A) \[\frac{a{{\omega }^{2}}}{2g}\]
B) \[\frac{2a{{\omega }^{2}}}{g}\]
C) \[\frac{3a{{\omega }^{2}}}{2g}\]
D) \[\frac{a{{\omega }^{2}}}{3g}\]
Correct Answer: C
Solution :
Let O is centre of rotation, then |
In above figure, |
T =tension in string when masses about to slip. |
Accelerations of A and B are |
\[{{a}_{A}}=a{{\omega }^{2}}\] |
\[{{a}_{B}}=2a{{\omega }^{2}}\] |
Frictions of A and B are |
\[{{f}_{A}}=\mu mg\] |
\[{{f}_{B}}=\mu mg\] |
Net force on A is |
\[{{F}_{A}}=\mu mg-T=ma{{\omega }^{2}}\] |
Net force on B is |
\[{{F}_{B}}=\mu mg+T=m(2a){{\omega }^{2}}\] |
For no slip; \[{{F}_{A}}+{{F}_{B}}\] |
\[{{f}_{A}}+{{f}_{B}}=3am{{\omega }^{2}}\]\[\Rightarrow \]\[\mu =\frac{3a{{\omega }^{2}}}{2g}\] |
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