| Math-Hammer; Let's Build a Catapult! | |
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| Tweet Topic Started: Oct 5 2011, 02:15 PM (379 Views) | |
| James Baillie | Oct 5 2011, 02:15 PM Post #1 |
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Righty, let's go to Tilea for our final problem. The Duke of the small city of Mantaro is considering two propositions for new pieces of lightweight ordinance from esteemed engineers locally. How to know which engineer is telling the truth that his machine is better? Rather than torturing them as usual, he's decided to see if he can work out which will fire further. Don Felipano of Tobaro is a cannon founder. His cannon can provide a force of 3000N to a lightweight 2kg cannonball in 0.05 seconds. It sits a metre off the ground. Nicolo Alattiano has built a catapult, and is offering the hire of an ogre called Mergug to hand-fire it. Mergug is strong enough to pull the lever round in a single movement, taking only a tenth of a second! From its pivot a metre and a half from the ground, the three and a half metre firing arm will then hurl a 5kg rock at any foes stupid enough to get in the way. The Duke thinks the cannon will go further due to being the more modern weapon, but wants to be sure. Despite these being two very different weapons, it's not too hard for us to work out how far they'll fire. First, we need to find their starting speeds for the projectiles... THE CANNON Now we're looking back at impulse and momentum, p=mv and I=Ft (http://s1.zetaboards.com/A_Call_to_Arms/topic/4328711 for a recap). The impulse given is force times time, which is 3000 times 0.05. That gives us 150kgm/s of impulse, which is then the momentum. The starting velocity can be found with p=mv; divide 150 by the mass, 2kg, and we have a cannonball firing seventy-five metres per second. THE CATAPULT This is last month's circular motion equations: http://s1.zetaboards.com/A_Call_to_Arms/topic/4526406 W = 2Pi/T The time period T is four times the quarter-circle through which the arm actually moves; 0.1 by 4 so it's 0.4 seconds. 2 Pi divided by 0.4 is 5 Pi (dividing by 2/5 is the same as multiplying by 5/2, so we can then cancel out the 2 from 2 Pi and the 2 on the underside of the 5/2 to leave 5 Pi / 1). That gives us our angular speed. V = Wr So our velocity is that times three and a half, or 17.5 Pi. (That's about 55 metres per second). Now we're just dealing with two projectiles, like the ones from Bryn Glas: http://s1.zetaboards.com/A_Call_to_Arms/topic/4311220 The maths is actually easier this time since we're assuming flat ground. CANNON Vertical: S = -1 U = 0 V = A = -9.8 T = Horizontal: S = U = 75 V = A = 0 T = The vertical displacement at the end will be -1: the ground is a metre below the cannon. We want to find how far it moves, so we need S = UT + 1/2A T^2 for the vertical to find how long it takes to hit the ground, then we can plug that value in to the horizontal. -1 = -4.9T^2 so T^2 is 0.204, and T is 0.452 seconds. Using the same equation the other way round, we can find the horizontal distance the cannonball travels. S = UT + 1/2A T^2 again, A is 0 so we can ignore that, S is then 0.452 times 75. The cannonball goes thirty-four metres; given that we haven't given it the benefit of being angled upwards at all or being on a slope, that's not too bad. Now... CATAPULT Vertical: S = -5 U = 0 V = A = -9.8 T = Horizontal: S = U = 55 V = A = 0 T = The firing arm is higher, but the projectile is slower. S = UT + 1/2A T^2 again, U still 0, -5 = -4.9T^2, 1.02 = T^2, T = 1.01. So the rock flies for longer before hitting the ground - but does it fly further? S = UT as A is 0, so S = 1.01 times 55... That's 55.56 metres fired! A resounding victory for Alattiano's device! And to think, the Duke might've chosen the cannon. Useful stuff, maths, really. Edited by James Baillie, Nov 18 2011, 06:32 PM.
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8:46 AM Jul 11