The pulley shown in figure has a radius 10 cm and moment of inertia 0.5 kg-m^2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

Answer

B

Explanation

`m_1gsintheta-T_1=m_1a`…….1
`T_1-T_2=I a/r^2`…..2
`T_2-m_2gsintheta=m_2a`…….3
Adding the equation 1 and 3 we will get
`m_1gsintheta+(T_2-T_1)-m_2gsin theta`
`=(m_1+m_2)a`

`rarr (m_1-m_2) g sintheta `
`=(m_1+m_2+I/r^2)a`
`rarr a=((m_1-m_2)g sin theta)/((m_1+m_2+I/r^2))`
`=((4-2)xx10xx(1sqrt2))/({(4+2)+(0.5/0.01)})`
`=((2xx10xx1/sqrt2))/(6+50)`
`=0.248=0.25m/s^2`