University Physics Al No. 2 Motion in two and Three dimensions Class Number ame I. Choose the correct answer An object moves in the xy plane with an acceleration that has a positive x component. At time t=0 the object has a velocity given by v=3i+0j. What can be concluded about the y component of the acceleration? (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 dt Solution: According to the general equations a 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. Which of the following expression are also ( B d(v2) / (B) Solution: According to the general equation a=-, we know the velocity 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 a=h=constant, Answer(B)is the right answer dt 3. An object is launched into the air with an initial velocity given by vo=(4.9i+9.8])m/s Ignore air resistance. At the highest point the magnitude of the velocity (B) (A)0.(B)√495ms.(C)√98ms.(D)√(49)2+(98)ms 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 v=(4.9i)m/s, the magnitude is 4.9m/s
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 t = 0 the object has a velocity given by v i j = 3ˆ + 0 ˆ 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 v v d d( / ) r r . 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 ˆ 9. 9ˆ (4. 0 v = i + j r . Ignore air resistance. At the highest point the magnitude of the velocity is ( B ) (A) 0. (B) 4.9 m/s 2 . (C) 9.8 m/s 2 . (D) (4.9) (9.8) m/s 2 2 + . 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 9ˆ v = (4. i r , the magnitude is 4.9m/s
I. Filling the blanks a gypsy moth caterpillar(Porthetria dispar) inches along a crooked branch to a tasty oak leaf, wriggling 15 Tree trunk ty cm horizontally and then 30 cm along a section of the branch inclined at 30 to the horizontal as shown in Figure 1.(a) The initial position vectors -0 15i and the final Branch position vectors 0.153i +0.15 of the caterpillar in Fig 1 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 0.0068i+0.0025 j(m/s), and the magnitude of its average velocity is 724×10-3m/s Solution (a)The coordinate system are shown in figure. The initial position vectors is P=-0 15i, and the final position vectors is F=0.3cos 30 i+0.3sin 30 j=0. 26i+0.15j s0.15+0.3 (b) Its average speed is v 60=0.0075m/s Its average velocity isvave-A A 10.261+015/+0.15i=000 0.0025jm/s The magnitude of its average velocity is m11008+0010090572103(ms) 2. Use Equation 4.28 for r(o: r(o=[rose(Oli+[rsin e(ol and Equation 4.26 for (t) o(0=@(k to evaluate the expression r(o)[o(txr(o=0 Explain why this result is another way to see that v() is always perpendicular to r(t for circular motion: solution(2). Solution: (O(0xF(=0, (kxfrcos e(o)+(rsin(li=rcos 0(), (0j-rsin 0(), (02 →F()-[o(x-cosi+rsno小 rose(o2(0)- rsin g(n)a2( r A(cos0(o(0) sin e(ocos 6(Do(0=0
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 30o to the horizontal as shown in Figure 1. (a) The initial position vectors i 15ˆ − 0. and the final position vectors i j 15 ˆ 0. 15 3ˆ 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 ˆ 0. 0068ˆ 0. i + j , and the magnitude of its average velocity is 7.24 10 m/s −3 × . Solution: (a) The coordinate system are shown in figure. The initial position vectors is r i 15ˆ = −0. v , and the final position vectors is r i j i j f 15 ˆ 0. 26ˆ 0. 30 ˆ = 0.3cos30 ˆ + 0.3sin = + v o o . (b) Its average speed is 0.0075 m/s 60 0.15 0.3 = + = = t s vave . Its average velocity is i j i j i t r r t r v f i ave 0025 ˆ 0. 