University Physics AI No. 2 Motion in Two and Three Dimensions Class Number Name I．Choose the Correct Answer 1. An object moves in the xy plane with an acceleration that has a positive x component. At time 0=t the object has a velocity given by jiv ? 0 ? 3 += r . What can be concluded about the y component of the acceleration? ( D ) (A) The y component must be positive and constant. (B) The y component must be negative and constant. (C) The y component must be zero. (D) Nothing at all can be concluded about the y component. Solution: According to the general equations t v a x x d d = and t v a y y d d = , we just know the initial velocity of the object, so nothing at all can be concluded about the y component. 2. An object moves with a constant acceleration a r . Which of the following expression are also constant? ( B ) (A) t v d d r . (B) t v d d r . (C) t v d )d( 2 . (D) t vv d )/d( rr . Solution: According to the general equation t v a d d v v = , we know the velocity v v is vary with time. So the magnitude and the direction of the velocity may not constant. so (A), (C) and (D) are not corrected. As constant d d == t v a v v , Answer (B) is the right answer.. 3. An object is launched into the air with an initial velocity given by m/s) ? 8.9 ? 9.4( 0 jiv += r . Ignore air resistance. At the highest point the magnitude of the velocity is ( B ) (A) 0. (B) m/s9.4 2 . (C) m/s8.9 2 . (D) m/s)8.9()9.4( 22 + . Solution: The y component of the velocity is zero at the highest point, and the x component of the velocity is 4.9m/s, thus the velocity is m/s) ? 9.4( iv = r , the magnitude is 4.9m/s. II. Filling the Blanks 1. A gypsy moth caterpillar (Porthetria dispar) inches along a crooked branch to a tasty oak leaf, wriggling 15 cm horizontally and then 30 cm along a section of the branch inclined at 30 o to the horizontal as shown in Figure 1. (a) The initial position vectors i ? 15.0? and the final position vectors ji ? 15.0 ? 315.0 + of the caterpillar in SI units using the crook in the branch as the origin for a set of horizontal and vertical coordinate axes. (b) If the caterpillar traverses the distance during 1.0 min, its average speed is 0.0075m/s , its average velocity is )m/s( ? 0025.0 ? 0068.0 ji + , and the magnitude of its average velocity is m/s1024.7 3? × . Solution: (a) The coordinate system are shown in figure. The initial position vectors is ir ? 15.0?= v , and the final position vectors is jijir f ? 15.0 ? 26.0 ? 30sin3.0 ? 30cos3.0 +=+= oo v . (b) Its average speed is m/s0075.0 60 3.015.0 = + == t s v ave . Its average velocity is ji iji t rr t r v if ave ? 0025.0 ? 0068.0 60 ? 15.0 ? 15.0 ? 26.0 += ++ = ? ? = ? ? = vv v v m/s. The magnitude of its average velocity is m/s)(1024.70025.00068.0 ? 0025.0 ? 0068.0 322 ? ×=+=+= jiv ave v 2. Use Equation 4.28 for )(tr r : jtritrtr ? )](sin[ ? )](cos[)( θθ += r and Equation 4.26 for )(tω r : ktt z ? )()( ωω = r to evaluate the expression =×? )]()([)( trttr rrr ω 0 . Explain why this result is another way to see that )(tv r is always perpendicular to )(tr r for circular motion: solution(2) . Solution: (1) {} ittrjttrjtritrkttrt zzz ? )()(sin ? )()(cos ? )](sin[ ? )](cos[ ? )()()( ωθωθθθωω ?=+×=× rv [][ ] [ ] 0)()(cos)(sin)()(cos)(sin ? )()(sin ? )()(cos ? )(sin ? )(cos)()()( 22 =?= ??+=×?? tttrtttr ittrjttrjtritrtrttr zz zz ωθθωθθ ωθωθθθω rvv 15cm 30cm Branch Tree trunk Fig.1 o 30 x y 0 (2) According to )()()( tvtrt vrr =×ω and 0)]()([)( =×? trttr rrr ω , we know 0)()( =? tvtr vr , that is )(tv r is always perpendicular to )(tr r . 