PHYSICS 2053 Homework problem set VII (due Friday 20 April 2018) Again I will al
ID: 2304837 • Letter: P
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PHYSICS 2053 Homework problem set VII (due Friday 20 April 2018) Again I will allow you to work in groups (on all three problems), and expect that on the sub you will identify which student did which part of the work. 1. (42 points) In the hammer throw (a field event), the athlete hurls the hammer (essentially an i mitted c opy attached to a handle by a steel wire) for maximum distance by spinning his body with the hammer at If you're unfamiliar with the event, here are a few videos. http://www.wired.com/playbook/2012/08/erin-gilreath-hammer-throw http://www.youtube.com/watch?v-vUuz7BK09x4 http://www.youtube.com/watch?v Jpbgg2TRCuw http://www.youtube.com/watch?v 4qAE2PrCVhY http://www.youtube.com/watch?v wvDoVZOPosY http://www.youtube.com/watch?v-ZLi4DjcvsKM The hammer head has a mass of 7.26 kg (ignore the mass of the cable); the length from the handle to the center of the ball is 115 cm; and the thrower has a mass of 138 kg (they're usually big!). To get the motion started, he first whirls it over his head, giving it an initial angular velocity of 4.7 rad's. At that time he quickly drops into a semi-seated position, so that the hammer head and athlete are rotating about their common center of mass. While he's in that position, you can treat the system of athlete+ ball as two point masses connected by a rigid rod rotating about their common center of gravity. In that position he is gripping the handle at a point 97 cm in front of his body's center of mass. He then makes four complete rotations before releasing the hammer, during which time the angular speed of the athlete+ ball rotating system increases to 23.1 rad/s a. When the athlete is whirling with the hammer at arms' length, the system can be treated as consisting of two point masses rotating about a common center of mass. How far from his body's center of mass is the center of mass of the combined body + hammer system? Be sure you understand exactly what constitutes the center of mass. Note also that the center of mass is now the axis of rotation. Find the moments of inertia of the athlete and the ball (separately) about their common center of mass, then add them to obtain the total moment of inertia. Note that the moment of inertia of the ball should be greater than that of the athlete, even though the athlete is many times more massive. Why? Calculate the final speed of the ball just before the hammer is released. Assume that the angular addition is constant during the spin-up. Find the angular acceleration needed to get the hammer up to release velocity over four revolutions, again assuming it is constant. How much torque must the thrower apply to the hammer to achieve that rate of angular acceleration? b. c. d. e. (30 points) To ensure an adequate water supply, a town has a huge water tower in which the maximum water depth is 22.4 m. Its top is sealed and pressurized to a gauge pressure of I Pa. From the bottom of the tank there emerges a horizontal pipe that feeds water to the rest of the system. It has a cross-sectional area of 0.0707 m2. Tapping into that pipe is a second, smaller horizontal pipe with 2. 1.66 x 10 a cross-sectional area of 0.0255 m2. Assume three significant figures throughout, and briefly explain what steps you're taking, and why a. The smaller pipe is then opened to the air. Find the speed of the water as it comes out of that pipeExplanation / Answer
1. given mass of hammer, Mh = 7.26 kg
length from handle to center of mass of hammer, l = 115 cm
thrower mass, M = 138 kg
wi = 4.7 rad/s
distance from handle to center of mass of the body, d = 97 cm
wf = 23.1 rad/s
distance of com of the rotating system from the hammer = x
x = (M(d + l))/(M + Mh) = 138(2.12)/(145.26)
x = 2.014043783 m
hence, linear velocity of the hammer before throwing
v = w*x = 46.52441140024 m/s
a. distance of combined center of mass form body's center of mass = (l + d) - x = 0.105956217 m
b. moment of inertia of athlete about common center of mass = M*((l + d) - x)^2 = 1.549287349091 kg m^2
moment of inertia of the ball about common center of mass, Ib = Mh(x)^2 = 29.449263 kg m^2
total moment of inertia = I = Ib + Ia = 30.998550681536304 kg m^2
total momnt of inertia is greater than individual moment of inertia, because it is a scalar and it adds or non negative numbers
c. v = 46.5244 m/s
d. alpha = ?
2*alpha*4*2*pi = wf^2 - wi^2
alpha = 10.176367 rad/s/s
e. torque = Itotal * alpha = 315.452630103 Nm
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