| \[x={{x}_{0}}+a\cos {{\omega }_{1}}t\] |
| \[y={{y}_{0}}+b\sin {{\omega }_{2}}t\] |
| The torque, acting on the particle about the origin, at t = 0 is : |
A) \[m(-{{x}_{0}}b+{{y}_{0}}a)\omega _{1}^{2}\hat{k}\]
B) \[+m{{y}_{0}}a\omega _{1}^{2}\hat{k}\]
C) \[-m({{x}_{0}}b\omega _{2}^{2}-{{y}_{0}}a\omega _{1}^{2})\hat{k}\]
D) Zero
Correct Answer: B
Solution :
\[F=-m(a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{i}+b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{j})\] \[\vec{r}=({{x}_{0}}+a\cos {{\omega }_{1}}t)\hat{i}+({{y}_{0}}+b\sin {{\omega }_{2}}t)\hat{j}\] \[\vec{T}=\vec{r}\times \vec{F}=-m({{x}_{0}}+a\cos {{\omega }_{1}}t)b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{k}\] \[+m({{y}_{0}}+b\sin {{\omega }_{2}}t)a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{k}\] \[=ma\omega _{1}^{2}{{y}_{0}}\hat{k}\]
You need to login to perform this action.
You will be redirected in
3 sec
