34 C HAPTER 1 KINEMATICS OF A PARTICLE 1 1.5 Curvilinear Motion: Rectangular Component Occaionall the motion of a particle can bet be decribed along a path that can be epreed in term of it,, coordinate. Poition. If the particle i at point (,, ) on the curved path hown in Fig. 1 17a,then it location i defined b the poition vector i k j r i j k Poition r = i + j + k (1 10) When the particle move, the,, component of r will be function of time; i.e., = 1t, = 1t, = 1t, o that r = r1t. At an intant the magnitude of r i defined from Eq. C 3 in Appendi C a r = 4 + + And the direction of r i pecified b the unit vector u r = r>r. Velocit. The firt time derivative of r ield the velocit of the particle. Hence, v v i v j v k v = dr dt = d dt 1i + d dt 1j + d dt 1k Velocit Fig. 1 17 When taking thi derivative, it i necear to account for change in both the magnitude and direction of each of the vector component. For eample, the derivative of the i component of r i d d 1i = dt dt i + di dt The econd term on the right ide i ero, provided the,, reference frame i fied,and therefore the direction (and the magnitude) of i doe not change with time. Differentiation of the j and k component ma be carried out in a imilar manner, which ield the final reult, v = dr dt = v i + v j + v k (1 11) where v = # v = # v = # (1 1)
1.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 35 The dot notation #, #, # repreent the firt time derivative of = 1t, = 1t, = 1t, repectivel. The velocit ha a magnitude that i found from 1 v = 4 v + v + v and a direction that i pecified b the unit vector u v = v>v. A dicued in Sec. 1 4, thi direction i alwa tangent to the path, a hown in Fig. 1 17b. Acceleration. The acceleration of the particle i obtained b taking the firt time derivative of Eq. 1 11 (or the econd time derivative of Eq. 1 10). We have a = dv dt = a i + a j + a k (1 13) Acceleration a a i a j a k (c) where a = v # = $ a = v # = $ a = v # = $ (1 14) Here a, a, a repreent, repectivel, the firt time derivative of v = v 1t, v = v 1t, v = v 1t, or the econd time derivative of the function = 1t, = 1t, = 1t. The acceleration ha a magnitude a = 4 a + a + a and a direction pecified b the unit vector u a = a>a. Since a repreent the time rate of change in both the magnitude and direction of the velocit, in general a will not be tangent to the path, Fig. 1 17c.
36 C HAPTER 1 KINEMATICS OF A PARTICLE 1 Important Point Curvilinear motion can caue change in both the magnitude and direction of the poition, velocit, and acceleration vector. The velocit vector i alwa directed tangent to the path. In general, the acceleration vector i not tangent to the path, but rather, it i tangent to the hodograph. If the motion i decribed uing rectangular coordinate, then the component along each of the ae do not change direction, onl their magnitude and ene (algebraic ign) will change. conidering the component motion, the change in magnitude and direction of the particle poition and velocit are automaticall taken into account. Procedure for Anali Coordinate Stem. A rectangular coordinate tem can be ued to olve problem for which the motion can convenientl be epreed in term of it,, component. Kinematic Quantitie. Since rectilinear motion occur along each coordinate ai, the motion along each ai i found uing v = d>dt and a = dv>dt; or in cae where the motion i not epreed a a function of time, the equation ad = vdvcan be ued. In two dimenion, the equation of the path = f() can be ued to relate the and component of velocit and acceleration b appling the chain rule of calculu. A review of thi concept i given in Appendi C. Once the,, component of v and a have been determined, the magnitude of thee vector are found from the Pthagorean theorem, Eq. 3, and their coordinate direction angle from the component of their unit vector, Eq. 4 and 5.
1.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 37 EXAMPLE 1.9 1 At an intant the horiontal poition of the weather balloon in Fig. 1 18a i defined b = 18t ft, where t i in econd. If the equation of the path i = >10, determine the magnitude and direction of the velocit and the acceleration when t =. SOLUTION Velocit. The velocit component in the direction i 10 v = # = d 18t = 8 ft> : dt A To find the relationhip between the velocit component we will ue the chain rule of calculu. (See Appendi A for a full eplanation.) 16 ft v = # = d dt 1 >10 = # >10 = 11618>10 = 5.6 ft> c When t =, the magnitude of velocit i therefore v = 4 (8 ft> + (5.6 ft> = 6.8 ft> The direction i tangent to the path, Fig. 1 18b,where u v = tan v -1-1 5.6 = tan v 8 Acceleration. The relationhip between the acceleration component i determined uing the chain rule. (See Appendi C.) We have a = v # = d dt 18 = 0 = 7.6 v 6.8 ft/ u v 7.6 a = v # = d dt 1# >10 = 1 # # >10 + 1 $ >10 = 18 >10 + 11610>10 = 1.8 ft> c Thu, a = 4 (0 + (1.8 = 1.8 ft> The direction of a, a hown in Fig. 1 18c,i -1 1.8 u a = tan 0 = 90 a 1.8 ft/ u a 90 (c) Fig. 1 18 NOTE: It i alo poible to obtain v and a b firt epreing = f1t = 18t >10 = 6.4t and then taking ucceive time derivative.
38 C HAPTER 1 KINEMATICS OF A PARTICLE 1 EXAMPLE 1.10 For a hort time, the path of the plane in Fig. 1 19a i decribed b = (0.001 ) m. If the plane i riing with a contant velocit of 10 m>, determine the magnitude of the velocit and acceleration of the plane when it i at = 100 m. SOLUTION When = 100 m, then 100 = 0.001 or = 316. m. Alo, ince v = 10 m>,then = v t; 100 m = (10 m>) t t = 10 Velocit. Uing the chain rule (ee Appendi C) to find the relationhip between the velocit component, we have v = # = d dt (0.001 ) = (0.00) # = 0.00v (1) 100 m 0.001 Thu 10 m> = 0.00(316. m)(v ) v = 15.81 m> The magnitude of the velocit i therefore v = 4 v + v = 4 (15.81 m>) + (10 m>) = 18.7 m> Acceleration. Uing the chain rule, the time derivative of Eq. (1) give the relation between the acceleration component. a = v # = 0.00 # v + 0.00v # = 0.00(v + a ) When = 316. m, v = 15.81 m>, v # = a = 0, 100 m a v v v 0 = 0.00((15.81 m>) + 316. m(a )) a = -0.791 m> The magnitude of the plane acceleration i therefore Fig. 1 19 a = 4 a + a = 4 (-0.791 m> + (0 m> = 0.791 m> Thee reult are hown in Fig. 1 19b.