Two uniform thin rods each of Length $L$ and mass $m$ are joined as shown in the figure. Find the distance of center of mass the system from point O.

Open in App

Solution

$Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$
$\frac{m \times 0 + M x \frac{L}{2}}{2 M}$
$= \frac{L}{4}$
so, $\frac{L}{4}$ from $0$

Suggest Corrections

Similar questions

Q. Two identical thin uniform rods of length

L

, each are joined to form T-shape as shown in the figure. The distance of centre of mass from

D

is:

Q. A thin uniform rod of length

L

is bent at its midpoint as shown in the figure. The distance of the center of mass from the point

^{'} O^{'}

Q. Two identical uniform rods

EF

and

GH

, each of length

L = 6 m

are joined as shown in figure. Locate the centre of mass of the frame.
[Point

O

, which is the point where the axes of the rods meet is the origin]

Q. Two identical thin uniform rods of each mass

m

and length

l

are joined to form

L

shape. Find the moment of inertia of this system about an axis passing through the point at which both rods are joined and perpendicular to their lengths.

Two uniform identical rods each of mass $M$ and length $l$ are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure.

Join BYJU'S Learning Program

Two uniform thin rods each of Length L and mass m are joined as shown in the figure. Find the distance of center of mass the system from point O.

Ycm=m1y1+m2y2m1+m2m×0+MxL22M=L4so, L4 from 0

Two uniform thin rods each of Length $L$ and mass $m$ are joined as shown in the figure. Find the distance of center of mass the system from point O.

$Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$
$\frac{m \times 0 + M x \frac{L}{2}}{2 M}$
$= \frac{L}{4}$
so, $\frac{L}{4}$ from $0$