A) \[Mg\ell \]
B) \[Mg\ell (1+{{\theta }_{0}}^{2})\]
C) \[Mg\ell (1-{{\theta }_{0}}^{2})\]
D) \[Mg\ell \left( 1+\frac{{{\theta }_{0}}^{2}}{2} \right)\]
Correct Answer: B
Solution :
Angular momentum conservation. \[{{V}_{0}}L={{V}_{1}}(L-\ell )\]\[{{V}_{1}}={{V}_{0}}\left( \frac{L}{L-\ell } \right)\] \[{{\text{w}}_{g}}+{{\text{w}}_{p}}=\Delta KE\] \[-mg\ell +{{\text{w}}_{p}}=\frac{1}{2}m\left( V_{1}^{2}-V_{0}^{2} \right)\] \[{{\text{w}}_{p}}=mg\ell +\frac{1}{2}mV_{0}^{2}\left( {{\left( \frac{L}{L-\ell } \right)}^{2}}-1 \right)\] \[=mg\ell +\frac{1}{2}mV_{0}^{2}\left( {{\left( 1-\frac{\ell }{L} \right)}^{-2}}-1 \right)\] Now, \[\ell <<L\] By, Binomial approximation \[=mg\ell +\frac{1}{2}mV_{0}^{2}\left( \left( 1+\frac{2\ell }{L} \right)-1 \right)\] \[=mg\ell +\frac{1}{2}mV_{0}^{2}\left( \frac{2\ell }{L} \right)\] \[{{W}_{p}}=mg\ell +m\text{v}_{0}^{2}\frac{\ell }{L}\] here, \[{{V}_{0}}=\]maximum velocity \[=\omega \times A\] \[=\left( \sqrt{\frac{g}{L}} \right)({{\theta }_{0}}L)\] \[{{V}_{0}}={{\theta }_{0}}\sqrt{gL}\] So,\[{{\text{w}}_{0}}=mg\ell +m{{\left( {{\theta }_{0}}\sqrt{gL} \right)}^{2}}\frac{\ell }{L}\] \[=mg\ell \left( 1+\theta _{0}^{2} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
