Question 706: Centre of Mass and Rotational Motion

Question

A particle of mass $$m$$ is allowed to fall freely under gravity from a point $$P$$ as shown in the figure. If the vector position $$Q$$ of the particle from the origin is represented by $$\mathbf{r}$$, the magnitude of torque acting on the particle at time $$t$$ with respect to the origin $$O$$ is.

Accepted Answer

The torque acting on the particle about O is given by
$$\begin{aligned}
	au & =\mathbf{r} 	imes \mathbf{F}=r m g \sin 	heta \
& =\frac{r m g \sqrt{r^2-y^2}}{r}=m g \sqrt{r^2-y^2}
\end{aligned}$$

Answer

$$m g r$$

Answer

$$m g y$$

Answer

$$m g \sqrt{r^2-y^2}$$

Answer

$$m g r \sqrt{1-\left(\frac{r}{y}\right)^2}$$

Centre of Mass and Rotational Motion

A particle of mass $m is allowed to fall freely under gravity from a point P as shown in the figure. If the vector position Q of the particle from the origin is represented by \mathbf{r}, the magnitude of torque acting on the particle at time t with respect to the origin O$ is.

Choose the correct answer: