Physics 100A Homework 11- Chapter 11 (part 1) The force passes through the point A, so there is no arm and the torque is zero.

Transcription

1 Physics A Homework - Chapter (part ) Finding Torque A orce F o magnitude F making an ange with the x axis is appied to a partice ocated aong axis o rotation A, at Cartesian coordinates (,) in the igure. The vector F ies in the xy pane, and the our axes o rotation A, B, C, and D a ie perpendicuar to the xy pane. A) What is the torque A about axis A due to the orce F? The orce passes through the point A, so there is no arm and the torque is zero. B) What is the torque B about axis B due to the orce F? (B is the point at Cartesian coordinates (, b), ocated a distance b om the origin aong the y axis.) The magnitude o the torque is rf where F is the component o the orce perpendicuar to r. Note that the torque can aso be cacuated with the concept o the moment arm rf, where r is the component o the radia vector perpendicuar to the direction o the orce. The answer is the same. In most probems it is easier to visuaize the perpendicuar component o the orce. We wi use what is simpest in a given probem. r b F Fcos The direction o the torque oows the convention: positive i it produces a countercockwise rotation and negative i it produces a cockwise rotation. B bf cos C) What is the torque C about axis C due to the orce F? (C is the point at Cartesian coordinates (, c), ocated a distance c om the origin aong the x axis.) r c F Fsin The torque produces a cockwise rotation about C. c cf sin D) What is the torque D about axis D due to the orce F? (D is the point ocated a distance d om the origin and making an ange φ with the x axis.) The ange between F and the r vector AD is ( φ ). F Fsin( φ ) and the rotation is countercockwise Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher.

2 Chapter : Rotationa Dynamics and Static Equiibrium James S. Waker, Physics, 4 th Edition D df sin( φ ) Torque Magnitude Ranking Task The wrench in the igure has six orces o equa magnitude acting on it. Rank these orces (A through F) on the basis o the magnitude o the torque they appy to the wrench, measured about an axis centered on the bot. The arger the radius, the arger the torque magnitude. The arger the perpendicuar projection onto the radia direction, the arger the torque. Then in decreasing order: D, (B,E), F, A, C Torque! The car shown in the igure has mass m (this incudes the mass o the whees). The whees have radius r, mass m W, and moment o inertia I km W r. Assume that the axes appy the same torque to a our whees. For simpicity, aso assume that the weight is distributed uniormy so that a the whees experience the same norma reaction om the ground, and so the same ictiona orce. A) I there is no sipping, a ictiona orce must exist between the whees and the ground. In what direction does the ictiona orce act? Take the positive x direction to be to the right. The ictiona orce acts in the positive x direction. B) Use Newton's aws to ind an expression or the net externa orce acting on the car. Ignore air resistance. The vertica orces baance. The net orce is in the horizonta direction: F 4 where iction orce in each whee. tota is the Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher.

3 Chapter : Rotationa Dynamics and Static Equiibrium James S. Waker, Physics, 4 th Edition C) Use Newton's aws to ind an expression or, the norma orce on each whee. Forces in the y direction: 4 N whee take to be equa since the weight is equay distributed. N mg /4 whee mg where is the norma orce on each whee that we N whee D) Now assume that the ictiona orce is not at its maximum vaue. What is the reation between the torque appied to each whee by the axes and the acceeration a o the car? Once you have the exact expression or the acceeration, make the approximation that the whees are much ighter than the car as a whoe. The torque orce at each whee is: r. The inear acceeration is obtained om the sum o the orces 4 / r and ma / 4, then ( ma /4 / r ) and ma 4 a mr A person sowy owers a.6 kg crab trap over the side o a dock, as shown in the igure. What torque does the trap exert about the person's shouder? Picture the Probem: The arm extends out either horizontay and the weight o the crab trap is exerted straight downward on the hand. Strategy:. In this case the downward gravitationa orce is perpendicuar to the radia direction. The rotation in countercockwise. Soution: The torque: rmg ( )( )( ).7 m.6 kg 9.8 m/s 5 N m Insight: I the man bent his ebow and brought his hand up next to his shouder, the torque on the shouder woud be zero but the orce on his hand woud remain 5 N or 7.9 b..) When a ceiing an rotating with an anguar speed o.9 rad/s is turned o, a ictiona torque o. N m sows it to a stop in 9. s. Picture the Probem: The ceiing an rotates about its axis, decreasing its anguar speed at a constant rate. Strategy: Determine the anguar acceeration using equation -6 and then use equation -4 to ind the moment o inertia o the an. Soution: Sove equation -4 or I: Insight: Friction converts the an s initia kinetic energy o in more detai in section -8. (. N m)(.5 s) (.75 rad/s) Δt I.98 kg m α Δω Δt Δω Iω. J into heat. Rotationa work wi be examined Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher.

