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A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other. The rope does not slip on the pulley. When the velocity of the heavier block is 2.0 m/s the total kinetic energy of the pulley and blocks is:

Respuesta :

Rau7star

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x [tex]10^{-3}[/tex] kgm²

mass on one side of rope '[tex]m_{1[/tex]'= 2kg

mass on other side of rope'[tex]m_{2[/tex]' =4kg

velocity'v' of mass [tex]m_{2[/tex]' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x [tex]10^{-2[/tex]

‎ω = 66.67 rad/s

For the rotating body, we have

KE = [tex]\frac{1}{2}[/tex] I ω²

[tex]KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )[/tex]

[tex]KE_p[/tex] = 10J

Next is to calculate kinetic energy of the blocks :

[tex]KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2[/tex]

[tex]KE_b[/tex]=12J

Therefore, the total kinetic energy will be

KE = [tex]KE_p + KE_b[/tex] =10 + 12

KE= 22J

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