A pendulum consists of a slender rod, AB , of mass M r = 4.95 kg and a wooden sp

Question

A pendulum consists of a slender rod, AB, of mass Mr = 4.95 kg and a wooden sphere of mass Ms = 13.2 kg .(Figure 1) The length of the rod is d1 = 4.35 m and the radius of the sphere is R = 0.100 m . A projectile of mass Mp = 0.250 kg strikes the center of the sphere at a velocity of v1= 402 m/s and becomes embedded in the center of the sphere.

Part C

What is , the maximum angle measured from the vertical that the pendulum will swing, after the projectile impacts the pendulum?

Express your answer numerically in degrees to three significant figures.

I got = 1.50  rad/s in part B and now I am stuck on part C because I am getting an answer that does not make sense. (cos = -12.77 which can not be solved). If you can help me I would really appreciate it.

Explanation / Answer

given
mass of slender rod, Mr = 4.95 kg
mass of sphere, Ms = 13.2 kg
length of rod, d1 = 4.35 m
radisu of sphere , R = 0.1 m
mass of projectile, Mp = 0.25 kg
velocity of projectile vr = 402 m/s

now, momnt of inertial of the sphere about the point A = 2(Ms)R^2/5 + Ms*(R + d1)^2 = I1 = 13.2(0.1+4.35)^2 + 2*13.2*0.1^2/5 = 261.4458 kg m^2
Moment of inertial of the rod about A = Mr*(d1)^2/3 = I2 = 4.95*4.35^2/3 = 31.222125 kg m^2
moment of inertial of the projectile about A = Mp*(d1 + R)^2 = I3 = 0.25(4.45^2) = 4.950625 kg m^2
let the angular velocity of the system about A just after collision be w2

from conservation of angular momentum abtou A
Mp*v1*(d1 + R) = (I1 + I2 + I3)w2
putting in values
w2 = 0.25*402(4.45)/(297.61855) = 1.5026 rad / sec

so initial KE of this system = 0.5(I1 + I2 + I3)(w2)^2 = 336.0176 J

now let the maximum angle be theta
at this point PE of the system is
PE = [Mr*d1/2 + (Ms + Mp)*((d1 + R)) ]g(1 - cos(theta))
from conservation of energy
[Mr*d1/2 + (Ms + Mp)*((d1 + R)) ]g(1 - cos(theta)) = 336.0176
(4.95*4.35/2 + (13.45(4.45)))g(1 - cos(theta)) = 336.0176
theta = 59 deg

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