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Definition of Torque Center of Gravity Finding the Center of Gravity Two

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Definition of Torque Center of Gravity Finding the Center of Gravity Two Definition of Torque Chapter 8: Rotational Equilibrium and Dynamics Torque, , is the tendency of a force to rotate an object about some axis Force vs. Torque Torque = (Magnitude of Force) * (Lever arm) = F d • Forces cause accelerations Lever arm is the perpendicular distance from the • Torques cause angular axis of rotation to a line drawn along the direction of accelerations the force. Torque is a vector quantity. Direction of torque: right-hand rule = r F sin points from the point of reference (rotational axis) to the location of the force. Obviously, the magnitude of a torque depends on where we assume the axis of rotation to be. F sin is the tangential component of the force. Unit of torque: N•m Multiple torques, net torque, Center of Gravity and an example The net torque produced by the force of gravity: • When two or more torques are acting on an object, the torques • The object is divided up into a large number of very small are added as vectors. The sum is the net torque. particles of weight (mig) • Each particle will have a set of coordinates indicating its location (xi, yi) Find the net torque (magnitude and • The torque produced by each direction) produced by the forces F1 particle is migxi and F2 about the rotational axis shown in the drawing. The forces are acting • The total (net) torque is on a thin rigid rod, and the axis is (migxi) perpendicular to the page. Now locate a point (xcg, ycg), so that the torque produced by the gravity of the whole mass on that point is equal to the total torque: (mig) xcg = (migxi) Finding the Center of Gravity Two Conditions for Equilibrium • The center of gravity of a homogenous, symmetric body must lie on the axis of A rigid body is in equilibrium if it has zero translational acceleration and symmetry. zero angular acceleration. In equilibrium, the sum of the externally • Often, the center of gravity of such an object applied forces is zero, and the sum of the externally applied torques is is the geometric center of the object. zero: • If a single force can be used to support an object in equilibrium, it must exactly and counteract the effect of gravitational force. The force must lie on a line through the Note that if the net force is zero, the net torque becomes independent of center of gravity of that object. the choice of rotational axis. In this case, we can use any rotational axis to sum up the net torque. The nature of the problem will often suggest a convenient location for the axis. A zero net torque does not mean the absence of rotational motion. An object that rotates at uniform angular velocity can be under the influence of a zero net torque. This is analogous to the translational situation where a zero net force does not mean the object is not in motion. CG could be in a “hollow”! 1 Solving Equilibrium Problems Example of Equilibrium • Draw a diagram of the system – Include coordinates and choose a rotation axis A woman of mass m=55.0 kg sits • Isolate the object being analyzed and draw a free body diagram on the left end of the seesaw, a showing all the external forces acting on the object plank of length L=4.00 m and mass m=12.0 kg, pivoted in the middle. – For systems containing more than one object, draw a separate A man of mass M=75.0 kg sits on free body diagram for each object the right side of the pivot to balance • Apply the Second Condition of Equilibrium the seesaw. (a) Where should the – This will yield a single equation, often with one unknown man sit? (b) Find the normal force which can be solved immediately exerted by the pivot. • Apply the First Condition of Equilibrium – This will give you two more equations • Solve the resulting simultaneous equations for all of the unknowns More Example Consequence of Net Torques: Angular Accel. If the 70-kg fireman goes any higher than as shown, the ladder, which is For a point object rotating about an axis a distance r away from the object, 8.0-m long and has an evenly distributed mass of 20 kg, will slip. The wall is smooth. What is the coefficient of static friction between the 2 FT = ma = mr rFT = = mr ladder and the floor? Now define Moment of Inertia of a point object going around an external: I = mr2 then: = I For a rigid body rotating about a fixed axis Net external torque = ( moment of intertia) * (angular acceleration) = I Note that a centripetal force leads The relationship is analogous to F = ma to no angular acceleration. Moment of Inertia Moments of Inertia For a rigid system consisting of many particles, 2 1 = (m1r1 ) 2 2 = (m2r2 ) 2 3 = (m3r3 ) . . . The moment of inertia for a (rigid) system consisting of many particles is found to be Note: distance to the axis of rotation (a line)! Distance parallel to the axis does not matter. Moment of inertia depends on the location of the rotational axis Obviously, the moment of inertia of an object depends on the chosen axis of rotation. 2 Example Rotational Kinetic Energy A solid, frictionless cylindrical reel of mass M=3.00 kg and radius R=0.400 m is used to draw water from a well. A bucket of mass The rotational kinetic energy KER of a rigid m=2.00 kg is attached to a cord that is object rotating with an angular speed about a wrapped around the cylinder. (a) Find the fixed axis and having a moment of inertia I is tension T in the cord and the acceleration a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? Total Mechanical Energy Conservation, and Work-Energy Theorem in a Rotating System The total kinetic energy of an object that moves with both a linear velocity and a rotational velocity is the sum of the two contributions. Here v is the linear velocity of the center of mass of the object and • Conservation of Mechanical Energy is the angular velocity about its center of mass. (KEt+ KEr+PE)i = (KEt+ KEr+PE)f – Remember, this is for conservative forces, no dissipative forces such as friction can be present – Potential energies of any other conservative forces of center of mass could be added The total mechanical energy of the object is then the sum of its kinetic energy and potential energy. In the absence of work due to • In the case where there are dissipative forces such as friction, use non-conservative forces, the total mechanical energy is conserved. the generalized Work-Energy Theorem instead of Conservation of of center of mass Energy: Wnc = KEt + KEr + PE Energy concepts can be useful for simplifying the analysis of rotational motion. Example Problems Angular Momentum A thin-walled hollow cylinder (mass = mh, radius = rh ) and a solid cylinder (mass = ms, radius = rs) start from rest at the top of an incline. Both cylinders start at the same vertical height h0. All heights are measured relative to an arbitrarily chosen zero level The angular momentum L of a rigid body rotating about a fixed axis is the that passes through the center of mass of a cylinder when it is at product of the body’s moment of inertia I and its angular velocity with respect the bottom of the incline. Ignoring energy losses due to retarding to the axis: forces, determine which cylinder has the greater translational speed upon reaching the bottom. L = I , and For a point object, L=I =mr2 =rmv. Changes in angular momentum are due to rotational impulse: t . Internal forces of a system cannot lead to a net torque on the system, because of the action and the reaction act in opposite directions, but ALONG THE SAME LINE. If the net external torque acting on the system is zero, the total angular momentum of a system remains constant, i.e. it is conserved. 3 Conservation of Angular Momentum, Examples Example Problems of Angular Momentum • With hands and feet drawn closer to the body, the Movie skater’s angular speed increases – L is conserved, I decreases, increases The puck has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and the puck is moving with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Hint: Consider the change of kinetic energy of the puck.) Rotational Analog of Translational Concepts Chapter 8 Summary Torque; Conditions for equilibrium; Center of gravity; Item Rotational Translational Newton’s second law for rotational motion; Moment of inertia; Rotational work; Rotational kinetic energy; Conservation of Displacement s total mechanical energy; Angular momentum; Conservation of Velocity v angular momentum. Acceleration a Cause of Accel. Torque Force F Inertia Moment of Inertia I Mass m Newton’s 2nd Law = I F = ma Work F s Kinetic Energy I2 / 2 mv2 / 2 Momentum Angular momentum Linear Momentum I mv 4 .
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