Chapter H - Problems

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1 Chapter H - Problem Blinn College - Phyic 45 - Terry Honan Problem H.1 A wheel rotate from ret to 1 ê in 3. Aume the angular acceleration i contant. (a) What i the magnitude of the wheel' angular acceleration? (b) Through what angle (in ian) doe the wheel rotate? Solution to H.1 HaL w = w 0 + a t ï 1 = 0 + a 3 ï a = 4 HbL Dq = 1 Hw 0 + wl t = 1 H0 + 1L 3 = 18 Problem H. (a) What i the angular velocity of the Earth in it orbit about the Sun. (b) What i the angular velocity of the Moon in it orbit about the Earth. Solution to H. A contant angular velocity i related to the period T by w = p T. HaL T = 1 year = 365.4µ4µ3600 = µ10 7 ï w = 1.99µ10-7 HbL T = 7.3 day = 7.3µ4µ3600 =.36 05µ10 6 ï w =.66µ10-6 Problem H.3 A grinding wheel rotating at 100 revêmin i turned off. It low with a contant angular acceleration of ê. (a) How long doe it take the wheel to top? (b) Through what angle, in ian, doe it turn while lowing? Solution to H.3 w = 0 and w 0 = 100 rev min µ p rev µ 1 min 60 = HaL w = w 0 + a t ï 0 = t ï 5.4 HbL w - w = a Dq ï Dq = = 7.4 µ Problem H.4 A car race around a circular track of iu 50 m at a contant peed of 45 mê. (a) What i it angular velocity? (b) What i the magnitude and direction of it acceleration?

1 HaL w = w 0 + a t ï 1 = 0 + a 3 ï a = 4 HbL Dq = 1 Hw 0 + wl t = 1 H0 + 1L 3 = 18 Problem H. (a) What i the angular velocity of the Earth in it orbit about the Sun.

2 Chapter H - Problem Solution to H.4 R = 50 m and v = 45 m HaL v = r w ï w = v R = 0.18 HbL a c = w r = v R = 8.1 m The direction of the acceleration i centripetal, meaning that i directed toward the center. Problem H.5 By rotating through 1.5 rev of a 1 m iu arc, a dicu thrower uniformly accelerate a dicu from ret to 5 mê. (a) What i the final angular velocity of the dicu? (b) What i it angular acceleration? (c) What i the total time of acceleration? Solution to H.5 HaL v = r w ï w = v r = 5 1 = 5 HbL w -w 0 = a Dq ï 5-0 = a H pµ1.5l ï a = 39.8 HcL Dq = 1 Hw 0 + wl t ï pµ1.5 = 1 H0 + 5L t ï t = 0.68 Problem H.6 Aume that a baeball i a uniform phere of iu 3.80 cm. If it move at a peed of 38 mê (the peed of the center of ma) and rotate at 15 ê then what i the ratio of the rotational kinetic energy to the tranlational kinetic energy. Solution to H.6 For a uniform olid phere I = 5 M R K rot = 1 I w = 1 K rot K tran = 5 M R w = 1 5 M R w 1 5 M R w R = w 1 M v 5 v Inert the number: R = m, v = 38 m and w = 15 giving K rot K tran = Problem H.7 Three mae glued onto a light rigid heet in the xy plane. The mae and poition are: 4 kg at H-3 m, 5 ml, 6 kg at H4 m, 0L and 8 kg at H0, -3 ml. (a) Suppoe thi heet i rotated about the y axi at an angular velocity of 1 ê. What i it total rotational kinetic energy? (b) Suppoe it rotate about the z axi at 1 /. What i it rotational kinetic energy?

5 rev of a 1 m iu arc, a dicu thrower uniformly accelerate a dicu from ret to 5 mê. (a) What i the final angular velocity of the dicu? (b) What i it angular acceleration?

