Question

A particle of mass m is moving along a trajectory given by
$x={{x}_{\circ }}+a\cos {{\omega }_{1}}t$
$y={{y}_{\circ }}+b\sin {{\omega }_{2}}t$
The torque acting on the particle about the origin, at t=0 is:
$\begin{aligned} & a)m({{x}_{\circ }}b+{{y}_{\circ }}a){{\omega }_{1}}^{2}\widehat{k} \\\ & b)+m{{y}_{\circ }}a{{\omega }_{1}}^{2}\widehat{k} \\\ & c)-m({{x}_{\circ }}b{{\omega }_{2}}^{2}+{{y}_{\circ }}a{{\omega }_{1}}^{2})\widehat{k} \\\ & d)Zero \\\ \end{aligned}$

Answer 1

The position of the particle with respect to the x axes and the y axes is given. In the question it is asked to determine the torque on the particle at t=0. Hence obtaining the required position vector of the particle will allow us to determine the torque about the origin.
Formula used:
$\overline{r}=\overline{x}\widehat{i}+\overline{y}\widehat{j}$
$a=\dfrac{{{d}^{2}}r}{d{{t}^{2}}}$
$\tau =r\times F$

Complete answer:
The position vector of any particle if its coordinates are known i.e. its x coordinate is ‘x’ and y coordinate is ‘y’ then the position with respect to origin is given by,
$\overline{r}=\overline{x}\widehat{i}+\overline{y}\widehat{j}$
Where $\widehat{i},\text{ }\widehat{j}$ are the unit vectors. Hence the position vector of the above particle is given by,
$\overline{r}=({{x}_{\circ }}+a\cos {{\omega }_{1}}t)\widehat{i}+({{y}_{\circ }}+b\sin {{\omega }_{2}}t)\widehat{j}$
The acceleration of the particle is given by the double derivative of the position vector. Hence the acceleration ‘a’ of the particle at any time t is
$\begin{aligned} & \dfrac{d\overline{r}}{dt}=(-a{{\omega }_{1}}\sin {{\omega }_{1}}t)\widehat{i}+({{\omega }_{2}}b\cos {{\omega }_{2}}t)\widehat{j} \\\ & a=\dfrac{{{d}^{2}}\overline{r}}{d{{t}^{2}}}=(-a{{\omega }_{1}}^{2}\cos {{\omega }_{1}}t)\widehat{i}+(-{{\omega }_{2}}^{2}b\sin {{\omega }_{2}}t)\widehat{j} \\\ \end{aligned}$
From Newton’s second law the force on the particle is given by the dot product of mass into acceleration. Hence the force (F)on the particle at any time ‘t’ is equal to
$\begin{aligned} & F=ma \\\ & \Rightarrow F=m\left[ (-a{{\omega }_{1}}^{2}\sin {{\omega }_{1}}t)\widehat{i}+(-b{{\omega }_{2}}^{2}\cos {{\omega }_{2}}t)\widehat{j} \right] \\\ \end{aligned}$
At time t=0,
$\begin{aligned} & \overline{r}=({{x}_{\circ }}+a\cos {{\omega }_{1}}t)\widehat{i}+({{y}_{\circ }}+b\sin {{\omega }_{2}}t)\widehat{j} \\\ & \Rightarrow \overline{r}=({{x}_{\circ }}+a\cos {{\omega }_{1}}(0))\widehat{i}+({{y}_{\circ }}+b\sin {{\omega }_{2}}(0))\widehat{j} \\\ & \Rightarrow \overline{r}=x{{\widehat{i}}_{\circ }}+a\widehat{i}+{{y}_{\circ }}\widehat{j}....(1) \\\ \end{aligned}$
And
$\begin{aligned} & F=m\left[ (-a{{\omega }_{1}}^{2}\cos {{\omega }_{1}}(0))\widehat{i}+(-b{{\omega }_{2}}^{2}\sin {{\omega }_{2}}(0))\widehat{j} \right] \\\ & \Rightarrow F=-ma{{\omega }_{1}}^{2}\widehat{i}.....(2) \\\ \end{aligned}$
The torque ($\tau$ )about the origin of a particle is given by the cross product between the position vector and the force acting on the particle i.e. $\tau =r\times F$
From 1 and 2,
$\begin{aligned} & \tau =r\times F \\\ & \Rightarrow \tau =({{x}_{\circ }}\widehat{i}+a\widehat{i}+{{y}_{\circ }}\widehat{j})\times (-ma{{\omega }_{1}}^{2}\widehat{i}),\text{ }\because \widehat{i}\times \widehat{i}=0,\text{ }\widehat{j}\times \widehat{i}=-\widehat{k} \\\ & \Rightarrow \tau =-ma{{\omega }_{1}}^{2}{{y}_{\circ }}(-\widehat{k})=ma{{\omega }_{1}}^{2}{{y}_{\circ }}\widehat{k} \\\ \end{aligned}$

Hence the correct answer of the above question is option b.

Note:
The position vector of the particle is taken with respect to origin. Hence the torque obtained will also be with respect to the origin. Also the force acting and the acceleration are in the same direction for the particle.