hw1sols - Homework 1 Solutions 1. The position of the plane...

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Unformatted text preview: Homework 1 Solutions 1. The position of the plane is given by b AAE 250 ! , p = 1000ex + 300tey + 4000^z  m r e b and the velocity of the plane is easily obtained by di erentiation as ! , p = d , p = 300^y m=s: v dt ! r e The position of the plane is given by and the velocity is given by ! , a = 100texm r ^ ! , a = 100^x m=s v e a The velocity of the plane relative to the velocity of the auto is simply the di erence of the two vectors, ! , p=a = , p , , a = 300^y , 100^xm=s !! v vv e e 1000 = 100t  t = 10 b The plane is directly over the auto when their position vectors only di er in the x-coordinate, i.e the x-coordinates are equal: and the y-coordinates are equal, i.e. 300t = 0  t = 0 Therefore, since the x-components are only equal at t = 10 and the y-components are only equal at t = 0, there is no time when the plane is directly over the auto. c The scalar speed of the auto relative to the plane is given by: ! , a=p = ,, p=a = p1002 + 3002  316:23m=s ! v v 1 !! d The minimum separation distance can be found by minimizing d2 = ,  , , dd ! rather than , . We can do this since , d distance is always positive, and the minimum of d2, is also the minimum of !. We obtain d d2 = 1  105t2 , 2  105t + 1:7  107 We minimize d2 by taking the derivative and setting equal to zero to nd that a minimum occurs at t = 1. We can check second derivative to verify that this is a minimum and not a maximum or saddle. Evaluating d2at t=1, and taking the positive square root, we nd that the minimum distance is given by dmin = dt = 1  4; 110:96m e To nd when the separation distance is d = 10km = 10; 000m, we use the quadratic equation from part d, setting d = 10; 000m. Solving this quadratic equation gives two solutions, t  29:83s and t  ,27:83s. 2. To start, we de ne another coordinate frame, with unit vectors er pointing along ^ the length of the bar, and e pointing normal to the bar. ^ j A C L/ 2 L i θ P eθ V 0 er 2 a To solve this problem, we use Equation 2-83 from Greenwood, which relates the velocity of two points in a body. The equation, as we shall use it here, is ! where P and A are two point in the body, and , is the position of point P relative to point A. For this problem, we shall set point P to be the point of contact with the ground, so that ,P = vo^ = vocos er + sin e  !i v ^ ^ where we have introduced unit vectors er which points along the length of the ^ bar, and e , which is normal to the bar. Point A is the point of the bar which ^ is in contact with the corner of the step. The velocity of this point must be along the length of the bar, so it must be directed from point A to point P. Also, the component of velocity of point A along the length of the bar must be the same as the component of velocity of point P along the length of the bar, so we have ! , A = vo cos er v ^ so that ,P = ,A + ,  , !!! v v !! !! , p , , A = v0 sin e vv ^ is the component of velocity of point P normal to the bar. Then, if we calculate ! , =  l csc ^r e 2 and write , = _k !^ ! we nd that Eqn 2-83 becomes l^ ^ v0 sin e = , _k  2 sin er ^ where the minus sign in front of _ denotes the fact that an angular velocity in ^ the positive k direction decreases the angle : We then conclude that _ = , 2v0 sin2 l 3 is the angular velocity of the bar, determined from part a, and ! , = , l cos ^ + l sin ^ r i2 j 2 is the position vector from point C to point P. Inserting these quantities into the above equation gives the velocity of the center of the bar as ! , c = v0 1 , sin3 ^ , sin2 cos ^ v i j 3. We use Eqn. 2-106 in Greenwood to nd the absolute acceleration of point P, which is in motion relative to the moving coordinate system r; z. h i b The second derivative   is found by simply di erentiating the result of part a, 2 2v0 2 sin cos _ = 8v0 sin3 cos   = , l l2 c The velocity of the center of mass is given by Equation 2-83 from Greenwood, !! !! , =,p+, , vv!r !! where, for our case, , = , c is the absolute velcoity of the center of the bar, vv i.e., what we want to nd. In addition, ! , p = v0^ v i is the velocity of point P, the point of contact between the bar and the ground, ! , = 2v0 sin2 k ^ ! l z O 30m y O’ ez x 10m/s eφ er a Top view 4 P vertical .8m 15 ez er O’ b Rear view of bike a This equation is: ! !!!!! !! ! ! , = , + ,  , + ,  ,  ,  + 2,  , r + , r aR! _ !! ! _ We shall explain and determine each of these terms in turn. ! b , is the absolute acceleration of point P, the tack at the highest point of the a wheels rotation. This is what we want to nd. ! , c R is the absolute acceleration of point O', relative to the inertial frame OXYZ. We have !^ , = Rer = 30^r R e ! , _ ! , d^ R = 30 dt er = 30 _ e  ^ But _ is the angular velocity of O' about O, which is for a circular motion of the bike about the origin _ = j!j = v = 10m=s = :33rad=s R 30m 5 d^ R = 30 e + 30 _ dt e = 30 e , 30 _ 2er ^ ^ ^ Then, since ! = _ is constant, = 0; and hence ! , R = ,3:33m=s2^r e We note that this term is essentially the centripetal acceleration of a point at O', due to the velocity of the bike around the track. ! d The second term on the right hand side is zero, since , = = 0 ! _ e The third term on the right hand side is the centripetal acceleration of the tack due to the angular velocity of O' about O. Thus, we use the angular !