Problem 1 Gravity Powered Catapult: Maximum Thrown Distance. A gravity powered c

Question

Problem 1 Gravity Powered Catapult: Maximum Thrown Distance.

A gravity powered catapult is released from its locked and loaded at rest position at ?0 = - 80 deg., rotates in a frictionless support at O located an elevation ht = 4.1 m above grade. Upon firing, the catapult arm rotates freely, counter-clockwise then (magically) releases a projectile at position ?1= + 45 degrees. Determine the maximum thrown distance (at first strike). Refer to Figures 1 and 2 below. Ignore the mass of the projectile and assume the aerodynamic losses are negligible. The size, mass and location parameters for catapult arm (rod, plate counter-weight and support) are as follows:

Rod: L = 4.50 m         massROD = 2.30 kg

Plate: a = 0.40 m           b = 0.80 m            mPLATE = 110 kg

Location of @ Pt. O: LAO = 4.00 m

1) Maximum Distance????

Problem 2 Gravity Powered Catapult: Maximum Force

The rotating catapult described above in Problem 1 exerts force on the support as it freely rotates about the axis at point O. Let

Position of Maximum Loading Locked and loaded, at angular position theta 0, at rest with projectile fixed to catapult pt A. Projectile released, at "optimum' angular position theta 1, with velocity at release, VA1, dependent on angular velocity of catapult and rA/ . Projectile Strike: at distance w / r / t origin at velocity of strike, V2, dependent on projectile motion in field of gravity ignoring aero effects.

Explanation / Answer

The momentum of inertia of the plate with respect to an axis through its center of mass (its center) is

I0 =(m/12)(a^2+b^2) =110/12*(0.4^2+0.8^2) =7.33 kg*m^2

The point of rotation O is at distance d=0.5+a/2 =0.5+0.2 =0.7 m from plate center

Moment of inertia of plate with respect to point O (see Steiner Theorem) is

I1 =I0 +m*d^2 =7.33+110*0.7^2 =61.23 kg*m^2

The moment of inertia of the rod can be neglected since its mass is much lower than that of the plate.

Now we need to find the position of mass center CM on the rod.

Taking point O as origin of x axis we have

X(CM)*(110+2.3) =X(rod)*M(rod) +X(plate)*M(Plate)

X(rod) =-(4-(4.5/2)) =-1.75 m

X(plate) =0.5+0.2 =0.7m

X0(cm) =(-1.75*2.3 +0.7*110)/112.3 =0.65 m with respect to point O

X1(cm) =0.65 +4 =4.65 m with respect to end of the rod

Now we apply energetic considerations. Intitially there is only potential energy. finally there is kinetic energy of rotation + potential energy.

H1 =ht + X0(cm) *sin(80) =ht +0.65*sin(80) =ht +0.64 m =4.1+0.64 =4.74 m

H2 =ht- X0(cm)*sin(45) =4.1 -0.65*sin(45) =3.64 m

(M+m)*H1 =(m+M)*H2 + I1*omega^2/2

I1*omega^2 = 2*(m+M)*(H2-H1) =2*(2.3+110)*1.1 =247.06 J

omega = sqrt(247.06/I1) =sqrt(247.06/61.23) = 2 rad/sec

the launch speed is

V =omega*R   where R =4 m

V = 2*4 =8 m/s at an angle of 45 degree.

Vx =V*cos(45) =5.66 m/s

Vy =V*sin(45) =5.66 m/s

Time to maximum height is

0 = Vy -g*t1   t1 = Vy/g = 5.66/9.81 =0.577 sec

Initial launch height is

H0 = ht +4*sin(45) =4.1+4*sin(45) =6.93 m

Max height is

Hmax =H0+ Vy^2/(2g) =6.93 +5.66^2/2/9.81 =8.56 m

Time from max height to ground is

Hmax =gt2^2/2

t2 =sqrt(2*Hmax/g) =(2*8.56/9.81) =1.32 sec

Horizontal range is

D =(t1+t2)*Vx = (1.32+0.577)*5.66 =10.74 m

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