Rotation Quest Answer key PDF

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jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

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∆s = r ∆θ ∆s r= ∆θ 2.06 m = 2.88 rad

001 (part 1 of 2) 10.0 points A record has an angular speed of 25 rev/min. What is its angular speed?

= 0.715278 m .

Correct answer: 2.618 rad/s. Explanation:

keywords:

Let : ω = 25 rev/min .

ω = (25 rev/min)



2 π rad rev



1 min 60 s



= 2.618 rad/s . 002 (part 2 of 2) 10.0 points Through what angle does it rotate in 2.34 s? Correct answer: 6.12611 rad. Explanation:

004 10.0 points What angular acceleration is necessary to increase the angular speed of a fan blade from 8.8 rad/s to 15.6 rad/s in 5 s? Correct answer: 1.36 rad/s2 . Explanation:

Let : ωo = 8.8 rad/s , ωf = 15.6 rad/s ∆t = 5 s .

and

Let : t = 2.34 s .

θ = ωt = (2.618 rad/s) (2.34 s)

ωf − ωo ∆t 15.6 rad/s − 8.8 rad/s = 5s

αavg =

= 6.12611 rad .

= 1.36 rad/s2 . 003 10.0 points A girl sitting on a merry-go-round moves counterclockwise through an arc length of 2.06 m. If the girl’s angular displacement is 2.88 rad, how far is she from the center of the merry-go-round? Correct answer: 0.715278 m. Explanation: Let :

∆s = 2.06 m and ∆θ = 2.88 rad .

keywords: 005 10.0 points A ladybug sits at the outer edge of a merrygo-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. What is the gentleman bug’s angular speed? 1. the same as the ladybug’s correct

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3)

2

From kinematics, 2. twice the ladybug’s 3. half the ladybug’s 4. impossible to determine Explanation: Angular speed is the same for every point on the merry-go-round. 006 10.0 points A tire placed on a balancing machine in a service station starts from rest and turns through 7 rev in 1.46 s before reaching its final angular speed. Find its angular acceleration. Correct answer: 41.2669 rad/s2 . Explanation: Let : ω0 = 0 rev/s , θ = 7 rev , and t = 1.46 s . There is a constant angular acceleration, so 1 1 θ = ω0 t + α t2 = α t2 2 2 2 (7 rev) 2 π rad 2θ · α= 2 = rev t (1.46 s)2 = 41.2669 rad/s2 .

ωf = ωi + α ∆t ωf − ωi α= ∆t 13.1 rad/s − 0.5 rad/s = 4.3 s = 2.93023 rad/s2 .

008 10.0 points A coin with a diameter of 1.13 cm is dropped on its edge onto a horizontal surface. It starts rolling with an initial angular speed of 17.6 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of 2.04 rad/s2 , how far does the coin roll before coming to rest? Correct answer: 0.428957 m. Explanation:

Let : ωi = 17.6 rad/s , α = −2.04 rad/s2 , and r = 0.565 cm = 0.00565 m . Since ωf = 0,

007 10.0 points A fish swimming behind an oil tanker gets caught in a whirlpool created by the ship’s propellers. The fish has an initial angular speed of 0.50 rad/s. After 4.3 s, the fish’s angular speed is 13.1 rad/s. If the water in the whirlpool accelerates at a constant rate, what is the angular acceleration? Correct answer: 2.93023 rad/s2 . Explanation:

ωf2 = ω i2 + 2 α ∆θ = 0 ∆θ =

−ωi2 2α

and the linear displacement is  −ωi2 ∆s = r ∆θ = r 2α −(17.6 rad/s)2 = (0.00565 m) 2 (−2.04 rad/s2 ) 

= 0.428957 m . Let : ωi = 0.50 rad/s , ωf = 13.1 rad/s , ∆t = 4.3 s .

and 009

10.0 points

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) Harry and Sue cycle at the same speed. The tires on Harry’s bike have a larger diameter than those on Sue’s bike. Which tires have the greater rotational speed? 1. Sue’s tires correct 2. It depends on the speed.

3. m L2 m L2 2 m L2 correct 5. 3 Explanation: By the parallel-axis theorem 4.

3. Harry’s tires

I = ICM + m d2 ,

4. The rotational speeds are the same. Explanation: v = r ω. Tires with a smaller radius needs a larger rotational speed to obtain the same linear speed. 010 10.0 points A small pebble breaks loose from the treads of a tire with a radius of 31 cm. If the pebble’s tangential speed is 54 m/s, what is the tire’s angular speed? Correct answer: 174.194 rad/s. Explanation: Let : r = 31 cm = 0.31 m vt = 54 m/s .

3

and

where

L 2 is the distance from the center-of-mass axis to the parallel axis, we have d=

I=

m L2 m L2 m L2 . = + 3 4 12

012 10.0 points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown below. 2F

3R

vt = r ω vt 54 m/s ω= = 0.31 m r = 174.194 rad/s .