3 Chapter H - Problem 3 Solution to H.7 (a) For a dicrete ditribution of ma: I = i m i r i where r i i the perpendicular ditance of the ma m i from the axi. If the axi i the y axi then r i = x i I = i m i r i = i m i x i = 4µ3 + 6µ4 + 8µ0 = 13 kmÿm The kinetic energy i K = 1 I w = 1 13µ1 = 9500 J. (b) If the axi i the z-axi then r i = x i + yi. I = i m i r i = i m i Ix i + yi M = 4µI3 + 5 M + 6µI4 + 0M + 8µI0 + 3 M = 304 kmÿm K = 1 I w = 1 304µ1 = J. Problem H.8 Two identical thin rod of length L and ma M are perpendicular and joined at their center. What i the moment of inertia about an axi at the end of one rod that i: (a) perpendicular to the plane, (b) parallel to the other rod. HaL axi HbL axi Solution to H.8 The rod connected to the axi ha, in both cae, a moment of I 1 = 1 3 M L. The other rod' moment varie. (a) Ue the parallel axi theorem to find the other moment. The ditance from the center of ma to the axi i D = Lê. I = I cm + M D = 1 1 M L + M J L N = 1 3 M L. The total moment i the um. I = I 1 + I = 3 M L. (b) In thi cae all the ma of the econd rod i a ditance Lê rom the axi. It' moment i I = M J L N = 1 4 M L. ï I = I 1 + I = J N M L = 7 1 M L

I = i m i r i = i m i Ix i + yi M = 4µI3 + 5 M + 6µI4 + 0M + 8µI0 + 3 M = 304 kmÿm K = 1 I w = 1 304µ1 = 1 900 J. Problem H.

4 4 Chapter H - Problem Problem H.9 What are the moment of inertia of the following flat uniform bodie of ma M about the axi mentioned? (a) an LµL quare about a diagonal (b) a circle of iu R about a diameter (c) a circle of iu R about a tangent HaL HbL HcL axi axi axi Solution to H.9 (a) The moment of a flat aµb rectangle of ma M about a perpendicular axi through the center i I = 1 1 M Ia + b M. Thi follow from the perpendicular axi theorem. (For a flat body in xy-plane, I z = I x + I y ) Here take the x and y axe to axe through the center that are parallel to the ide. For I x and I y the moment i equivalent to a uniform rod about a perpendicular axi through the center. It follow that for a quare I z = 1 6 M L. Now take the x and y axe to be the diagonal. By ymmetry I x = I y = I and by the perpendicular axi theorem I z = I x + I y ï I = 1 I z = 1 1 M L (b) I z = 1 M R. Now take the x and y axe to be the perpendicular diameter.. By ymmetry I x = I y = I and by the perpendicular axi theorem I z = I x + I y ï I = 1 I z = 1 4 M R (c) Ue the parallel axi theorem to find thi moment in term of the previou.. The ditance from the center of ma to the axi i D = R. The axi parallel to the tangent through the center i the diameter, o I cm i the anwer to part (b). I = I cm + M D = 1 4 M R + M R = 5 4 M L. Problem H.10 (a) A uniform olid cylinder of ma m and iu R rotate about an axi at it rim parallel to the central axi. What i it moment of inertia. (b) What i the moment of inertia of a uniform olid phere of ma m and iu R about an axi tangent to it urface. (c) What i the moment of inertia of a uniform thin-helled hollow phere of ma m and iu R about an axi tangent to it urface. Solution to H.10 For all three cae we ue the parallel axi theorem.

9 (a) The moment of a flat aµb rectangle of ma M about a perpendicular axi through the center i I = 1 1 M Ia + b M. Thi follow from the perpendicular axi theorem.

5 Chapter H - Problem 5 (a) I = I cm + M D = 1 m R + m R = 3 m R. (b) I = I cm + M D = 5 m R + m R = 7 5 m R. (c) I = I cm + M D = 3 m R + m R = 5 3 m R.

PHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013

PHYSICS 111 HOMEWORK SOLUTION #9 April 5, 2013 0.1 A potter s wheel moves uniformly from rest to an angular speed of 0.16 rev/s in 33 s. Find its angular acceleration in radians per second per second.