Question

The catapult shown below is used to lay siege to your neighbor's cubicle. You fire 1-g peanut M&Ms at their desk until they surrender. The catapult arm moves from horizontal to 30° in 0.1 s, when it hits the crossbar and releases the projectile. Assume air resistance is negligible and the arm velocity is constant until it hits the crossbar. If the M&M lands on their desk 3 meters away in 1 s, determine: A. The velocity of the projectile as it leaves the bucket. B. The angular velocity of the bucket at the end of the arm just before it hits the crossbar. C. The force on the M&h as the arm hits the crossbar. (Hint: at of the bucket = a of the projectile.) D. The torsional spring constant of the bands in N.m/deg. E. The energy of the M&M just before impact on the desk. F. The velocity of the M&M after impact with the desk. Assume a coefficient of restitution of 0.8. 120 MM

Answer 1

A. Let the angular accelenation be x and ut Oy final angulare velocity of the catapault is w, and final velocity of projectile is v. svitid So, 8-12 [here, 0 = 30°, 4-0.15 oc entowe o popuz 100h Pad/8² => X2 3 االت 101 3 radſe stran again, wext Lerestori alt store 2(1000 + (0.1)) radset Loreste Now the length of the catapault arm is or o.12 m. So velocity of the brojectile as it leaves the bucket is- Vzw. p [where r= 0.12 m] = /109 x (0.12)) m/s ~ 1.26 m/s. 120 mm

lo 3 " Tako [k- torsional Spring B. As done in the previous bout, the angular velocity of the bucket at the end of the arem just before it hits the cross bar is radio x 10.4.7 rad/s c. If the moment of inertia of the Mem is I and force on M&M es f, then [r=20mm => mp² x= PF [m=19] kg Ian 1000 madla ñ 0.0126 N. D. We know I da PF 1000 = Fz mra 120 loor に 1000 1000 3 here Tz Ixamp? constant (P) [23] amp² o => k=amp2

putting the values-rocks - kg, N=2 120 m, tzols ma 1000 1000 1120 12 1000 1000 KZ N.m/rad feitution 3 = 2.88 X10 N.m/rad E. Assuming air resistance is negligible, we can conclude that energy is conserved at the time of flight. So before energy of the M&M before it leaves the catabault is- K.E= I w² 2 (1ory [wa lon rad/s 2 Tooo (1004) (lon) 2 J ~ 7.9 x 10-9 J So energy of the M&M just before et hits the ground is 7.9x1077 -

F. Let the velocity of the Mem after it hits the desk is N. Then Imve [e-coefficient of restitution] 7.9 x 10-4 txj xv² 1000 7.9x10-4 = 0.8 gider -> V = 0.8% 7.9 2 AN = 1 m/s

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