Question

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at 100 rpm. What is its rotational kinetic energy?

Answer 1

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= $100\times \dfrac{2\pi}{60}$

                         =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

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

 KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J