Static Equilibrium In Chap 6 we studied the equilibrium of pointobjects mass m with the application of Newton s Laws F x 0 F y 0 Therefore no linear translational acceleration a 0 For rigid bodies non point like objects we can apply another condition which describes the lack of rotational motion 0 If the net of all the applied torques is zero we have no rotational angular acceleration 0 don t need to know moment of inertia We can now use these three relations to solve problems for rigid bodies in equilibrium a 0 0 Example Problem The wheels axle and handles of a wheelbarrow weigh 60 0 N The load chamber and its contents weigh 525 N It is well known that the wheelbarrow is much easier to use if the center of gravity of the load is placed directly over the axle Verify this fact by calculating the vertical lifting load required to support the wheelbarrow for the two situations shown FL FD L1 FL FD Fw L2 Fw L2 L3 L3 L1 0 400 m L2 0 700 m L3 1 300 m First draw a FBD labeling forces and lengths from the axis of rotation FL FD Choose a direction for the rotation CCW being axis Fw positive is the convention L 1 a 0 L2 L3 D W L 0 FD L1 FW L2 FL L3 0 FD L1 FW L2 FL L3 525 N 0 400 m 60 0 N 0 700 m FL 1 300 m FL 194 N Apply to case with load over wheel Torque due to dirt is zero since lever arm is zero axis b 0 FL FD Fw L2 D W L 0 FD L1 FW L2 FL L3 0 FD L1 FW L2 FL L3 525 N 0 m 60 0 N 0 700 m FL 1 300 m FL 32 3 N Who What is carrying the balance of the load Consider sum of forces in y direction F 0 FL FD FW FN 0 FN FD FW FL FN 525 60 194 391 N FN 525 60 32 3 553 N FD y a b FL Fw FN We did not consider the Normal Force when calculating the torques since its lever arm is zero Center of Gravity FD The point at which the weight of a rigid body can be considered to act when determining the torque due to its weight Consider a uniform rod of length L Its center of gravity cg corresponds to its geometric L center L 2 L 2 cg Each particle which makes up the rod creates a torque about cg but the sum of all torques due to each particle is zero So we treat the weight of an extended object as if it acts at one point Consider a collection of point particles on a massless rod The sum of the torques Mg xcg m1 gx1 m2 gx2 m g 1 m2g m3g m3 gx3 Mgxcg x2 M m1 m2 m3 x1 m1 x1 m2 x2 m3 x3 xcg xcm M x3 Angular Momentum In Chapter 9 we defined the linear momentum Analogously we can define Angular Momentum p mv L I Since is a vector L is also a vector L has units of kg m2 s The linear and angular momenta are related vt L I mr rmv t rpt r 2 pt r L gives us another way to express the rotational motion of an object For linear motion if an external force was applied for some short time duration a change in linear momentum resulted F t p p ext f Similarly if an external torque is applied to a rigid body for a short time duration its angular momentum will change t L L If ext f i i ext 0 then L f Li This is the Principle of Conservation of Angular Momentum How to interpret this Say the moment of inertia of an object can decrease Then its angular speed must increase Ii I f L f Li Ii I f f I i i f i i If Example Problem For a certain satellite with an apogee distance of rA 1 30x107 m the ratio of the orbital speed at perigee to the orbital speed at apogee is 1 20 Find the perigee distance rP Not uniform circular motion Satellites generally move in elliptical orbits Kepler s 1st Law Also A vA the tangential velocity is not constant vP P If the satellite rA rP is circling the Earth the furthest point in its orbit from the Earth is called the apogee The closest point the perigee For the Earth circling the sun the two points are called the aphelion and perihelion Given rA 1 30x107 m vP vA 1 20 Find rP Method Apply Conservation of Angular Momentum The gravitational force due to the Earth keeps the satellite in orbit but that force has a line of action through the center of the orbit which is the rotation axis of the satellite Therefore the satellite experiences no external torques LA LP I A A I P P 2 vA 2 vP mrA mrP rA rP rA v A rP v P rP rA v A v P 7 1 30 x10 1 1 20 7 1 08x10 m Summary Translational Rotational x displacement v velocity a acceleration F cause of motion m inertia I F ma 2nd Law I Fdd work 1 2mv2 KE 1 2I 2 p mv momentum L I
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