0068ˆ 0. 60 15ˆ 0. 15 ˆ 0. 26ˆ 0. = + + + = ∆ − = ∆ ∆ = v v v v m/s. The magnitude of its average velocity is 0.0068 0.0025 7.24 10 (m/s) 0025 ˆ 0. 0068ˆ 0. 2 2 −3 vave = i + j = + = × v 2. Use Equation 4.28 for r(t) r : r t r t i r t j ˆ [ sin ( )] ˆ ( ) = [ cosθ ( )] + θ r and Equation 4.26 for ω(t) r : t t k z ˆ ω( ) = ω ( ) r to evaluate the expression r(t)⋅[ (t)× r(t)] = r r r ω 0 . Explain why this result is another way to see that v(t) r is always perpendicular to r(t) r for circular motion: solution(2) . Solution: (1) t r t t k { } r t i r t j r t t j r t t i z z z ˆ sin ( ) ( ) ˆ cos ( ) ( ) ˆ [ sin ( )] ˆ [ cos ( )] ˆ ω( ) × ( ) = ω ( ) × θ + θ = θ ω − θ ω v r [ ] [ ] [ ] sin ( ) cos ( ) ( ) sin ( ) cos ( ) ( ) 0 ˆ sin ( ) ( ) ˆ cos ( ) ( ) ˆ sin ( ) ˆ ( ) ( ) ( ) cos ( ) 2 2 = − = ⇒ ⋅ × = + ⋅ − r t t t r t t t r t t r t r t i r t j r t t j r t t i z z z z θ θ ω θ θ ω ω θ θ θ ω θ ω v v r 15cm 30cm Branch Tree trunk Fig.1 o 30 x y 0
(2)According to o(oxr(t=v( and r(r. o(xr(o=0, we know r(0)v(0=0 that is v(t is always perpendicular to r(o) 3. The angle turned through by the flywheel of a generator during a time interval I is given by 0=at+br'-cr, where a, b, and c are constant. Its angular velocity is a+3b12-4cr.And its angular acceleration is 6bt-12ct Solution de Its angular velocity is a at+br'-ct*)=a+ 3bt-4ct And its angular acceleration is B-do d(a+ 3br2-4c1)=6bt-12ct2 4. The flywheel of an engine I rotating at 25.2 rad/. When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after 197s. The angular acceleration(in rad/s2)of the flywheel is 12.8 rad/s2, 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 @(0=@o+at, the angular acceleration is 25.2 =-128rad/ (2) According to the equation 0()=0+o+a, the change angle is △=t+-at2=252×197-×128×19.72=248rad (3)The number of revolutions is \.1 39.5 2x2×3.14 5 A car is traveling around a banked, circular curve of radius 150 m on a test track. At the instant when f=Os, 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 FOs. The angular speed of the car when fOs is 0.2rad/s, the angular speed 5.0 s later is _ 0. 15rad/s, the magnitude of the centripetal acceleration
(2) According to (t) r(t) v(t) r r v ω × = and r(t)⋅[ (t)× r(t)] = 0 r r r ω , we know r(t)⋅ v(t) = 0 r v , that is v(t) r is always perpendicular to r(t) r . 3. The angle turned through by the flywheel of a generator during a time interval t is given by 3 4 θ = at + bt − ct , where a, b, and c are constant. Its angular velocity is 2 3 a + 3bt − 4ct . And its angular acceleration is 2 6bt −12ct . Solution: Its angular velocity is 3 4 2 3 ( ) 3 4 d d d d at bt ct a bt ct t t = = + − = + − θ ω And its angular acceleration is 2 3 2 ( 3 4 ) 6 12 d d d d a bt ct bt ct t t = = + − = − 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/s2 ) of the flywheel is 12.8 rad/s2 , 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 ω t = ω +αt 0 ( ) , the angular acceleration is 0 2 1.28 rad/s 19.7 25.2 = − = − = − t ω α (2) According to the equation 2 0 2 1 (t) t t θ = θ +ωo + α , the change angle is 1.28 19.7 248.1 rad 2 1 25.2 19.7 2 1 2 2 ∆θ = ωot + αt = × − × × = (3) The number of revolutions is 39.5 2 3.14 248.1 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 fOs is 60m/s-, the magnitude of the centripetal acceleration of the car when f500s is_3.375m/- the magnitude of the angular acceleration is_0.0lrad/s-, the magnitude of the tangential acceleration is 1.5m/s olutions ()The angular speed of the car when os 00-30 0.2 rad/ r150 (2) The angular speed 5.0 s later is a==Oo=0.15 rad/s (3)The magnitude of the centripetal acceleration of the car when fOs is 302 60m/s2 150 (4)The magnitude of the centripetal acceleration of the car when f500s is a.