning tendency of the force is counterclockwise, the torque is positiveo if the turning tendency is clockwise, the torque is negative- when two or more torques are acting on an object, the torques are added as vectors (can be positive or negative)- if the net torque applied to an object is zero, the object’s rate of rotation doesn’t change: it either does not rotate or does constant speed rotation (the object is said to bein rotational equilibrium)Example- two people are pushing a revolving door at the same timeo F1 = 625 N; r1 = 1.2 mo F2 = 850N; r2 = 0.8m- Both forces are perpendicular to the position vectors- Calculate the net torque on the revolving door- Analysis: torque can be positive (CCW) or negative (CW)- Solution: torque 1: negative (CW)o τ 1 = -r1F1 = 1.2 x 625 = -750 N·m- Torque 2: Positive (CCW)o τ2 = r2F2 = 0.8 x 850 = 680 N·m - total torque:o τ= τ1 + τ2 = -750 + 680 = -70N·m8.2 Torque & Rotational Equilibrium- we have discussed the conditions for equilibrium: the net force applied on the object is zero- actually this condition alone is not sufficient for an object in equilibrium- consider the revolving door again- suppose we have 2 Forces (F1,F2) equal in magnitude, opposite in direction applied to the dooro F1 + F2 = 0- The door keeps spinning faster and faster – not in


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UB PHY 101 - Chapter 8: Rotational Equilibrium and Dynamics

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