\"A trebuchet is a device used during the middle ages to throw rocksat castles a

Question

"A trebuchet is a device used during the middle ages to throw rocksat castles and sometimes now used to fling pianos as a sport. Asimple trebuchet is shown in the textbook (figure 10.19 page 327fourth edition)" *Picture a seasaw, with one side only0.14m long and a 60kg mass attached. The other side 3m long with a0.12kg mass attached. *
"Model it as a stiff rod of negligible mass, 3m long, joingingparticlles of mass 60kg and 0.12kg at its ends. It turns on a frictionless horizontal axle perpendiclar to therod and 0.14m from the large mass particle. The rod is released from rest in a horizontalorientation. Find the maximum speed that the small mass objectattains"
I have started with a diagram with all the forces acting andcalculated the net forces. Could someone please give me a clue asto where to go from here?
Thanks Ben
*Picture a seasaw, with one side only0.14m long and a 60kg mass attached. The other side 3m long with a0.12kg mass attached. *
"Model it as a stiff rod of negligible mass, 3m long, joingingparticlles of mass 60kg and 0.12kg at its ends. It turns on a frictionless horizontal axle perpendiclar to therod and 0.14m from the large mass particle. The rod is released from rest in a horizontalorientation. Find the maximum speed that the small mass objectattains"
I have started with a diagram with all the forces acting andcalculated the net forces. Could someone please give me a clue asto where to go from here?
Thanks Ben
"Model it as a stiff rod of negligible mass, 3m long, joingingparticlles of mass 60kg and 0.12kg at its ends. It turns on a frictionless horizontal axle perpendiclar to therod and 0.14m from the large mass particle. The rod is released from rest in a horizontalorientation. Find the maximum speed that the small mass objectattains"
I have started with a diagram with all the forces acting andcalculated the net forces. Could someone please give me a clue asto where to go from here?
Thanks Ben

Explanation / Answer

I = 60 * .142 + .12 * 9 =2.256   kg-m2 the moment of inertiaof the catapult T = 60 * 9.8 * .14 - .12 * 9.8 * 3 = 78.8 N-m   thenet torque acting on the catapult = T / I = 34.9 m/s2 I don't have a copy of the book (diagram) but will assume herethat = 90 deg the maximum angle that can be utilized by the catapult = / 2 = 1/2 t2   andt = .300 sec using the values above = t = 34.9 * .300 = 10.47 / sec v = R = 10.47 * 3 = 31.4 m/s I don't have a copy of the book (diagram) but will assume herethat = 90 deg the maximum angle that can be utilized by the catapult = / 2 = 1/2 t2   andt = .300 sec using the values above = t = 34.9 * .300 = 10.47 / sec v = R = 10.47 * 3 = 31.4 m/s

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