A metre stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end figure. A particle of mass 100 g is attached to the upper end of the stick through a light sting of length 1 m. Initially the rod is kept veritcal and the string horizontal when the system is released from rest. The particle colides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.

Answer

A, D

Explanation

`1/2Iomega^2-0=0.1xx10xx1`
`rarr omega=sqrt20`
for collision `0.1xx12xxsqrt20x+0`
`=[((0.24)/3)xx12+(0.1)^2(1)^2]omega`
`rarr omega=sqrt(20)/([10xx(0.18)])`
rarr 0-1/2Iomega^2` =-m_1gI(1-costheta)`
`-m_2g1/2 (1-costheta)`
`rarr 1/2Iomega^2=-mgI(1-costheta)`
`-m-2g 1/2(1-costheta)`
`=0.1x10(1-costheta)`
`=0.1xx10(1-costheta)`
`-0.24xx10xx0.5(1-costheta)`
`rarr 1/2x0.18xx(20/324)=2.2x9(1-costheta)`
`rarr (1-costheta)=1/((2.2xx1.8))`
`rarr 1-costheta=0.252`
`rarr costheta=0.252`
`costheta=1-0.252=0.748`
`rarr W=cos^-1(0.748)=41^@`