Name: Rotational Kinetic Energy
Uploaded: 2019-07-19
Duration: 10 min 15 s
Description: Do you think that rotational objects possess kinetic energy? Yes, as it is in motion. Then how do you calculate it? Using conventional formula! But there is...

Question

Given the centre of mass. Obtain an expression for the position vector of centre of mass.

Accepted Answer

When a rigid body undergoes translational and rotational motion simultaneously. We can consider the motion of the rigid body as a combination of a trajectory of a special point in the rigid body due to translational motion only a rotation about the point due rotational motion only. That special point is called the centre of mass.

Let the two particles $m_{1}$ and $m_{2}$ are located at position $r_{1}$ and $r_{2}$ mass $m$ is given by

$r_{c m} = \frac{r_{1} m_{1} + r_{2} m_{2}}{m_{1} + m_{2}}$

$x_{c m} = \frac{x_{1} m_{1} + x_{2} m_{2}}{m_{1} + m_{2}}$

$v_{c m} = \frac{v_{1} m_{1} + v_{2} m_{2}}{m_{1} + m_{2}}$

If a system consists of $n$ particles of masses $m_{1}, m_{2}, m_{3} . . . . . . . . . m_{n}$ whose positions vectors are.

$\vec{r_{1}}, \vec{r_{2}}, \vec{r_{3}} . . . . . . . . . . \vec{r_{n}}$ respectively then position vector of centre of mass.

$\vec{r} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}} + m_{3} \vec{r_{3}} + . . . . . . . . m_{n} \vec{r_{n}}}{m_{1} + m_{2} + m_{3} . . . . . . . m_{n}}$

If two masses are equal i.e. $m_{1} = m_{2}$ then position vector of centre of mass.

$\vec{r} = \frac{\vec{r_{1}} + \vec{r_{2}}}{2}$

Given the centre of mass. Obtain an expression for the position vector of centre of mass.

Important Questions on Rotation

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