2R F F

keywords: 011 10.0 points The moment of inertia about the center of mass of a rod of length L is I = m L2/12 . Its moment of inertia about its end point is m L2 1. 6 m L2 2. 4

F What is the magnitude of the net torque on the system about the axis? 1. τ = 5 F R 2. τ = F R 3. τ = 2 F R correct 4. τ = 0

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) d = 0.34 m , θ = 90.0◦ .

5. τ = 14 F R Explanation: The three forces F apply counter-clockwise torques while the other force 2 F applies a clockwise torque, so X τ= Fi R i = (−2 F ) (3 R) + F (3 R) + F (3 R) + F (2 R) = 2F R.

013 10.0 points A friend incorrectly says that a body cannot be rotating when the net torque acting on it is zero. What is the correct statement? 1. A body’s angular velocity cannot change if the net torque acting on it is zero and the moment of inertia does not change. correct

4 and

τ = F d sin θ = (4 N) (0.34 m) sin 90.0◦ = 1.36 N · m . 015 10.0 points A circular-shaped object of mass 7 kg has an inner radius of 6 cm and an outer radius of 30 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 10 N, 24 N, and 16 N act on the object, as shown. The force of magnitude 24 N acts 25 ◦ below the horizontal. 10 N

ω

25◦ 24 N

2. Once a body starts rotating the net torque is zero.

16 N

3. A body can have an angular velocity only when a non-zero net torque is acting on it.

Find the magnitude of the net torque on the wheel about the axle through the center of the object.

4. The original statement made by the friend is actually correct.

Correct answer: 6.36 N · m.

Explanation: The rate and direction of rotation of a body cannot change when a zero net torque acts on it. Once started rotating, a body will continue rotating even when no torque acts on it. Again, emphasize change.

Explanation: Let : a = 6 cm = 0.06 m , b = 30 cm = 0.3 m , F1 = 10 N , F2 = 24 N , F3 = 16 N , and θ = 25◦ . F1

014 10.0 points Find the magnitude of the torque produced by a 4.0 N force applied to a door at a perpendicular distance of 0.34 m from the hinge. Correct answer: 1.36 N · m. Explanation: Let : F = 4.0 N ,

θ

ω

F2 F3

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) The total torque is τ = a F2 − b F1 − b F3 = (0.06 m) (24 N) − (0.3 m) (10 N + 16 N) = −6.36 N · m , with a magnitude of 6.36 N · m . 016 10.0 points A 346 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 4.7 rad/s in 3.4 s? Correct answer: 691.135 N · m. Explanation: Let : M = 346 kg , R = 1.7 m , ωf = 4.7 rad/s , and ∆t = 3.4 s . ωf − ωi ωf Since ωi = 0 , α = = ∆t ∆t τ = Iα   ωf 1 2 MR = ∆t 2   (346 kg)(1.7 m)2 4.7 rad/s = 2 3.4 s

Explanation:

Let : M = 1.82 kg , m = 4.59 kg , and g = 9.81 m/s2 . a = α R and the torque is τ = Iα   1 a 2 TR= MR 2 α 1 T = M R2 2 The net force is Fnet = m g − T 1 ma = mg − M a 2 2ma + M a = 2mg 2mg a= 2m+ M 2 (4.59 kg)(9.81 m/s2 ) = 2 (4.59 kg) + 1.82 kg = 8.18689 m/s2 .

018 (part 2 of 3) 10.0 points How far does it drop? Correct answer: 88.5105 m. Explanation:

= 691.135 N · m .

Let : ∆t = 4.65 s Since vi = 0 m/s ,

017 (part 1 of 3) 10.0 points A cylindrical 1.82 kg pulley with a radius of 0.893 m is used to lower a 4.59 kg bucket into a well. The bucket starts from rest and falls for 4.65 s. What is the linear acceleration of the falling bucket? The acceleration of gravity is 9.81 m/s2 . Correct answer: 8.18689 m/s2 .

∆y = vi∆t +

1 1 a (∆t)2 = a (∆t)2 2 2

1 (8.18689 m/s2 )(4.65 s)2 2 = 88.5105 m .

∆y =

019 (part 3 of 3) 10.0 points

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jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) What is the angular acceleration of the cylindrical pulley? Correct answer: 9.16785 rad/s2 . Explanation:

022 (part 3 of 3) 10.0 points Find the linear acceleration of the airplane tangent to its flight path. Correct answer: 0.775115 m/s2 .

Let : R = 0.893 m . a = αR a 8.18689 m/s2 α= = 0.893 m R = 9.16785 rad/s2 . 020 (part 1 of 3) 10.0 points A model airplane whose mass is 1.085 kg is tethered by a wire so that it flies in a circle 38.7 m in radius. The airplane engine provides a net thrust of 0.841 N perpendicular to the tethering wire. Find the torque the net thrust produces about the center of the circle. Correct answer: 32.5467 N · m. Explanation: Let : T = 0.841 N r = 38.7 m . The torque is

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Explanation: The linear acceleration is

a = rα   = (38.7 m) 0.0200288 rad/s2 = 0.775115 m/s2 .

