King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 13

Rectangular Coordinates

Normal and Tangential Coordinates

Cylindrical Coordinates Resultant force components causing a particle to move with a known acceleration

Polar Coordinates Radial coordinate r Transverse coordinate   and r are perpendicular Theta  in radians 1 rad = 180 o /  Direction u r and u 

Directions of Forces Wight force mg – vertical downward. Normal force N – perpendicular to the tangent of path Frictional force F f – along the tangent in the opposite direction of motion. Acting force F  – along  direction FfFf

Free-Body Diagram Select the object Establish r,  coordinate system Set a r always acts in the positive r direction Assume a  acts in the positive q Set W = mg always acts in a vertical directio n–downward in vertical problem neglect we ight in horizontal problem Set the tangent line Set normal force F N perpendicular to tangent line Set frictional force F f opposite to tangent Set acting force F along  direction Calculate the  angle using Calculate other angles Apply r  arar aa FNFN tangent FfFf F mg 

(psi)  angle Establish r = f(  ) Example : r = 10t 2 and  = 0.5t r = 40   From geometry (Psi) is defined between the extended radial line and the tangent to the curve Positive – counterclockwise sense or in the positive direction of . Negative – opposite direction to positive .  r 2 

W = 2 Ib Smooth horizontal r = 10 t 2 ft  = 0.5t rad F = ? Tangent force at t = 1 s. Example 13-10

Problem m=2 kg Smooth horizontal r = 0.4  P tangent = ? N=? At  = 45 o P N r 

Problem Smooth Horizontal m = 0.5 kg r = (0.5  )  = 0.5 t 2 rad Applied force Normal force t = 2 s.  r   

Problem

Problem 13-95

Review Example Example 13-12