Two identical solid spheres with radius R = 6.00 cm and mass M = 2.00 kg are con

Question

Two identical solid spheres with radius R = 6.00 cm and mass M = 2.00 kg are connected by a strong massless rod. The centres of the spheres are separated by a distance d = 0.50 m. The origin is precisely midway between the two spheres. What is the moment of inertia for rotation around the x-axis? What is the moment of inertia for rotation around the y-axis? If omega_y = 3.0 rad/s is the angular velocity for rotation around the y-axis, find the angular velocity omega_x for rotation around the x-axis with the same amount of rotational energy. At time t = 0 the spheres are at rest in the position shown in the figure. From t = 0 onwards the spheres begin to rotate around the y-axis due to a constant torque of 2.0 N m. Calculate the time it takes for the first turn to be completed.

Explanation / Answer

a)
Mx = (2/5)*M*R^2 + 2/5*M*R^2
= (4/5)*M*R^2
= (4/5)*2*(0.06)^2
= 5.76*10^-3 Kgm^2
Answer: 5.76*10^-3 Kgm^2

b)
I along axis of a sphere,Iz = (2/5)*M*R^2
Use:
Ix + Iy = Iz
2*Iy = (2/5)*M*R^2
Iy = (1/5)*M*R^2

distance of sphere from y axis, r = d/2 = 0.5/2 = 0.25 m

My1 = Iy + M*r^2
= (1/5*M*R^2) + (M*0.25^2)
= (1/5*2*0.06^2) + (2*0.25^2)
= 0.12644 Kgm^2

This will add up for both sphere.
My = 0.12644*2 =0.25288 Kgm^2
Answer: 0.25288 Kgm^2

c)
KEy = KEx
0.5*Iy*wy^2 = 0.5*Ix*wx^2
Iy*wy^2 = Ix*wx^2
0.25288*3^2 = (5.76*10^-3)*wx^2
wx = 44.71 rad/s
Answer: 44.71 rad/s

d)
I = 0.25288 Kgm^2
T = 2 Nm
a = T/I
= 2 / (0.25288)
= 7.91 rad/s^2

for 1 revolution, d = 2*pi rad

yse:
d = 0.5*a*t^2
2*pi = 0.5*7.91*t^2
t = 0.2006 s
Answer: 0.2006 s

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