A) \[{{\left[ \frac{{{\mu }^{2}}{{g}^{2}}-{{a}^{2}}}{{{r}^{2}}} \right]}^{\frac{1}{2}}}\]
B) \[{{\left[ \frac{\mu g-a}{3r} \right]}^{\frac{1}{2}}}\]
C) \[{{[({{\mu }^{2}}{{g}^{2}}-{{a}^{2}}){{r}^{2}}]}^{\frac{1}{4}}}\]
D) \[{{\left[ \frac{{{\mu }^{4}}{{g}^{2}}-{{a}^{4}}}{{{r}^{2}}} \right]}^{\frac{1}{4}}}\]
Correct Answer: C
Solution :
The centripetal acceleration \[{{a}_{c}}=\frac{{{v}^{2}}}{r},\]where \[v=\]speed in a direction and tangential acceleration (as there is non uniform circular motion) \[{{a}_{t}}=\frac{dv}{dt}=a\] The resultant acceleration is \[{{a}_{R}}=\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}+{{a}^{2}}}\] Thus force on the vehicle \[F=m\,{{a}_{R}}=m\sqrt{{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}+{{a}^{2}}}\] At the condition when vehicle will skid \[F=\text{friction}\,=f\] \[\Rightarrow \] \[m\sqrt{{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}+{{a}^{2}}}=\mu \,mg\] \[\Rightarrow \] \[v={{[({{\mu }^{2}}{{g}^{2}}-{{a}^{2}}){{r}^{2}}]}^{1/4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
