CHAPTER 12 SOLUTIONSProblem 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=!02+2"(#f$#0)=0+2(18 rev!60=min2)(2 rev)=65.7 rpm,and (b) Equation 12-10 gives !f"!0=0+12#t2,or t=2(2 rev)=(18 rev=60 s2)=3.65 s.Problem 6. You switch a food blender from its high to its low setting; the blade speed drops from 3600 rpmto 1800 rpmin 1.4 s.How many revolutions does it make during this time? Solution At constant angular acceleration, Equations 12-1 and 8 give !"=12(#0+#f) !t=12(3600+1800)(rev=60 s)(1.4 s)=63.0 rev. Problem 15. The angular acceleration of a wheel in rad/s2is given by 24t2!16t3,where tis the time in seconds. The wheel starts from rest at t=0.(a) When is it again at rest? (b) How many revolutions has it turned between t=0and when it is again at rest? Solution (a) Integrating Equation 11-5 (with !0=0),we find !(t)="0t#($t )d$t ="0t(24$t 2%16$t 3)d$t =8t3%4t4.The wheel is at rest when !(t)=(8"4t)t3=0,or at t=0and t=2 s.(b) Integrating Equation 11-2 for the angular displacement (in radians), we find !(t)"!0=#0t$(%t )d%t =#0t(8%t 3"4%t 4)d%t =2t4"4t5=5.For t=2 s,!"=25(1#4=5)=6.4 radians=1.02 rev.Problem 17. A torque of 110 N!mis 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°.)
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