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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. |
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Answer» Correct Answer - C `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` |
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