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Chapter 12

by: Jacob Edwards

Chapter 12 80218 - PHYS 1220 - 001

Jacob Edwards
GPA 3.2

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About this Document

These notes cover torque
Physics with Calculus I
Lih-sin The
Class Notes
25 ?




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This 2 page Class Notes was uploaded by Jacob Edwards on Sunday April 3, 2016. The Class Notes belongs to 80218 - PHYS 1220 - 001 at Clemson University taught by Lih-sin The in Winter 2016. Since its upload, it has received 32 views. For similar materials see Physics with Calculus I in Physics 2 at Clemson University.


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Date Created: 04/03/16
Chapter 12 A rigid object moves with all it’s components locked so that the shape does not change. When a rigid object is rotating each particle on the object moves in a circular path that is center on the same axis called the rotational axis. A fixed axis is a rotational axis that does not move in relation to the observer. Polar coordinates are used for rotational motion. The angular position is shown in the below equation where s is distance (m) and r is a measurement in radians θ= s r The change in angular position is called angular displacement and is found by the following: ∆ θ=θ fθ i The average angular speed is shown by: θ fθ i ∆θ ω av = t ft i ∆t The instantaneous velocity is the taking the limit of the average angular speed as the change in time approaches zero. The instantaneous angular speed is the magnitude of the instantaneous velocity vector. Angular accelerations is given by the following: a= ω fω i= ∆ω t −t ∆t f i Similar to translational kinematic equations there are rotational kinematic equations. For these velocity is replaced by ω and x is replaced by θ. 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 (ωr) 2 a c r =+ω r Torque causes the angular acceleration of a rigid object. Torque is given as the lowercase Greek letter τ. The formula for torque is below: τ=rFsinφ φ is the angle between F and r. The unit for torque is an Nm. The vector equation for torque is found by taking the cross product of two vectors; A and B. The magnitude of the cross product is given by: R=ABsinφ The inertia of an object can be found by taking the torque vectors and dividing them by the angular acceleration vectors. If given a lever problem the force of the lever is equal to the torque of the load divided by the torque of the applied force times the normal force.


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