One of the common descriptions of curvilinear motion usespath variables, which are measurements made along thetangent t and normal n to the path of the particles.These coordinates provide a very natural description forcurvilinear motion and are frequently the most direct andconvenient coordinates to use; they move along the pathwith the particle. Therefore no position vector is required.The positive direction for n at any position is alwaystaken toward the center of curvature of the path.The positive t direction is in the direction of motion(direction of velocity).ds= d where is in radians.We now use the coordinates n and t to describe the velocity and acceleration. For this purpose, we introduce unit vectors in the n-direction and in the t-direction. nerterDuring a differential increment of time dt, theparticle moves a differential distance ds alongthe curve fromA to A.It is unnecessary to consider thedifferential change in between Aand A.The acceleration of the particle was defined as , and we observe that the acceleration is a vector which reflects both the change in magnitude and the change in direction of velocity. dtv da rr=We now differentiate the velocity applying the ordinary rule for the differentiation of the product of a scalar and a vector and get The unit vector now has a nonzero derivative because its direction changes.terTo find we analyze the change in during a differential increment of motion as the particle moves from A to A. terte&re dr{{12 /2sin?12= == = =de dee ddde ddtdde ddte dettttt ttdr4 4 3 4 4 2 1 rrr3 2 1rr r&r&The direction of is given by ..2te dr2 dn t e e r&&r =2 2 = te dd 4 4 3 4 4 2 1 r{ {n t t tnatat tevv e v e v e v avv e v e v e v e v antr r&&r r&r& &r&r&&r r&r + = + == = + = + =n t t t eve v e v e v a r r&&r r&r2+ = + =2 2n t a a a + =(Change in direction of the velocity)(Change in magnitude of the velocity time rate of change of the speed)22 & &= = = vvans v at & && && = = = Magnitude of the accelerationRadius of Curvature (Erilik Yarap): If trajectory is given as y=f(x), the radius of curvature can be determined with the following formula:2 / 321dxdy(((

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\|+= 22dxy d= The absolute value sign is used to guarantee that will always be positive +. The normal component of accelerationanis always directed toward the center ofcurvature. The tangential component ofacceleration a , on the other hand, will be acceleration at, on the other hand, will bein the positive t-direction of motion if speedv is increasing and in the negative t-direction if the speed is decreasing.Figure shows the variation in the accelerationvector for a particle moving fromA to B with(a) increasing speed and (b) decreasing speed.02 2=== v vanAt an inflection point on the curve,the normal acceleration goes to zerobecause becomes infinite. & & v rrvan = = = 22Circular motion is an important special case of plane curvilinear motionwhere the radius of curvature becomes the constant radius r of thecircle and the angle is replaced by the angle measured from anyconvenient radial reference to OP.=r=constant& && r v a rt = = &= =dtd & &= =dtd(Angular velocity) (Angular acceleration)1. Six acceleration vectors are shown for the car whose velocityvector is directed forward. For each acceleration vector describein words the instantaneous motion of the car.2. A train enters a curved horizontal section of track at a speed of100 km/h and slows down with constant deceleration to 50 km/h in12 seconds. An accelerometer mounted inside the train records ahorizontal acceleration of 2 m/s2 when the train is 6 seconds intothe curve. Calculate the radius of curvature of the track for thisinstant.3. The design of a camshaft-drive system of a four cylinderautomobile engine is shown. As the engine is revved up, the beltspeed v changes uniformly from 3 m/s to 6 m/s over a two-secondinterval. Calculate the magnitudes of the accelerations of point P1and P2 halfway through this time interval.4. A ball is thrown horizontally from the top of a 50 m cliff at Awith a speed of 15 m/s and lands at point C. Because of stronghorizontal wind the ball has a constant acceleration in the negativex-direction. Determine the radius of curvature of the path of theball at B where its trajectory makes an angle of 45o with thehorizontal. Neglect any effect of air resistance in the verticaldirection.5. The pin P is constrained to move in the slotted guides whichmove at right angles to one another. At the instant represented, Ahas a velocity to the right of 0.2 m/s which is decreasing at therate of 0.75 m/s each second. At the same time, B is moving downwith a velocity of 0.15 m/s which is decreasing at the rate of 0.5m/s each second. For this instant, determine the radius ofcurvature of the path followed by P.6. Pin P in the crank PO engages the horizontal slot in the guide Cand controls its motion on the fixed vertical rod. Determine thevelocity and the acceleration of guide C for a given value ofthe angle is(a) and(b) if =0 and =& =& &&y& y& &0 = & &