3. The angle turned through by the flywheel of a generator during a time interval t is given by 43 ctbtat ?+=θ , where a, b, and c are constant. Its angular velocity is 32 43 ctbta ?+ . And its angular acceleration is 2 126 ctbt ? . Solution: Its angular velocity is 3243 43)( d d d d ctbtactbtat tt ?+=?+== θ ω And its angular acceleration is 232 126)43( d d d d ctbtctbta tt ?=?+== ω β 4. The flywheel of an engine I rotating at 25.2 rad/s. When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after 19.7s. The angular acceleration (in rad/s 2 ) of the flywheel is 12.8 rad/s 2 , the angle (in rad) through which the flywheel rotates in coming to rest is 248.1 rad , and the number of revolutions made by the flywheel in coming to rest is 39.5 . Solution: (1) According to the equation 4.49 tt αωω += 0 )( , the angular acceleration is 20 rad/s28.1 7.19 2.25 ?=?=?= t ω α (2) According to the equation 2 0 2 1 )( ttt o αωθθ ++= , the change angle is rad1.2487.1928.1 2 1 7.192.25 2 1 22 =××?×=+=? tt o αωθ (3) The number of revolutions is 5.39 14.32 1.248 2 = × = ? π θ 5 A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when t=0s, the car is moving north, and its speed is 30.0 m/s but decreasing uniformly, so that after 5.0 s its angular speed will be 3/4 that it was when t=0s. The angular speed of the car when t=0s is 0.2rad/s , the angular speed 5.0 s later is 0.15rad/s , the magnitude of the centripetal acceleration of the car when t=0s is 6.0m/s 2 , the magnitude of the centripetal acceleration of the car when t=5.00s is 3.375m/s 2 , the magnitude of the angular acceleration is 0.01rad/s 2 , the magnitude of the tangential acceleration is 1.5m/s 2 . Solution: (1) The angular speed of the car when t=0s is rad/s2.0 150 30 0 0 === r v ω (2) The angular speed 5.0 s later is rad/s15.0 4 3 0 == ωω (3) The magnitude of the centripetal acceleration of the car when t=0s is 2 22 m/s0.6 150 30 === r v a c (4) The magnitude of the centripetal acceleration of the car when t=5.00s is 222 m/s375.315015.0 =×== ra c ω (5) According to the equation 4.49 tt αωω += 0 )( , the angular acceleration is 2 0 rad/s01.0 5 2.015.0)( ?= ? = ? = t t ωω α (6) The magnitude of the tangential acceleration is 2 m/s5.101.0150 =×== α τ ra . 6. A projectile is launched at speed v 0 at an angle θ (with the horizontal) from the bottom of a hill of constant slope β as shown in Figure 2. The range of the projectile up the slope is β βθθ 2 2 0 cos )sin(cos2 g v r ? = Solution: The coordinate system is shown in figure. Assume the range is r, the time is t, we have ? ? ? ? ? ?= = 2 0 0 2 1 sinsin:direction coscos:direction gttvry tvrx θβ θβ Solve the two equations, we have β βθθ 2 2 0 cos )sin(cos2 g v r ? = 0 v r β θ Range Fig.2 y x III. Give the Solutions of the Following Problems 1. A particle leaves the origin at 0=t with an initial velocity iv ? m/s)6.3( 0 = r . It experiences a constant acceleration jia ? )m/s4.1( ? )m/s2.1( 22 ??= r . (a) At what time does the particle reach its maximum x coordinate? (b) What is the velocity of the particle at this time? (c) Where is the particle at this time? Solution: (a) According to the problem, we have 2 m/s2.1?= x a and m/s6.3 0 = x v . Then the position of the particle is 4.5)3(6.06.06.3 2 1 222 0 +??=?=+= ttttatvx xx , when max xx = , 0 d d = t x , thus we get t=3s, the particle reach its maximum x coordinate x=5.4. (b) Using t v a d d v v = , we have jtitjtititjivv tav t tv v ? 