Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m^2. Find the tension in the part of the string joiningk the pulleys.

Answer

C

Explanation

`m=2kg, I_1=0=0.10 kg-m^2`
r_1=5 cm=0.05m`
`I_2=2.20 kg-m^2`
`r_2=10xm=0.1m`
Therefore
`mg-T_1=ma`…..1
`(T_1-T_2)r_1=I_1alpha`………..2
`T_2r_2=I_2alpha`…….3
substituting the value of `T_2` in the equation 2 we get
`rarr (T_1-I_2alpha/r_1)r_2=I_1alpha`
`(T_1-I_2a/r_1^2)=I_1a/r_2^2`
`rarr T_1={(I_1/r_1^2)+(I_2/r_2^2)}a`
LSubstituting the value of `T_1` in the equation 1 we get
`rarr mg-{(I_1/r_1^2)+(I_2/r_2^2)}a=ma`
`rarr (mg)/({(I_1/r_1^2)+(I_2/r_2^2)}+m+=a`
`rarr a=(2xx9.8)/(0.1/0.0025+0.2/0.01+2)`
`=0.316m/s^2`
`rarr T_2=I_2 a/r_2^2` ltbr.gt `=(0.20xx0.316)/0.01=6.32N`