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# PSU PHYS 250 - Rotational Dynamics and Static Equilibrium

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p250c2:110: Rotational Dynamics and Static EquilibriumTorque: the rotational analogue of forceTorque = force x moment armτ = Fr┴=F r sin θmoment arm = perpendicular distance through which the force actsconvention: counterclockwise torques are positive (i.e., a torque that would cause a counterclockwise rotation if it were the only torque)r┴Fr┴Fr┴Fθθθ=90°p250c2:2Example: Two forces act on a wheel at rest as shown. The wheel is free to rotate without friction , and has a radius of 0.42 m, and is initially at rest. Given that F1 is 12 N and F2 is 9.5 N, determine the torques created by the individual forces as well as the net torque.90°50°F2F1p250c2:3Torque, Moment of Inertia and angular accelerationFor a single mass:FT= maT(tangential acceleration)FTr =maTrτ = mαr r = mr2 αrecall: moment of inertia I = mr2τ= I α (looks like F = ma!)for a system of objects ( a rigid object) I = Σmiri2FaTLinear Quantity Angular Quantitym Ia F p250c2:4Example: A light rope is wrapped around a uniform disk acting used as a pulley. The tension in the rope is 0.53 N. The mass of the pulley is 1.30 kg and its radius is 0.11 m. What is the angular acceleration of the pulley? If the end of the rope is pulled for a distance of 1.50 m, What is the final angular velocity of the pulley?What is the pulley's final rotational kinetic energy?How much work was done by the person pulling the end of the rope?p250c2:5A meter stick fixed to rotate about one end is initially held horizontally and then released. What is the initial angular acceleration of the stick? What is the tangential acceleration of the free end of the stick? How far from the axis of rotation of the stick is the acceleration equal to the acceleration of gravity?p250c2:6Equilibrium: stability, steadiness, balance etc.Mechanical Equilibrium: absence of change in motion=> Net Force = 0 !(usually, no motion)With Rotational EquilibriumRotational Equilibrium: absence of change in rotation (usually: no rotation)=> net torque is zero Στi = 0about any axis!planesametheinlyingtorquesallfori 0=∑τPositive torque for counterclockwise rotation: τ = F r┴Watch signs for torquer┴Fr┴FNegative torque for clockwise rotation: τ = − F r┴0= components force of sum0= components force of sum0===∑∑∑yxFyFxFp250c2:7example: A 20 kg child sits on a light plank 4.00 m long, supported on each end by the child's parents. If the child sits 1.00 m from the mother (and hence 3.00 m from the father) determine the force exerted by each parent supporting the plank.alternate example: opening a door with a .357 magnump250c2:8Example: A 60 kg woman stands at the end of a uniform 4m, 50 kg diving board supported as shown. Determine the forces exerted by the two supports..800 m4.00 mp250c2:9m1m2Center of Gravity: Balance pointsbalance: net torque is zeroxcg= m1 g x1+m2 g x2 +… extended object's weight “located” at their center of gravitymg mgcg kit demosp250c2:10Dynamic Applications of torqueCombining torques and forcestaking into account rotational inertia Example: analyze the system at rightmIRp250c2:11Angular MomentumL = I ω (like p = m v )(the tendency of an object to keep rotating)+ Angular momentum is conserved in the absence of external torquesLstart = Lend note: for a point mass moving in a circle L = mr2ω= mvr = prp250c2:12A child (m=21.2 kg) runs at the edge (i.e. in a path tangent to the rim)of a merry-go-round which has a radius of 510 kg m2 and a radius of 2.31 m. As the child jumps on, the entire system begins to rotate. What is the angular speed of the system?p250c2:13Rotational work: FW = F  x x=R W = F  x=FR W =  Example: From the example on slide 4, determine the torque exerted on the pulley by the rope and and the work done by that torque.p250c2:14Gyroscope

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