^ velocity , = _ ez = :33rad=s^z , and the position vector of the tack relative ! e to O', = :8mcos 15^z , sin 15^r  e e and, calculating the cross product, we obtain ! !! ,  ,  ,  = :023m=s2^r !! e Comparing this term to the term from part b, we note that both terms are centripetal acceleration terms due to the velocity of the bike around the circle of radius 30m. Whereas the term from part b was the centripetal acceleration of point O', this term is an adjustment accounting for the fact that the tack is actually closer to point O than 30m due to the inward banking of the bike. f The fourth term in the right hand side of the equation above represents the ! Coriolis acceleration, and depends on the term , r , which is the velocity of _ the tack as seen by an observer at point O'. Since the tack is at the top of the wheel, its velocity is completely in the tangential or e direction, and is given ^ by ! , r = 10m=s^ _ e and thus the fourth term can be found as ! 2!  , r = ,6:67m=s2^r _ e The Coriolis term results from the motion of a point in a rotating coordinate system. ! g The nal term is , r , which is the acceleration of the tack seen by an observer rotating with point O'. Since the wheel is rotating about its center, 6 this acceleration is simply a centripetal acceleration, due to the rotation of the wheel. This term is di erent from the centripetal acceleration terms in parts b and d, as those terms involved the velocity around the track, while this term involves the velocity around the wheel. This centripetal acceleration is directed towards the center of the wheel, and is given by 2 ! e e , r = vtack , cos 15^z + sin 15^r  r 2 = 10:m=s , cos 15^z + sin 15^r  e e 4m = ,241:48^z + 64:70^r m=s2 e e Where we have used the fact that the velocity of a point on the wheel is the same as the velocity of the bike around the track. Putting all of these terms together, we obtain the nal result ! , = 54:73^r , 241:5^z m=s2 a e e 4. For this problem, we use equations 2-83 and 2-106 from Greenwood φ Ω O P2 θ P1 VP = VP 1 O’ P 2 a To nd the angular rate _ ; we rst note that since the two wheels roll against each other without slipping, the velocities at their point of contact must be identical, i.e. if P1 is the point on wheel 1 which is in contact with wheel 2, and P2 is the point on wheel 2 which is in contact with wheel 1, then !! , P1 = , P2 v v 7 First, we note that the velocity of point O is ! , 0 = r^ v i Then, since the angular velocity of wheel 1 is ! , 1 = , k; ^ ! then the velocity of point P1 is given by 1 ! where , 1 is the position vector of point P1 relative to point 0, i.e. ! , 1 = rcos j + sin ^ ^ i Then, nally, the velcoity of point P1 is ! , P1 = r^ + r, sin ^ + cos ^ v i j i We can also nd the velocity of point O', using the same formula, but with the position vector being the position vector from O to O', ! , 2 = 2rcos ^ + sin ^ j i and the angular velocity vector being the angular velocity of the bar, ! ! !! ,P = ,0 +, ,1 v v! ! , 2 = ,!k ^ ! Then the velocity of point O is given by 0 0 ! ,O = ,O +,2, !!! V V ! 2 ^ + 2!r, sin ^ + cos ^ =ri j i 0 Finally, we apply this formula , last time to nd the velocity of point P2 on one ^ the second wheel. This time, !O is the known velocity, , _ k is the angular V velocity of the second wheel, and the position vector is the vector from O' to P2. Thus, ! , P = , O + , _ k  ,r sin ^ , r cos ^ ! ^ V V i j = r ^ +  _ , 2!r sin ^ , r cos ^ i j i 2 0 8 Then, as we said earlier, due to the no-slip condition, the velocities of points P1 and P2 are the same, so equating these two velocities, we obtain r ^ , r sin ^ , r cos ^ = r ^ +  _ , 2!r sin ^ , r cos ^ i j i i j i from which we obtain the nal result _ = 2! , b To solve this part of the problem, we again use equation 2-106 ! !!!!! !! ! ! , = , + ,  , + ,  ,  ,  + 2,  , r + , r aR! _ !! ! _ In this case, point O is located at the center of the lower wheel rotating at rate ; and point O' is located at the center of the second wheel. The vector ! , is the vector from point O to point O', and the vector , is the vector from ! R point O' to the point P whose acceleration we want to calculate. The rotating coordinate system at O' is such that one axis is always in the direction from O' to P, so the angular rate of the coordinate system at O' is the same as the angular rate calculated in part a, i.e. ! , = ,2! , k ^ ! In this case, since the, position of point P relative to the ! coordinate system at ! O does not change, !r and , r are both zero, and , = 0; and equation _ ! _ 2-106 becomes ! The position vector , is given by ! , = rcos ^ + sin ^ j i Performing the cross product, we end up with the result ! , = ,2r!2 sin , r2! , 2 sin ^ + ,2r!2 cos , r2! , 2 cos ^ a i j h i h i ! The position vector , is given by R ! , = 2rcos ^ + sin ^; R j i Then, if we di erentiate twice, and use the fact that = ! = 0; we obtain _ ! , R = 2r,!2 cos ^ , !2 sin ^ j i !!! !! , = , + ,  ,  ,  aR! ! 9 ...
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This document was uploaded on 04/25/2011.

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