MasteringPhysics: Chapter 8 HW - Google Chrome @ https://session.masteringphysic
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MasteringPhysics: Chapter 8 HW - Google Chrome @ https://session.masteringphysics.com/myct/itemView?assignmentProblemID-60840906 SPRING-2016- PHYSICS 13-MANCINI Signed in as Javeion Lee Hel Close Chapter 8 HProblem 8.25 ResourcesY previous | 7 of 14 | next» Problem 8.25 Part A A pendulum 1.50 m long is released (from rest) at an angle 80 30.00 (Figure 1) Determine the speed of the 900 g bob at the lowest point (8-0) Express your answer with the appropriate units. Im UI= 3.94 Submit My Answers Give Up Incorrect, one attempt remaining; Iry Again Part B Determine the speed of the 900 g bob at 15.0 Express your answer with the appropriate units. Figure 1 of 1 v2 = Value Units ahie Submit My Answers Give Up Part C Determine the speed of the 90.0 g bob at =-15.0° (i.e., on the opposite side) Express your answer with the appropriate units. Value Units ahie My Answers Give Un 2:43 PM 6/4/2016Explanation / Answer
8.25 a) We apply conservation of energy from the release point to the lowest point:
E = K1 + U1 = K2 + U2;
0 + mgL(1- cosqo) = 1/2mva2 + 0;
0 + (9.80m/s2)(1.50m)(1-cos30.0°) = 1/2va2,
which gives va = .50 m/s
(b) We apply conservation of energy from the release point to the given point:
E = K1 + U1 = K3 + U3;
0 + mgL(1-cosqo) = 1/2mvb2 + mgL(1 - cosqb);
0 + (9.80m/s2)(1.50m)(1 - cos30.0°)
= 1/2 vb2 + (9.80m/s2)(1.50m)(1 - cos15.0°),
which gives vb = .5835 m/s
(c) We apply conservation of energy from the release point to the given point:
E = K1 + U1= K4+ U4;
0 + mgL(1 - cosqo) = 1/2mvc2 + mgL(1 -cosqc);
0 + (9.80m/s2)(1.50m)(1 -cos30.0°)
= 1/2vc2 + (9.80m/s2)(1.50m)[1 -cos(-15.0°)],
which gives vc = .5835
Because this is the same elevation as in part (b), the answer is the same.
d)e) and f)
he net force along the cord must provide the radial acceleration:
FT - mgcosq = mv2/L, or FT = m[(v2/L) + gcosq].
Thus we have
FTa = m[(va2/L) + gcosqa]
= (0.090kg){[(.5m/s)2/(1.50m)] + (9.80m/s2)cos0°} = 0.897N.
FTb = m[(vb2/L) + gcosqb]
= (0.090kg){[(.5m/s)2/(1.50m)] + (9.80m/s2)cos15} = 0.867N.
FTc = m[(vc2/L) + gcosqc]
= (0.090kg){[(.5m/s)2/(1.50m)] + (9.80m/s2)cos15} = 0.867N.