PHYS 1401 Chapter 11 Solutions

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PHYS 40 Chapter Solutions 3. Picture the Problem: The arm etends out either horizontall or at some angle below horizontal, and the weight o the troph is eerted straight downward on the hand.

In this case the moment arm is the horizontal distance between the shoulder and the hand, and the orce is the downward weight o the troph.

1 PHYS 40 Chapter Solutions 3. Picture the Problem: The arm etends out either horizontall or at some angle below horizontal, and the weight o the troph is eerted straight downward on the hand. Strateg: The torque equals the moment arm times the orce according to equation -3. In this case the moment arm is the horizontal distance between the shoulder and the hand, and the orce is the downward weight o the troph. Find the horizontal distance in each case and multipl it b the weight o the troph to ind the torque. In part (b) the horizontal r r cos m cos m. distance is Solution:. (a) Multipl the moment arm b the weight: r mg m.6 kg 9.8 m/s 9.56 N m. (b) Multipl the moment arm b the weight: r mg m.6 kg 9.8 m/s 8.83 N m Insight: The torque on the arm is reduced as the arm is lowered. The torque is eactl zero when the arm is vertical. 5. Picture the Problem: The biceps muscle, the weight o the arm, and the weight o the ball all eert torques on the orearm as depicted at right. Strateg: Use equation 0-3 to determine the torques produced b the biceps muscle, the weight o the orearm, and the weight o the ball. Sum the torques together to ind the net torque. According to the sign convention, torques in the counterclockwise direction are positive, and those in the clockwise direction are negative.

Solution:. (a) Compute the individual torques using equation 0-3 and sum them: biceps orearm ball ball rf 0.075 m.6 N 0.347 N m r mg 0.70 m.0 kg 9.8 m/s.00 N m rw 0.340 m.4 N 0.

2 Solution:. (a) Compute the individual torques using equation 0-3 and sum them: biceps orearm ball ball rf m.6 N N m r mg 0.70 m.0 kg 9.8 m/s.00 N m rw m.4 N N m biceps orearm ball N m.4 N m. (b) Negative net torque means the clockwise direction; the orearm and hand will rotate downward. 3. (c) Attaching the biceps arther rom the elbow would increase the moment arm and increase the net torque. Insight: The biceps would need to eert a orce o at least 90.3 N in order to prevent the arm rom rotating downward (see problem 5).. Picture the Problem: The CD rotates about its ais, increasing its angular speed at a constant rate. Strateg: Determine the angular acceleration o the CD using equation 0- and its moment o inertia (treat it as a disk) rom Table 0-. Then use equation -4 to ind the torque eerted on the CD. Solution:. Solve equation 0- or : 450 rev/min rad rev min 60 s rev rad rev 59 rad/s. Use Table 0- to ind I MR I MR 0.07 kg m 3.0 kg m 5 : 5 3. Appl equation -4 directl: I 3.0 kg m 59 rad/s Nm Insight: The CD takes 0.80 s to accelerate to its inal angular velocit, accounting or some o the dela between when ou press pla and when ou irst hear the music. 6. Picture the Problem: The person lies on a lightweight plank that rests on two scales as shown in the diagram at right. Strateg: Write Newton s Second Law in the vertical direction and Newton s Second Law or rotation to obtain two equations with two unknowns, m and. Solve each to ind m and. Using the let side o the

3 plank as the origin, there are two torques to consider: the positive torque due to the right hand scale and the negative torque due to the person s mass. Solution:. (a) Write Newton s Second Law in the vertical direction to ind m: F F F mg 0 F F 90 N m 4 kg g 9.8 m/s. (b) Write Newton s Second Law or r F mg 0 rotation and solve or : F.50 m N mg 4 kg9.8 m/s 0.74 m Insight: The equation in step does not depend on the ais o rotation that we choose, but the equation in step does. Nevertheless, we ind eactl the same near her eet, to be the ais o rotation. i we choose the other scale, 38. Picture the Problem: The stick rests on the bowling ball as shown in the diagram at right. Strateg: Write Newton s Second Law or torque (let the pivot point be the point o contact between the stick and the loor) and then write Newton s Second Law in the vertical and, and horizontal directions to ind the magnitudes o N,. Use the dimensions indicated in the igure, together with the deinition that the overall length o the stick. B careull studing the geometr we ind that H R Rcos. θ L H R θ Solution:. (a) Set and solve or N: 0 N 0 r mg NL cos mg N sin cosmg mg sincos H R cos 0.6 m cos 30.0 H sin 0.4 kg 9.8 m/s m sin 30.0cos N. Set F 0 to ind F : F N sin 0 N sin N sin N 3. Set F 0 FJ, : to ind F mg N cos 0 mg N cos 0.4 kg 9.8 m/s N cos N 3

Insight: The two components o the loor s contact orce combine to produce a orce o.34 N inclined at 68.5 above the horizontal and directed toward the positive direction. 4.