4 Chapter : Rotationa Dynamics and Static Equiibrium James S. Waker, Physics, 4 th Edition.9) A ish takes the bait and pus on the ine with a orce o. N. The ishing ree, which rotates without iction, is a uniorm cyinder o radius.55 m and mass.99 kg. A) What is the anguar acceeration o the ishing ree? B) How much ine does the ish pu om the ree in.5 s? Picture the Probem: The ish exerts a torque on the ishing ree and it rotates with constant anguar acceeration. Strategy: Use Tabe - to determine the moment o inertia o the ishing ree assuming it is a uniorm cyinder ( MR ). Find the torque the ish exerts on the ree by using equation -. Then appy Newton s Second Law or rotation (equation -4) to ind the anguar acceeration and equations - and - to ind the amount o ine pued om the ree. Soution:. (a) Use Tabe - to ind I: ( )( ). Appy equation - directy to ind : I MR.99 kg.55 m.5 kg m rf ( )( ).55 m. N. N m. Sove equation -4 or α :. N m α I.5 kg m 8 rad/s 4. (b) Appy equations - and -: s r r( αt ) (.55 m) ( 8 rad/s )(.5 s).4 m Insight: This must be a sma ish because it is not puing very hard;. N is about.49 b or 7.9 ounces o orce. Or maybe the ish is tired?.) A string that passes over a puey has a. kg mass attached to one end and a.65 kg mass attached to the other end. The puey, which is a disk o radius 9.5 cm, has iction in its axe. What is the magnitude o the ictiona torque that must be exerted by the axe i the system is to be in static equiibrium?. Picture the Probem: The two masses hang on either side o a puey. Strategy: Use Newton s Second Law or rotation (equation -4) to ind the ictiona torque that woud make the anguar acceeration o the system equa to zero. In each case the torque exerted on the puey by the hanging masses is the weight o the mass times the radius o the puey. Let m.65 kg and m. kg. The torque due to m is cockwise and thereore taken to be in the negative direction. Soution: Write Newton s r( mg) + r( mg) + Second Law or rotation and sove or : rg( m m ).94 m 9.8 m/s.65. kg ( )( )( ).9 N m Insight: This ictiona torque represents a static iction orce. I a itte bit o mass were added to m, the system woud begin acceerating cockwise and the ictiona torque woud be reduced to its kinetic vaue. Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher. 4

5 Chapter : Rotationa Dynamics and Static Equiibrium James S. Waker, Physics, 4 th Edition.5) An 85 kg person stands on a uniorm 5. kg adder that is 4. m ong, as shown in the igure. The oor is rough; hence, it exerts both a norma orce,, and a ictiona orce,, on the adder. The wa, on the other hand, is ictioness; it exerts ony a norma orce,. Using the dimensions given in the igure, ind the magnitudes o,, and. To sove the probem using the technique emphasized in cass In which we ook or the perpendicuar component o the orce we can cacuate the ange that the adder makes with the oor. a Lsin sin ( a/ L) sin (.8 / 4) L a 7.8 (7.85) To ind the ength that the person is up the adder b cos b/ cos.7 / cos(7.8).4 m (.479) The torque about point : b sin cos L mg sin mg cos sin L For the weight o the person: m cos pg mpg For the weight o the adder ( L/) m gcos + + m g m g p mg Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher. 5 mg

6 Chapter : Rotationa Dynamics and Static Equiibrium James S. Waker, Physics, 4 th Edition L sin m g cos ( L / ) m g cos p And ( mp + ( L /) m) g cos Lsin ((.4)(85) + ()(5.))(9.8)cos(7.8) (4)sin(7.8) ((.4)(85) + ()(5.))(9.8)cos(7.8) (4)sin(7.8) 6 N Forces in the x-direction F x 6 N Forces in the y-direction Fy mpg mg ( mp + m) g ( )(9.8) 885 N.6) A rigid, vertica rod o negigibe mass is connected to the oor by a bot through its ower end, as shown in the igure. The rod aso has a wire connected between its top end and the oor. 6. Picture the Probem: The horizonta orce F is appied to the rod as shown in the igure at right. Strategy: Let L the rod ength and write Newton s Second Law or torques (et the bot be the pivot point) in order to determine the wire tension T. The pivot point is convenient since the orce F b going through it produces no torque. Then write Newton s Second Law in the horizonta and vertica directions to determine the components o the bot orce F b. Soution:. (a) Set and LT ( cos 45 ) ( LF ) sove or T : F F T cos45. (b) Set F x and sove or F b,x : Fx F + Fb, x T cos 45 F F Tcos 45 F cos 45 F cos 45. (c) Set F and sove or F : Fy Fb, y Tsin 45 y Insight: The bot orce has a magnitude o F b, y b, x F F sin 45 cos 45 b, y and points 45 above the horizonta and to the et. F F Copyright Pearson Education, Inc. A rights reserved. This materia is protected under a copyright aws as they currenty exist. No portion o this materia may be reproduced, in any orm or by any means, without permission in writing om the pubisher. 6