023 10.0 points The sketch shows the top view of a merrygo-round which is rotating clockwise. A boy jumps on the merry-go-round in three different ways: I) from the left, II) from the top, and III) from the right. For all three cases, the boy lands on the same spot.

and

τ = r T = (38.7 m) (0.841 N) = 32.5467 N · m . 021 (part 2 of 3) 10.0 points Find the angular acceleration of the airplane when it is in level flight. Correct answer: 0.0200288 rad/s2.

Compare the final angular momenta of the merry-go-round for the three cases. 1. LI > LII > LIII correct 2. LIII > LII > LI

Explanation: 3. LI = LIII > LII Let : m = 1.085 kg . The angular acceleration is τ τ α= = I m r2 32.5467 N · m = (1.085 kg) (38.7 m)2 = 0.0200288 rad/s2 .

4. LII > LI = LIII 5. LI = LII = LIII Explanation: With respect to the center of the merrygo-round, the angular momentum of the boy

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) is: I) clockwise, II) 0, and III) counterclockwise. The angular momentum of the merrygo-round is along the clockwise direction. By adding the two angular momenta in each case, we get LI > LII > LIII . 024 10.0 points A light, rigid rod l = 6.78 m in length rotates in the xy plane about a pivot through the rod’s center. Two particles of masses m1 = 8.8 kg and m2 = 3.8 kg are connected to its ends.

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Before

After

ω

3ω

At this instant, what is the total angular momentum of the system in terms of its original value Li? 1. Lf = Li correct 2. Lf =

Li 9

3. Lf = 3 Li 4. Lf =

Li 3

5. Lf = 9 Li Determine the angular momentum of the system about the origin at the instant the speed of each particle is v = 1.15 m/s. Correct answer: 49.1211 kg m2 /s. Explanation: From the definition of angular momentum l l v + m2 v 2 2 l = (m1 + m2 ) v 2 = 49.1211 kg m2 /s

L = m1

025 10.0 points A mouse hangs on the rim of a spinning turntable. The rotational inertia of the mouse about the center of the turntable is comparable to rotational inertia of the turntable by itself. Mouse and turntable are freely spinning with constant angular velocity ω. The mouse starts to crawl toward the center of the turntable. At a particular position the angular velocity of mouse and turntable has increased to 3 ω .

Explanation: No external torques act on the system of mouse and turntable, so as long as the mouse stays on the turntable, the total angular momentum cannot change. dL X = τi . dt 026 10.0 points Strictly speaking, as more and more skyscrapers are built on the surface of the Earth, does the day tend to become longer or shorter? And strictly speaking, does the falling of autumn leaves tend to lengthen or shorten the 24-hour day? What physical principle supports your answers? 1. shorten; lengthen; conservation of inertia 2. lengthen; shorten; conservation of angular momentum correct 3. lengthen; shorten; conservation of kinetic energy 4. shorten; lengthen; conservation of angular torque Explanation:

jonnalagadda (saj2436) – Quest 6: Rotation – ayida – (4355-3) Applying conservation of angular momentum, as the radial distance of mass increases, the angular speed decreases. The mass of material used to construct skyscrapers is lifted, slightly increasing the radial distance from the Earth’s spin axis. This would tend to slightly decrease the earth’s rate of rotation, which in turn tends to lengthen days slightly. The opposite effect occurs for falling leaves, since their radial distance from the earth’s axis decreases. As a practical matter, these effects are entirely negligible! 027 10.0 points A student holds two lead weights, each of mass 14 kg. When the students’ arms are extended horizontally, the lead weights are 1.1 m from the axis of rotation and the student rotates with an angular speed of 2 rad/sec. The moment of inertia of student plus stool is 9.5 kg m2 and is assumed to be constant; i.e., the student’s arms are massless! Then the student pulls the lead weights horizontally to a radius 0.24 m from the axis of rotation.

Calculate the final angular speed of the system. Correct answer: 7.80721 rad/s. Explanation: Basic Concepts: X

~L = 0

1 I ω2 2 Solution: The initial moment of inertia of the system is Ii = Is + 2 m R 2 = 9.5 kg m2 + 2 (14 kg) (1.1 m)2 = 43.38 kg m2 . Krot =

8

The final moment of inertia of the system is If = Is + 2 m r 2 = (9.5 kg m2) + 2 (14 kg) (0.24 m)2 = 11.1128 kg m2 . From conservation of the angular momentum it follows that Ii ω i = If ω f . Therefore, ωf = ω

Ii If

(43.38 kg m2 ) (11.1128 kg m2) = 7.80721 rad/s . = (2 rad/sec)...