Physics 2 LT2: Vectors - Rotational Motion
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Physics 2 LT2: Vectors - Rotational Motion

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Physics 2 LT2: Vectors - Rotational Motion Physics 2 LT2: Vectors - Rotational Motion Document Transcript

  • 566547031750Questionnaire No.:_________00Questionnaire No.:_________8382000PHILIPPINE SCIENCE HIGH SCHOOL – Main Campus<br />Physics 2 – Advanced Topics in Physics I<br />1ST QUARTER - LONG TEST 2: VECTORS-ROTATIONAL KINEMATICS<br />
    • CONCEPTUAL QUESTIONS
    Direction:<br />Analyze carefully each question and determine the correct answer. Write your answer in your answer sheet.<br />
    • (2 points) The magnitudes of two vectors are 12 units and 8 units.
    • What is the largest possible value for the magnitude of the resultant vector?
    • What is the smallest possible value for the magnitude of the resultant vector?
    • (1 points) If A=5.00 m, 15o south of east, what is the negative (opposite) of that vector?
    • (1 points) True or False: Component method is a solution of vector addition which is less accurate than graphical method.
    • (2 points) If the x and y components are positive and negative respectively, the vector is in the direction between ______ and _____. (Choices: east, north, west, south)
    • (1 points) What is the minimum number of unequal vectors (different magnitude and direction) to have a resultant equal to zero?
    (Choices for 07-10: twice, the same as, half of)<br /> (4 points) Harry and Hermione are riding on a merry-go-round. Harry rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Hermione, who rides on an inner horse. The merry-go-round is rotating at a constant angular speed.<br />
    • Angular displacement of Harry is ____ Hermione’s.
    • Angular speed of Harry is ____ Hermione’s.
    • Tangential speed of Harry is ____ Hermione’s.
    • Centripetal acceleration of Harry is _____ Hermione’s.
    True or False:(4 points) An object moves rotates with constant tangential speed. <br />
    • There is an angular velocity and is changing.
    • There is a centripetal acceleration and varies depends on the location from the axis of rotation.
    • There is a tangential acceleration and is constant.
    • There is no angular acceleration.
    • PROBLEM SOLVING
    Direction:<br />Solve all the problems completely and correctly in your answer sheet (Follow the format: Illustration: -> Given:-> Find: -> Derivation: -> Substitution: -> Final Answer: Box your final equation and answer. Follow significant digits rules.<br />
    • (10 points) While exploring a cave, a spelunker starts at the entrance and moves the following distances: 75.0 m, north; 250 m, east, 125 m, 34o south of east, and 180 m, 71o west of north. He then finds the cave. (a) Find the components of these three vectors. (b) What must be his final displacement so that he can return immediately to the entrance without using his previous displacements?
    • (7 points) A grindstone of radius 4.0 m is spinning uniformly (constant angular speed) with an angular speed of 8.0 rad/s. After 5.0 s, (a) what is its angular acceleration in rev/min; (b) what is its tangential speed; (c) what is its centripetal acceleration from the edge to the center (axis of rotation); (d) what is its tangential acceleration; and (e) how many revolutions does it make?
    • (8 points) On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are uniformly accelerated for 21 s to a maximum frequency of 14 rpm. (a) How many revolutions does the ride turn during the 21-s interval? (b) What is the tangential acceleration of the passengers? (c) What is the tangential speed of the passengers 15 s after the acceleration begins?
    GOD Bless!!! John 3:16<br />Tract: sirdq_07/21/2011_2020h<br />PHILIPPINE SCIENCE HIGH SCHOOL – Main Campus<br />Physics 2 – Advanced Topics in Physics I<br />1ST QUARTER - LONG TEST 2: VECTORS-ROTATIONAL KINEMATICS<br />ANSWER KEY<br />
    • CONCEPTUAL QUESTIONS
    • The magnitudes of two vectors are 12 units and 8 units.
    • What is the largest possible value for the magnitude of the resultant vector? - 20 units
    • What is the smallest possible value for the magnitude of the resultant vector? – 4 units
    • If A=5.00 m, 15o south of east, what is the negative (opposite) of that vector? - -A=5.00 m, 15o north of west
    • True or False: Component method is a solution of vector addition which is less accurate than graphical method. – False
    • If the x and y components are positive and negative respectively, the vector is in the direction between ______ and _____. – east & south
    • What is the minimum number of unequal vectors (different magnitude and direction) to have a resultant equal to zero? - 3
    Harry and Hermione are riding on a merry-go-round. Harry rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Hermione, who rides on an inner horse. The merry-go-round is rotating at a constant angular speed.<br />
    • Angular displacement of Harry is the same as Hermione’s.
    • Angular speed of Harry is half of Hermione’s.
    • Tangential speed of Harry is double Hermione’s.
    • Centripetal acceleration of Harry is double Hermione’s.
