Half of the rectangular plate shown in figure is made of a material of density $p_{1}$ and other half of density $p_{2}$ . The length of the plate is L. The distance of centre of mass of the plate from origin O is:-

Open in App

Solution

Let the width be $d$ .
$∴$ Mass of the left part of the plate, $M = \frac{L d}{2} p_{1}$
Mass of the right part of the plate, $M^{'} = \frac{L d}{2} p_{2}$
Measurement fo $C M$ from the origin,
$\begin{matrix} n = \frac{\frac{M L}{4} + M^{'} (\frac{3 L}{4})}{\frac{L d}{4} p_{1} + \frac{L d}{2} p_{2}} = \frac{L (p_{1} + 3 p_{2})}{4 (p_{1} + p_{2})} \end{matrix}$

Suggest Corrections

Similar questions

Half of the rectangular plate shown in figure is made of a material of density $p_{1}$ and the other half of density $p_{2}$ . The length of the plate is $L$ . Locate the centre of mass of the plate.

Q. Shows a rectangular plate of length L, the half of which is made of material of density di and another half of density

d_{2}

. The distance of centre of mass from the origin 0 is :

Q. A semicircular portion of radius

^{'} r^{'}

is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass

^{'} C^{'}

of remaining plate, from point

^{'} O^{'}

is:

Join BYJU'S Learning Program

Half of the rectangular plate shown in figure is made of a material of density p1 and other half of density p2. The length of the plate is L. The distance of centre of mass of the plate from origin O is:-

Let the width be d.∴ Mass of the left part of the plate, M=Ld2p1Mass of the right part of the plate, M′=Ld2p2Measurement fo CM from the origin,n=ML4+M′(3L4)Ld4p1+Ld2p2=L(p1+3p2)4(p1+p2)

Half of the rectangular plate shown in figure is made of a material of density $p_{1}$ and other half of density $p_{2}$ . The length of the plate is L. The distance of centre of mass of the plate from origin O is:-