Answer:
KE = 1.75 J
Explanation:
given,
mass of ball, m₁ = 300 g = 0.3 Kg
mass of ball 2, m₂ = 600 g = 0.6 Kg
length of the rod = 40 cm = 0.4 m
Angular speed = 100 rpm= [tex] 100\times \dfrac{2\pi}{60}[/tex]
=10.47\ rad/s
now, finding the position of center of mass of the system
r₁ + r₂ = 0.4 m.....(1)
equating momentum about center of mass
m₁r₁ = m₂ r₂
0.3 x r₁ = 0.6 r₂
r₁ = 2 r₂
Putting value in equation 1
2 r₂ + r₂ = 0.4
r₂ = 0.4/3
r₁ = 0.8/3
now, calculation of rotational energy
[tex]KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2[/tex]
[tex]KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)[/tex]
[tex]KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)[/tex]
[tex]KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)[/tex]
KE = 1.75 J
the rotational kinetic energy is equal to 1.75 J
The total rotational kinetic energy of the balls is 1.78 J.
The given parameters;
The angular speed of the balls is calculated as follows;
[tex]\omega = 100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s } \\\\\omega = 10.47 \ rad/s[/tex]
The radius of the balls is calculated as;
[tex]r_1 + r_2 = 0.4[/tex]
The torque on the rod due to each is calculated as;
[tex]F_1r_1 = F_2 r_2\\\\m_1 gr_1 = m_2 g r_2\\\\m_1 r_1 = m_ 2r_2\\\\r_2 = \frac{m_1r_1}{m_2} \\\\r_2 = \frac{0.3 r_1}{0.6} \\\\r_2 = 0.5r_1[/tex]
solve for the radius;
[tex]r_1 + 0.5r_1 = 0.4\\\\1.5r_1 = 0.4\\\\r_1 = \frac{0.4}{1.5} \\\\r_1 = 0.267 \ m\\\\r_2 = 0.5(0.267)\\\\r_2 = 0.133 \ m[/tex]
The moment of inertia of each ball is calculated as follows;
[tex]I_1 = m_1 r_1^2 = 0.3 \times (0.267)^2 = 0.0214 \ kgm^2\\\\I_2 = m_2 r_2^2= 0.6 \times (0.133)^2 = 0.011 \ kgm^2[/tex]
The total rotational kinetic energy of the balls is calculated as follows;
[tex]K.E = \frac{1}{2}I_1 \omega^2 \ + \ \frac{1}{2}I_2 \omega^2 \\\\K.E = \frac{1}{2} \omega^2(I_1 + I_ 2)\\\\K.E = 0.5 \times (10.47)^2 (0.0214 + 0.011)\\\\K.E = 1.78 \ J[/tex]
Thus, the total rotational kinetic energy of the balls is 1.78 J.
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