    True or False:An object rotates with constant tangential speed. <br />
    • There is an angular velocity and is changing. - False
    • There is a centripetal acceleration and varies depends on the location from the axis of rotation. - True
    • There is a tangential acceleration and is constant. - False
    • There is no angular acceleration. – True
    • PROBLEM SOLVING
    • While exploring a cave, a spelunker starts at the entrance and moves the following distances: 75.0 m, north; 250 m, east, 125 m, 34o south of east, and 180 m, 71o west of north. He then finds the gold. (a) Find the components of each vector. (b) What must be his final displacement so that he can return immediately to the entrance without using his previous displacements?
    Illustration:<br />Given: LetA=75.0 m, north (90o)B=250 m, east (0o)<br />
    • C=125 m, 34o south of east (326o)D=180 m, 71o west of north (161o)
    a. Find:components of each vector (Ax, Ay, Bx, By, Cx, Cy, Dx, Dy)<br />Derivation:for x-component: Ax=AcosθA. For y-component: Ay=AsinθA where θ is in polar angle.<br />Substitution:Ax=75.0 mcos90o Ax=0.00 mAy=75.0 msin90oAy=75.0 m<br />Bx=250 mcos0oBx=250 mBy=250 msin0oBy=0.0 m<br />Cx=125 mcos326oCx=104 mCy=125 msin326oCy=-69.9 m<br />Dx=180 mcos161oDx=-170 mDy=180 msin161oDy=59 m<br />b. Find: -R=?- opposite of the resultant vector so that he can return back without using his previous displacements. <br />Derivation:-Rx=-Ax+Bx+Cx+Dx-Ry=-Ay+By+Cy+Dy<br />
    • R=-Rx2+-Ry2θ=tan-1-Ry-Rx
    Substitution:-Rx=-250 m+104 m-170 m-Rx=-184 m<br />
    • -Ry=-75.0 m-69.9 m+59 m-Ry=-64.1 m
    • R=-184 m2+-64.1 m2R=195 m
    • θ=tan-1-64.1 m184 mθ=19o∴ R=195 m, 19o south of west (199o)
    • A grindstone of radius 4.0 m is spinning uniformly (constant angular speed) with an angular speed of 8.0 rad/s. After 5.0 s, (a) what is its angular acceleration in rev/min2; (b) what is its tangential speed; (c) what is its centripetal acceleration from the edge to the center (axis of rotation); (d) what is its tangential acceleration; and (e) how many revolutions does it make?
    Illustration:<br />Given: R=4.0 m- radius<br />ω=8.0 rad/s- constant angular speed<br />t=5.0 s- time interval<br />a) Find:α=?- angular acceleration in rev/min2<br />Solution: since it is in constant angular speed, α=0.0 rev/min2 <br />b) Find: v=?- tangential speed (constant since it is in uniform rotation)<br />Derivation:v=ωR<br />Substitution:v=8.0 rad/s4.0 mv=32 m/s<br />c) Find: ac=?- centripetal acceleration from the edge to the center<br />Derivation:ac=ω2R<br />Substitution:ac=8.0 rad/s24.0 mv=260 m/s2<br />d) Find:at=?- tangential acceleration <br />Solution: since it is in constant angular speed, at=0.0 m/s2 <br />e) Find: θ=?- angular displacement (no. of revolutions) after 5.0 s<br />Derivation:θ=ωt<br />Substitution:θ=8.0 rad/s5.0 s1 rev2π radθ=6.4 rev<br />
    • On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are uniformly accelerated for 21 s to a maximum frequency of 14 rpm. (a) How many revolutions does the ride turn during the 21-s interval? (b) What is the tangential acceleration of the passengers? (c) What is the tangential speed of the passengers 15 s after the acceleration begins?
    Illustration:<br />Given: R=7.5 m- radius<br />ωi=0 rpm- initial angular speed<br />t=21 s- time interval from ωi=0 to ωf=14 rpm<br />t1=15 s- time interval from ωi=0 to another ωf=?<br />a) Find:θ=?- angular displacement (no. of revolutions during t)<br />Derivation:θ=ωi+ωf2tθ=12ωft<br />Substitution:θ=1214 rpm21 s1 min60 sθ=2.5 rev<br />b) Find:at=?- tangential acceleration (uniform)<br />Derivation:ωf-ωi=αt becomes α=ωft and at=αR combining, at=ωfRt<br />Substitution:at=14 rpm7.5 m21 s2π1 rev1 min60 sat=0.52 m/s2<br />c) Find:vf=?- tangential speed after t1=15 s<br />Derivation:vf-vi=att1 becomes vf=att1 since the time is in the uniform acceleration, at=ωfRt <br />combining, vf=ωfRt1t<br />Substitution:vt=14 rpm7.5 m15 s21 s2π1 rev1 min60 svf=7.9 m/s<br />Tract: sirdq_07/23/2011_1737h<br />