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Exam 3 Study Guide

by: Jacob Edwards

Exam 3 Study Guide 80218 - PHYS 1220 - 001

Jacob Edwards
GPA 3.2

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Physics with Calculus I
Lih-sin The
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This 4 page Study Guide was uploaded by Jacob Edwards on Saturday April 9, 2016. The Study Guide belongs to 80218 - PHYS 1220 - 001 at Clemson University taught by Lih-sin The in Winter 2016. Since its upload, it has received 53 views. For similar materials see Physics with Calculus I in Physics 2 at Clemson University.


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Date Created: 04/09/16
Exam 3 Study Guide Momentum, Rotational Motion, Torque Momentum moves in the same direction as velocity and is given by the  equation: p=mv The initial momentum is always the same as the final momentum.  In the problems that ask for recoil motion use the following formula often  times in momentum problems they are paired with translational  kinematic equations: m1v 1m v2 2 An open systems is a system that gains or loses mass.  This is the  formula to us for the rocket or a similar system: ∆ M F thrust ∆t ) The angular position is given in polar coordinates and is found with the  following: s θ= r Angular displacement is the change in angular position and is given by  the following: ∆ θ=θ fθ i The average angular speed is shown by: θ −θ ∆θ ω av f = tf−ti ∆t Angular Acceleration is given in the following equation: a= ω fω i= ∆ω tf−ti ∆t Similar to translational kinematic equations there are rotational  kinematic equations. For these velocity is replaced by ω and x is  replaced by θ.  These act the same way as translational kinematic  equations. To find tangential velocity use: v=ωr Where r is the radius. To find tangential velocity a similar formula is used: a=ar Don’t forget the formula for centripetal acceleration is: 2 a = (ωr) =+ω r2 c r Torque is a part of Newtons second law.  It is given by the following: τ=rFsinφ φ  is the angle between F and r.  The unit for torque is an Nm. Torque can also be found by taking the cross product of two vectors.   The magnitude of Torque is given by: R=ABsinφ Inertia is the torque divided by the angular acceleration. In translational motion the more massive particle has the most rotational  inertia Mass distribution with respect to the rotational axis also impacts  rotational inertia; the farther away the mass is from the rotational axis  the more rotational inertia will exist. Rotational inertia can be found by: n I= ∑ m r2 i=1 ii M is the mass of the particle r is the distance from the rotational axis On a continuous object the rotational inertia is: I= ∫ dm The parallel axis theorem where M is mass, h is the perpendicular  distance between the new axis and the axis through the center of mass  and I CMs the rotational inertia around the center of mass I=ICM +M h 2 The center of mass is found by: m xCM=( 2 )x2 m 1m 2 For center of mass multiply the associated mass with the associated x  coordinate. Kinetic Energy for rotating object: 1 2 K r I ω 2 Conservation of energy holds true for rotational motion similar to  translational motion K iK +U ri=K +i +U +∆E f rf f th K­ is the translational kinetic K ­ris rotational kinetic Newtons second law is shown through torque and is the sum of all  torque. If the system is conserved torque initial is equal to torque final


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