Question

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:

Answer 1

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x $10^{-3}$ kgm²

mass on one side of rope '$m_{1$'= 2kg

mass on other side of rope'$m_{2$' =4kg

velocity'v' of mass $m_{2$' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x $10^{-2$

‎ω = 66.67 rad/s

For the rotating body, we have

KE = $\frac{1}{2}$ I ω²

$KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )$

$KE_p$ = 10J

Next is to calculate kinetic energy of the blocks :

$KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2$

$KE_b$=12J

Therefore, the total kinetic energy will be

KE = $KE_p + KE_b$ =10 + 12

KE= 22J