=02r=0.152×150=3.375m/s (5)According to the equation 4.49 @(0)=@o+at, the angular acceleration is 0.15-0.2 0.01rad/s2 150×001=1.5m/s2 ched at speed vo at an angle 0(with the horizontal) from the bottom of a hill of constant slope B as Figure 2. The Osin (0-B) FI g cos B The coordinate system is shown in figure. Assume the range is r;, the time is t, we have B os 0t sin B=vo sin At Solve the 2vo cos8sin(8-B)
of the car when t=0s is 6.0m/s2 , the magnitude of the centripetal acceleration of the car when t=5.00s is 3.375m/s2 , the magnitude of the angular acceleration is 0.01rad/s2 , the magnitude of the tangential acceleration is 1.5m/s2 . Solution: (1) The angular speed of the car when t=0s is 0.2 rad/s 150 0 30 0 = = = r v ω (2) The angular speed 5.0 s later is 0.15 rad/s 4 3 ω = ω0 = (3) The magnitude of the centripetal acceleration of the car when t=0s is 2 2 2 6.0 m/s 150 30 = = = r v ac (4) The magnitude of the centripetal acceleration of the car when t=5.00s is 2 2 2 a = r = 0.15 ×150 = 3.375 m/s c ω (5) According to the equation 4.49 ω t = ω +αt 0 ( ) , the angular acceleration is 0 2 0.01 rad/s 5 ( ) 0.15 0.2 = − − = − = t ω t ω α (6) The magnitude of the tangential acceleration is 2 = α =150× 0.01 =1.5m/s τ a r . 6. A projectile is launched at speed v0 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 θ θ β 2 2 0 cos 2 cos sin( ) 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 direction : sin sin direction : cos cos y r v t gt x r v t β θ β θ Solve the two equations, we have β θ θ β 2 2 0 cos 2 cos sin( ) g v r − = 0 v r θ β Range Fig.2 y x
I. Give the solutions of the following problems 1. A particle leaves the origin at t=0 with an initial velocity vo=(3.6m/s)i. It experiences a constant acceleration a=(1. 2m/s )i-(1. 4m/s ).(a)At what time does the particle reach its maximum x coordinate? (b)What is the velocity of the particle at this time?()Where is the particle at this time? Solution:(a) According to the problem, we have a, =-1. 2m/s. and vo =3.6m/s.Then the position of the particle is x= -Xmaxdi-0, thus we get =3s, the particle reach its maximum x coordinate x=5.4. dy (b )Using a= dr ,we have dv= adt →节=0+(-1.21-14/dr=361+(-1.2n-141)=(36-1.2)-14 When t=3s, the velocity of the particle is v=4.2(m/s) (c) Using v=dr, we ha d →F=36-12)-140d=(36-062)2-072 When f3s the position vector of the particle is F=(361-0612)-0.72j=541-63j(m) 2. A young basketball player is attempting to make a shot. the ball leaves the hand 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 2.0m the hoop as indicated in Figure 3 loud cheers, calculate the speed at which Fig 3
