S08P2AChap12Sol

S08P2AChap12Sol - CHAPTER 12 SOLUTIONS Problem 5. A wheel...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 12 SOLUTIONS Problem 5. A wheel turns through 2.0 revolutions while being accelerated from rest at 18 rpm/s. (a) What is the final angular speed? (b) How long does it take to turn the 2.0 revolutions? Solution For constant angular acceleration, (a) Equation 12-11 gives ! f = 0 2 + 2 " ( # f $ 0 ) = + 2 ( 18 rev ! 60 = min 2 )( 2 rev ) = 65 . 7 rpm , and (b) Equation 12-10 gives f " 0 = 0 + 1 2 t 2 , or t = 2 ( 2 rev ) = ( 18 rev = 60 s 2 ) = 3 . 65 s . Problem 6. You switch a food blender from its high to its low setting; the blade speed drops from 3600 rpm to 1800 rpm in 1.4 s. How many revolutions does it make during this time? Solution At constant angular acceleration, Equations 12-1 and 8 give ! = 1 2 ( 0 + f ) ! t = 1 2 (3600 + 1800)(rev = 60 s)(1.4 s) = 63.0 rev. Problem 15. The angular acceleration of a wheel in rad/s 2 is given by 24 t 2 ! 16 t 3 , where t is the time in seconds. The wheel starts from rest at t = 0. (a) When is it again at rest? (b) How many revolutions has it turned between t = 0 and when it is again at rest? Solution (a) Integrating Equation 11-5 (with 0 = 0 ), we find ( t ) = " 0 t ( $ t ) d $ t = " 0 t ( 24 $ t 2 % 16 $ t 3 ) d $ t = 8 t 3 % 4 t 4 . The wheel is at rest when ( t ) = ( 8 " 4 t ) t 3 = 0 , or at t = 0 and t = 2 s . (b) Integrating Equation 11-2 for the angular displacement (in radians), we find ( t ) " 0 = # 0 t $ ( % t ) d % t = # 0 t ( 8 % t 3 " 4 % t 4 ) d % t = 2 t 4 " 4 t 5 = 5 . For t = , ! = 2 5 ( 1 # 4 = 5 ) = 6.4 radians = 1.02 rev. Problem 17. A torque of 110 N ! m is required to start a revolving door rotating. If a child can push with a maximum force of 90 N, how far from the door’s rotation axis must she apply this force? Solution If the force is applied perpendicular to the door, the radial distance should be r = = F = 110 N " m = 90 N = 1.22 m from the axis. (See Equation 12-12 with = 90 ° .)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHAPTER 12 279 Problem 23. A 1.5-m-diameter wheel is mounted on an axle through its center. (a) Find the net torque about the axle due to the forces shown in Fig. 12-42. (b) Are there any forces that don’t contribute to the torque? Solution The displacements from the center of the wheel to the points of application of the forces, and the angles between these displacements and the forces, are shown on Fig. 12-42. (a) If we take positive torques counterclockwise, the net torque (sum of Equation 12-12) is ! net = " r i F i sin # i = r 1 F 1 sin0 ° + (0.25 m)(3.0 N)sin45 ° $ (0.50 m)(2.0 N)sin90 ° + (0.75 m) % (1.8 N)sin60 ° + 0 ! F 5 = 0.699 N " m. (b) Evidently, F 1 and F 5 , whose lines of action pass through the center of the wheel, produce no torque about the center. FIGURE 12-42 Problem 23 Solution. Problem 28. The full diameter of a wheel is 92 cm, and its rotational inertia is 7.8 kg ! m 2 . (a) What is the minimum mass it could have? (b) How could it have more mass? Solution (a) Any part of the wheel has a distance from the center less than or equal to the maximum radius, so I = ! m i r i 2 " ( ! m i ) r max 2 . Thus M = ! m i " I = r max 2 = M min = ( 7 . 8 kg # m 2 ) = ( 1 2 $ 0 . 92 m ) 2 = 36 . 9 kg . (b) If not all the mass of the wheel is concentrated at the rim, the total mass is greater than this minimum.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.