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 phy131studiof15:lectures:m2p2sol [2015/10/26 09:18]mdawber [Question 1 Solution] phy131studiof15:lectures:m2p2sol [2015/10/26 09:35]mdawber [Question 3 Solution] Both sides previous revision Previous revision 2015/10/26 09:38 mdawber 2015/10/26 09:35 mdawber [Question 3 Solution] 2015/10/26 09:22 mdawber [Question 2 Solution] 2015/10/26 09:18 mdawber [Question 1 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/20 11:53 mdawber [Question 2 Solution] 2015/10/20 11:51 mdawber 2015/10/20 11:51 mdawber [Midterm 2 Q2 Solution (Ave Score: 29.1/35)] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:48 mdawber created Next revision Previous revision 2015/10/26 09:38 mdawber 2015/10/26 09:35 mdawber [Question 3 Solution] 2015/10/26 09:22 mdawber [Question 2 Solution] 2015/10/26 09:18 mdawber [Question 1 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/20 11:53 mdawber [Question 2 Solution] 2015/10/20 11:51 mdawber 2015/10/20 11:51 mdawber [Midterm 2 Q2 Solution (Ave Score: 29.1/35)] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:48 mdawber created Last revision Both sides next revision Line 60: Line 60: {{phy141f12mid2fig2.png}} {{phy141f12mid2fig2.png}} - a) $I=2\times2\times1.2^{2}+2\times3\times(\frac{1}{12}\times1.1^{2}+0.65^{2})+\frac{1}{2}\times50\times0.1^2$ ​ + A robot iceskater is initially at rest and then fires rockets on the end of it's arms, each of which exerts a force of 100N for 0.5s. Once the rockets have turned off and the robot has reached a constant angular velocity, the robot raises its arms, leading to a change in its angular velocity. + + (a) (10 points) Find the moment of inertia of the robot when it's arms are extended considering that the rockets weigh 2 kg each, each arm weighs 3 kg, the body of the robot is a 50 kg cylinder with radius 10cm. The arms have length 1.1m and are attached to the edge of the cylinder. The rockets may be considered as point masses. You may neglect the moment of inertia of the robot'​s head. + + $I=2\times2\times1.2^{2}+2\times3\times(\frac{1}{12}\times1.1^{2}+0.65^{2})+\frac{1}{2}\times50\times0.1^2$ ​ $I=5.76+3.14+0.25=9.15\,​\mathrm{kg\,​m^{2}}$ $I=5.76+3.14+0.25=9.15\,​\mathrm{kg\,​m^{2}}$ - b) $\sum \tau =2 \times 100 \times 1.2 =240\,​\mathrm{N\,​m}$ + (b) (5 points) What is the angular acceleration of the robot during the 0.5s that the rockets are on? + + $\sum \tau =2 \times 100 \times 1.2 =240\,​\mathrm{N\,​m}$ $\alpha=\frac{\sum \tau}{I}=\frac{240}{9.15}=26.23\,​\mathrm{s^{-2}}$ $\alpha=\frac{\sum \tau}{I}=\frac{240}{9.15}=26.23\,​\mathrm{s^{-2}}$ - c)$\omega=26.23\times0.5=13.115\mathrm{s^{-1}}=125.24\mathrm{rpm}$ + (c) (5 points) What is the angular velocity achieved when the rockets have been turned off but the robot has not yet raised it's arms? Give your answer in both s$^{-1}$ and rpm. - d)$L=I\omega=9.15\times13.115=120\,\mathrm{kg\,m^{2}\,s^{-1}}$ + $\omega=26.23\times0.5=13.115\mathrm{s^{-1}}=125.24\mathrm{rpm}$ - e)$I=2\times5\times0.1^{2}+\frac{1}{2}\times50\times0.1^{2}=0.35\,\mathrm{kg\,m^{2}}$ + (d) (5 points) What is the angular momentum when the rockets have been turned off but the robot has not yet raised it's arms? Give magnitude and direction (ie. left, right, up, down, in to page, out of page)? - f) Angular momentum is conserved so + $L=I\omega=9.15\times13.115=120\,​\mathrm{kg\,​m^{2}\,​s^{-1}}$ + + (e) (5 points) What is the moment of inertia of the robot after it has raised it's arms? + + $I=2\times5\times0.1^{2}+\frac{1}{2}\times50\times0.1^{2}=0.35\,​\mathrm{kg\,​m^{2}}$ + + (f) (5 points) What is the angular velocity of the robot once it has raised it's arms? Give your answer in both s$^{-1}$ and rpm. + + Angular momentum is conserved, so $\omega=\frac{120}{0.35}=342.9\mathrm{s^{-1}}=3274\mathrm{rpm}$ $\omega=\frac{120}{0.35}=342.9\mathrm{s^{-1}}=3274\mathrm{rpm}$ - g) $v=\omega r=342.9\times0.1=34.29\,​\mathrm{m\,​s^{-1}}$ + (g) (5 points) What is the magnitude of the linear velocity of one of the rockets once the robot has raised it's arms? + + $v=\omega r=342.9\times0.1=34.29\,​\mathrm{m\,​s^{-1}}$ ===== Question 3 Solution ===== ===== Question 3 Solution ===== Line 84: Line 100: {{m2fig2f11.png}} {{m2fig2f11.png}} - A. Sum of torques on beam around hinge + A Jack-O'​-Lantern of mass 4kg is to be suspended as shown in the diagram using a hinged uniform beam of mass 2kg and length 0.8m and a massless string. The  beam should be level and the string at 30º to the horizontal. + + (a) (10 points) If the maximum tension the string can support without breaking is 50N, what is the furthest distance from the hinge, $x$, that I can hang the Jack-O'​-Lantern. + + Sum of torques on beam around hinge $0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ $0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ Line 90: Line 110: $x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$ $x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$ - B. Sum of horizontal forces on beam + (b) (10 points) If I hang the Jack-O'​-Lantern at the distance found in part (a), (ie. when the tension in the string is 50N), what is the magnitude of the net force on the hinge? What is the direction of the force (give your answer in terms of the angle $\theta$ from the $y$ axis shown in the diagram)? + + Sum of horizontal forces on beam $F_{HH}-T\cos30^{o}=0$ $F_{HH}-T\cos30^{o}=0$ Line 108: Line 130: But the force **on** the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$ But the force **on** the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$ - C. Sum of torques on beam around hinge + + (c) (10 points) If I would like to hang the Jack-O'​-Lantern at the far end of the beam from the hinge using the same string as was used in parts (a) and (b), what is the minimum angle the string should make with the horizontal instead of the $30^{o}$ angle it makes in parts (a) and (b). + + Sum of torques on beam around hinge $0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ $0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$