JEE Main 2023 — Rotational Motion Question with Solution

Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $$\vec{F}=-P\hat{k}$$.

To find the torque, we first find the position vector of point A with respect to point B:

$${\vec{r}}_{AB}={\vec{r}}_A-{\vec{r}}_B=(0-2)\hat{i}+(0-(-3))\hat{j}+(0-0)\hat{k}=-2\hat{i}+3\hat{j}$$

To calculate the cross product, we can use the determinant method with a 3x3 matrix:

$$\vec{\tau }={\vec{r}}_{AB}\times \vec{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 3& 0\\ 0& 0& -P\end{array}\right|$$

Now, we will calculate the cross product components by expanding the determinant along the first row:

1. $${\tau }_i=\hat{i}\left|\begin{array}{cc}3& 0\\ 0& -P\end{array}\right|=\hat{i}((3)(-P)-(0)(0))=-3P\hat{i}$$

2. $${\tau }_j=-\hat{j}\left|\begin{array}{cc}-2& 0\\ 0& -P\end{array}\right|=-\hat{j}((-2)(-P)-(0)(0))=-2P\hat{j}$$

(Notice the negative sign in front of the $$\hat{j}$$ term, as it comes from the expansion of the determinant.)

3. $${\tau }_k=\hat{k}\left|\begin{array}{cc}-2& 3\\ 0& 0\end{array}\right|=\hat{k}((-2)(0)-(3)(0))=0\hat{k}$$

Now, combine the components to get the torque vector:

$$\vec{\tau }=-3P\hat{i}-2P\hat{j}+0\hat{k}=-3P\hat{i}-2P\hat{j}$$

Comparing this to the given torque vector $$\vec{\tau }=P(a\hat{i}+b\hat{j})$$, we find that:

$$a=-3$$

$$b=-2$$

Thus, the ratio $$\frac{a}{b}=\frac{-3}{-2}=\frac{x}{2}$$.

Therefore, $$x=3$$.