King Fahd University of Petroleum & Minerals

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

2. Rectangular Coordinates

3. Normal and Tangential Coordinates

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

5. Polar Coordinates • Radial coordinate r • Transverse coordinate q • q and r are perpendicular • Theta q in radians • 1 rad = 180o/p • Direction ur and uq

6. Directions of Forces Wight force mg – vertical downward. Normal force N – perpendicular to the tangent of path Frictional force Ff – along the tangent in the opposite direction of motion. Acting force Fq – along q direction Ff

7. Free-Body Diagram • Select the object • Establish r, q coordinate system • Set ar always acts in the positive r direction • Assume aq acts in the positive q • Set W = mg always acts in a vertical direction–downward in vertical problem neglect weight in horizontal problem • Set the tangent line • Set normal force FN perpendicular to tangent line • Set frictional force Ff opposite to tangent • Set acting force F along q direction • Calculate the y angle using • Calculate other angles • Apply tangent F ar r y FN aq Ff mg q

8. r 2 q (psi) y angle • Establish r = f(q) • Example : r = 10t2 and q = 0.5t • r = 40q2 • 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 q. • Negative – opposite direction to positive q. q

9. Example 13-10 W = 2 Ib Smooth horizontal r = 10 t2 ft q = 0.5t rad F = ? Tangent force at t = 1 s.

10. Problem m=2 kg Smooth horizontal r = 0.4 q Ptangent = ? N=? At q = 45o q N P r

11. Problem 13-89 Smooth Horizontal m = 0.5 kg r = (0.5 q) q = 0.5 t2 rad Applied force Normal force t = 2 s. r y y y q

12. Problem

14. Review • Example 13-11 • Example 13-12

15. Thank you