4.1 ? )2.16.3() ? 4.1 ? 2.1( ? 6.3d) ? 4.1 ? (-1.2 dd 0 0 0 0 ??=??+=?+=? = ∫ ∫∫ vv v v v When t=3s，the velocity of the particle is )m/s( ? 2.4 jv = v . (c) Using t r v d d v v = , we have jtitttjtitr tvr t tr ? 7.0 ? )6.06.3(d] ? 4.1 ? )1.2-[(3.6 dd 22 0 00 ??=?=? = ∫ ∫∫ v vv v When t=3s the position vector of the particle is )m( ? 3.6 ? 4.5 ? 7.0 ? )6.06.3( 22 jijtittr ?=??= v 2. A young basketball player is attempting to make a shot. The ball leaves the hands the player at angle of 60° to the horizontal at an elevation of 2.0 m above the floor. The skillful player makes the shot with the ball traveling precisely through the center of the hoop as indicated in Figure 3. To loud cheers, calculate the speed at which 8.0m 3.0m 2.0m 60° Fig.3 y x the ball left the hands of the player. Solution: Establish the Cartesian coordinate system given in figure. According to the problem, we have gayxyx y ?===== m3,m8,m2,m0 00 .Using the general equations ? ? ? ? ? ++= += 2 00 00 2 1 tatvyy tvxx yy x Then ? ? ? ? ? ?°+= °= )2( 2 1 60sin23 )1(60cos8 2 0 0 gttv tv Solving equation (1) and (2), we have the speed at which the ball left the hands of the player is v 0 =9.88 m/s 3. A pilot flies a plane at a speed of 700 km/h with respect to the surrounding air. A wind is blowing to the northeast at a speed of 100 km/h. In what direction must the plane be aimed so that the resulting direction of the plane is due north? What is the ground speed of the aircraft? Solution: Establish the Cartesian coordinate system given in figure. According to WGPGPW vvv rrr ?= From the problem, we have ? ? ? ? ? +?= = °+°= jiv jvv jiv PW PGPG WG ? sin700 ? cos700 ? ? 45sin100 ? 45cos100 θθ r r r jijvji PG ? 45sin100 ? 45cos100 ?? sin700 ? cos700 oo ??=+?? θθ y x 0 45 ° PG v v PW v v WG v v θ North East ? ? ? ?= ?=? ? o o 45sin100sin700 45cos100cos700 PG vθ θ ? ? ? ? ? =+= === ? m/s1.76745sin100sin700 2.84 14 2 arccos) 14 2 arccos( o o θ θ PG v Thus the direction which the plane must be aimed is West to North 84.2? or North to West 5.8?. The ground speed of the plane is 767.1m/s. 4. A particle moving clockwise in a circle with a radius of 3.00 m has a total acceleration at some instant of magnitude 15.0 m/s 2 directed as indicated in Figure 4. (a) Find the magnitude of the centripetal acceleration at this instant. (b) Find the speed of the particle at this instant. (c) Find the angular speed of the particle at this instant. (d) Determine the magnitude of the tangential acceleration at this instant. (e) Determine the magnitude of the angular acceleration at this instant. (f) Make a sketch indicating the directions of ω r , α r , r r and v r . (g) Is the particle speeding up or slowing down in its circular motion? Solution: (a) The magnitude of the centripetal acceleration at this instant is 5.1140cos1540cos total =°=°= aa c (m/s 2 ) (b) Using r v a c 2 = , We have the speed of the particle at this instant 87.5== rav c (m/s) (c) The angular speed of the particle at this instant is 96.1=== r a r v c ω rad/s (d) The magnitude of the tangential acceleration at this instant is 6.940sin1540sin total =°=°= aa t (m/s 2 ) (e) Using rara tt αα =×= , rrr We have the magnitude of the angular acceleration at this instant 2.3== r a t α (rad/s 2 ) (f) The directions of ω r , α r , r r and v r shown in figure. (g) Because the angle between a v and v v is 60?, it is less than 2/π , so the particle is speeding up. 40.0° total a r Fig.4 v vr v a v ω v