4 Insight: The two components o the loor s contact orce combine to produce a orce o.34 N inclined at 68.5 above the horizontal and directed toward the positive direction. 4. Picture the Problem: The cat is perched on the plank as shown in the igure at right. Active Eample - indicates that the uniorm plank is 4.00 m long and has a mass o 7.00 kg. The plank is supported b two sawhorses, one m rom the let end o the board and the other.50 m rom its right end. Strateg: Set the net torque, with the right hand sawhorse acting as the pivot point, equal to zero and solve or the distance r o the cat rom the right sawhorse. Then determine the distance o the cat dcat rom the right end o the plank. The center o mass o the plank is a distance 4.00 m.50 m 0.50 m rom the right hand sawhorse. Solution:. Set 0 and solve or : cat r r mplank g rcat mcat g 0 mplank r r cat 7.00 kg 0.50 m.5 m m.8 kg cat. Now ind d cat : d cat.50.5 m 0.5 m Insight: I the cat were twice as massive it could onl walk to within 0.88 m o the end. We bent the rules or signiicant igures a bit in step in order to retain two digits in the answer. 5. Picture the Problem: The weight o the empt paint can pulls downward on one end o the meter stick and the weight o the meter stick pulls down at the mark, the center o mass. The sstem is balanced when suspended at the 0.0- mark. Strateg: Write Newton s Second Law or torque about the 0.0 mark and Newton s Second Law or orce in the vertical direction. Combine the two equations to ind the mass M o the empt paint can and the mass m o the meter stick. Let S represent the upward scale orce o.54 N, let rpc the 0.0- moment arm o the paint can, stick s center o mass. r the and let moment arm o the meter Solution:. (a) Set m: F 0 and solve or F S Mg mg 0 m S g M 4

5 0. Set or the pivot point and solve or M: 3. Substitute the epression or M into step : r M r m 0 pc pc M r r m m S g r r m pc S.54 N m g r r pc 9.8 m/s kg 4. (b) Now use the equation rom step : M r r m pc kg 0.56 kg Insight: Veri or oursel that adding 0.00 kg o paint to the can shits the balance point to 4.4 and increases the scale orce to 3.53 N. 60. Picture the Problem: The windmill rotates about its ais with constant angular acceleration due to the applied torque. Strateg: Use equation -4 to ind the torque required to change the angular momentum o the windmill b the speciied amount during the given time interval. Solution: Appl equation -4 directl: L L kg m /s 00 Nm 0.0 kn m t 5.86 s Insight: Note that the 00 kg m /s change in angular momentum limits the answer to onl two signiicant igures. When the wind is blowing at a constant speed, the net torque on the windmill is zero and it rotates at constant speed. 7. Picture the Problem: The student catches the mass awa rom the ais o rotation o the stool upon which she sits, and as a result o the collision she rotates with constant angular speed. Strateg: Because there is no eternal torque on the sstem, the angular momentum o the.5-kg mass equals the angular momentum o the student plus mass sstem ater the catch. Use equation -5 together with equations - and - and a calculation o the inal moment o inertia to determine the inal angular speed o the student. Solution: Set Li L and solve or mvr Istudent-stool mr : mvr 0 I.5 kg.7 m/s0.40 m mvr Istudent-stool mr 4. kg m.5 kg 0.40 m 0.37 rad/s 5

6 Insight: Doubling the initial speed o the.5 kg mass will double the inal angular speed o the student even though the moment o inertia o the.5 kg mass (0.4 kg m ) is small compared to the student-stool sstem (4. kg m ). This is because the total angular momentum depends linearl upon the initial speed o the mass. 84. Picture the Problem: The saw blade rotates on its ais and gains rotational kinetic energ due to the torque applied b the electric motor. Strateg: The torque applied through an angular displacement gives the blade its rotational kinetic energ. Use equations -7 and 0-7 to relate the kinetic energ to the torque applied b the motor. Then use equation -7 again to ind the kinetic energ and angular speed ater the blade has completed hal as man revolutions. Solution:. (a) Find in units o rad/sec:. Set W K and solve or : rev rad min rad/s min rev 60 s W I and I mr kg0.5 m 379 rad/s mr 5.8 N m 6.30 rev rad/rev 3. (b) The time to rotate the irst 3.5 revolutions is greater than the time to rotate the last 3.5 revolutions because the blade is speeding up. So more than hal the time is spent in the irst 3.5 revolutions. Thereore, the angular speed has increased to more than hal o its inal value. Ater 3.5 revolutions, the angular speed is greater than 80 rpm. 4. (d) Set W K and solve or : I mr 4 4 mr kg0.5 m N m 3.5 rev rad/rev 60 s rev 68 rad/s 560 rev/min min rad Insight: The angular speed increases linearl upon time ( 0 t t ) but depends upon the 0 square root o the angular displacement:. 6

ROTATIONAL DYNAMICS AND STATIC EQUILIBRIUM

ROTATIONAL DYNAMICS AND STATIC EQUILIBRIUM Chapter 11 Units of Chapter 11 Torque Torque and Angular Acceleration Zero Torque and Static Equilibrium Center of Mass and Balance Dynamic Applications of Torque