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 — phy131studiof17:lectures:m2p2sol [2017/08/04 13:41] (current) Line 1: Line 1: + ~~SLIDESHOW~~ + ====== Midterm 2 Practice Exam 2 Solutions ====== + + + ===== Question 1 Solution===== + + {{phy141f12mid2fig1.png}} + + A 100g solid ball of radius 2 cm on a string is raised to an angle of 50º and then released. The distance from the point at which the string is attached to the center of the ball is 1m. The ball swings down and hits an identical ball. + + A. (5 points) How much kinetic energy does the ball on the string have just before it hits the second ball? + + $\Delta h= 1-\cos50^{o}=0.357\mathrm{m}$ + + $KE=-mg\Delta h=0.1\times 9.8\times0.357=0.35 \mathrm{J}$ + + B. (5 points) The collision is elastic. How much kinetic energy does the second ball have just after the collision? + + All of the kinetic energy is transferred to the second ball so + + $KE=0.35\mathrm{J}$ + + C. (5 points) The second ball then rolls up a 15º incline to a height of 10 cm. What is the magnitude of the velocity at the top of the slope. + + KE at top of ramp is $0.35-0.1\times9.8\times0.1=0.252\mathrm{J}$  ​ + + For a rolling ball $KE=\frac{1}{2}m(1+\frac{2}{5})v^{2}$ so + + $v^{2}=\frac{0.252\times 2\times 5}{0.1\times7}=3.6$ + + $v=1.9\mathrm{m\,​s^{-1}}$ + + D. (5 points) What is the angular momentum of the ball at this point? Give  magnitude and direction (ie. left, right, up, down, in to page, out of page). + + $L=I\omega$ + + $I=\frac{2}{5}mr^{2}=\frac{2}{5}\times0.1\times.02^{2}=0.000016\mathrm{kg\,​m^{2}}$ + + $\omega=\frac{v}{r}=\frac{1.9}{0.02}=95\mathrm{s^{-1}}$ + + $L=0.00152\mathrm{kg\,​m^{2}\,​s^{-1}}$ in to the page + + E. (10 points) The ball then flies off the ramp. At what distance $x$ from the bottom of the ramp doe it strike the ground? ​ + + + First we need to find the time it takes until the ball hits the ground + + $0=0.1+1.9\sin 15^{o}t-\frac{1}{2}gt^{2}$ + + $4.9t^{2}-0.49t-0.1=0$ + + $t=0.2 \mathrm{s}$ + + $x=1.9\times\cos 15^{o}t$ + + $x=0.37\mathrm{m}$ + + + ===== Question 2 Solution ===== + + {{phy141f12mid2fig2.png}} + + 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. + + $I=2\times2\times1.2^{2}+2\times3\times(\frac{1}{12}\times1.1^{2}+0.65^{2})+\frac{1}{2}\times50\times0.1^2$ ​ + + $I=5.76+3.14+0.25=9.15\,​\mathrm{kg\,​m^{2}}$ + + B. (5 points) What is the angular acceleration of the robot during the 0.5s that the rockets are on? + + $\sum \tau =2 \times 100 \times 1.2 =240\,​\mathrm{N\,​m}$ + + $\alpha=\frac{\sum \tau}{I}=\frac{240}{9.15}=26.23\,​\mathrm{s^{-2}}$ + + 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. + + $\omega=26.23\times0.5=13.115\mathrm{s^{-1}}=125.24\mathrm{rpm}$ + + 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)? + + $L=I\omega=9.15\times13.115=120\,​\mathrm{kg\,​m^{2}\,​s^{-1}}$ + + 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 + + $\omega=\frac{120}{0.35}=342.9\mathrm{s^{-1}}=3274\mathrm{rpm}$ + + 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 ===== + + {{m2fig2f11.png}} + + 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 + + $0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ + + $x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$ + + B. (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 + + $F_{HH}-T\cos30^{o}=0$ + + $F_{HH}=43.3\mathrm{N}$ in the opposite direction to the tension. + + Sum of vertical forces on beam + + $F_{HV}-4g-2g+T\sin30^{o}=0$ + + $F_{HV}=33.8\mathrm{N}$ up + + $F_{NET}=\sqrt{43.3^2+33.8^{2}}=54.93\mathrm{N}$ + + Direction of force on beam by hinge $\tan^{-1}=\frac{-43.3}{33.8}=-52^{o}$ + + But the force **on** the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$ + + + 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 + + $0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ + + $\sin\phi=\frac{4\mathrm{kg}\times g+\frac{2\mathrm{kg}}{2}\times g}{50\mathrm{N}}$ + + $\phi=78.5^{o}$