III. Give the Solutions of the Following Problems 1. A particle leaves the origin at t = 0 with an initial velocity v i 6m/s)ˆ (3. 0 = r . It experiences a constant acceleration a i j ˆ (1.4m/s ) ˆ ( 1.2m/s ) 2 2 = − − 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 = −1.2m/s ax and 3.6m/s 0 =x v . Then the position of the particle is 3.6 0.6 0.6( 3) 5.4 2 1 2 2 2 x = v0xt + ax t = t − t = − t − + , when max x = x , 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 v v i j t i ti tj t i tj v a t t v t v 4 ˆ 1. ˆ ) (3.6 1.2 ) 4 ˆ 1. 2 ˆ ( 1. 6ˆ )d 3. 4 ˆ 1. ˆ (-1.2 d d 0 0 0 0 ⇒ = + − = + − − = − − = ∫ ∫ ∫ v v v v v When t=3s，the velocity of the particle is (m/s) 2 ˆ v = 4. j v . (c) Using t r v d d v v = , we have r t i tj t t t i t j r v t t r t 7 ˆ 0. ˆ ]d (3.6 0.6 ) 4 ˆ 1. ˆ [(3.6 -1.2 ) d d 2 2 0 0 0 ⇒ = − = − − = ∫ ∫ ∫ v v v v When t=3s the position vector of the particle is (m) 3 ˆ 6. 4ˆ 5. 7 ˆ 0. ˆ (3.6 0.6 ) 2 2 r = t − t i − t j = i − j 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 Establish the Cartesian coordinate system given in figure. According to the problem, we have x,=0m, yo =2 m, x=8m, y=3 m a, =-g Using the general equations x=xo+vo y=y Then 8=voc0s60°1(1) Solving equation (1)and (2), we have the speed at which the ball left the hands of the player is 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 North 45° From the problem, we have vnG=100c0s4521+100sm45°j -700cos 0i+700sin 6 →-70c0+700iO)=vmo3-100c0s451-100n45
the ball left the hands of the player. Solution: Establish the Cartesian coordinate system given in figure. According to the problem, we have x0 = 0 m, y0 = 2 m, x = 8 m, y = 3 m ay = −g .Using the general equations ⎪ ⎩ ⎪ ⎨ ⎧ = + + = + 2 0 0 0 0 2 1 y y v t a t x x v t y y x Then ⎪ ⎩ ⎪ ⎨ ⎧ = + ° − = ° (2) 2 1 3 2 sin 60 8 cos60 (1) 2 0 0 v t gt v t Solving equation (1) and (2), we have the speed at which the ball left the hands of the player is v0=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 PW PG WG v v v r r r = − From the problem, we have ⎪ ⎩ ⎪ ⎨ ⎧ = − + = = ° + ° v i j v v j v i j PW PG PG WG ˆ 700cos ˆ 700sin ˆ 45 ˆ 100cos45 ˆ 100sin θ θ r r r i j v j i j PG 45 ˆ 700cos ˆ 700sin ˆ ˆ 100cos45 ˆ 100sin o o ⇒ − θ + θ = − − y x 0 45° PG v v v PW v WG v v θ North East
700cos6=-100cos45° 17005i=-100545° 8=arccos()=arccos=84.2 vma=700sin6+100sin45°=767.1m/s 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. Im/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- 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 @, a, r and v ( 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 an= a cOS40°=15cos40°=1.5(m/s2) (b)Using We have the speed of the particle at this instant v=var=5.87(m/s) (c)The angular speed of the particle at this instant is @===1-C=1.96 rad/s (d) The magnitude of the tangential acceleration at this instant a1= a sin40°=15sin40°=96(m/s2) (e)Using d,=a×F,a1=ar We have the magnitude of the angular acceleration at this instant a=t=3.2(rad/s2) (f The directions of a, a, r and v shown in figure (g) Because the angle between a and v is 60%, it is less than T/2, so the particle is speeding
⎩ ⎨ ⎧ = − − = − ⇒ o o 700sin 100sin 45 700cos 100cos45 PG θ v θ ⎪ ⎩ ⎪ ⎨ ⎧ = + = = = = ⇒ 700sin 100sin 45 767.1m/s 84.2 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/s2 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 ac = atotal cos40° =15cos40° =11.5 (m/s2 ) (b) Using r v ac 2 = , We have the speed of the particle at this instant v = ac r = 5.87 (m/s) (c) The angular speed of the particle at this instant is = = = 1.96 r a r v c ω rad/s (d) The magnitude of the tangential acceleration at this instant is at = atotalsin 40° =15sin 40° = 9.6 (m/s2 ) (e) Using a r a r t = α × , t = α r r r We have the magnitude of the angular acceleration at this instant = = 3.2 r at α (rad/s2 ) (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 π / 2 , so the particle is speeding up. 40.0° atotal r Fig.4 v r v v a v ω v