MasteringPhysics: Week 11: Rotations II - Google Chrome ! https://session.master

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MasteringPhysics: Week 11: Rotations II - Google Chrome ! https://session.masteringphysics.com/myct/itemView?offset-next&assignmentProblemID-82500368; Resources previous 10 of 13 next Item 10 Part A In the figure the lower disk, of mass 460 g and radius 3.8 cm, is rotating at 180 rpm on a frictionless shaft of negligible radius. The upper disk, of mass 280 g and radius 2.2 cm, is initially not rotating. It drops freely down onto the lower disk, and frictional forces bring the two disks to a common rotational speed. (Figure 1) Find the final common frequency in rpm Express your answer using two significant figures regf rpm Submit My Answers Give Up Part B Find the percentage of the initial kinetic energy lost to friction. Express your answer using two significant figures Figure 1 percent = of! Submit My Answers Give Up dbac Continue

Explanation / Answer

(a)m1= 460g= 0.46 kg

r1=3.8 cm= 0.038 m

moment of inertia= i1= 1/2 m1r12=1/2 (0.46)(0.038)2=3.32x10-4 kgm2

w1=180 rpm= 180 * (2 pi /60)= 18.8 rad/s

m2= 280g= 0.28 kg

r2=2.2 cm= 0.022 m

moment of inertia= i1= 1/2 m2r22=1/2 (0.28)(0.022)2=6.78x10-5 kgm2

w2=0

using law of conservation of angular momentum

i1 w1 + i2 w2= (i1+i2)w

3.32x10-4(18.8 )+6.78x10-5(0)= (3.32x10-4+6.78x10-5)w

w= 15.62 rad/s= 15.62 *60 /2 pi rpm

so w=150 rpm

(b) loss in KE due to friction= 1/2 i1w12+ 1/2 i2w22- 1/2 (i1+i2)w2

loss in KE=1/2 (3.24x10-4)(18.8)2+1/2 (6.78x10-5)(0)2=1/2 (3.32x10-4+6.78x10-5)(15.62)2

loss in KE=8.48x10-3j

percentage=( loss/ initial KE)x100

percentage= (8.48x10-3/0.057)x100

percentage=15 %

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