Question

Find the magnitude of the horizontal force exerted by the hinge on the body.

Answer 1

Center of mass is the point where the whole mass of the system is concentrated. In an equilateral triangle, the length of the median is $\sqrt 3 /2$ times the side of the triangle. The centroid, where the entire mass of the equilateral triangle lies, is equal to the 2/3 times of the length of the median.

Complete step-by-step answer:
We know that the center of mass is a point where the whole mass of the system is assumed to be concentrated. Let the mass at each corner be ‘m’.
Therefore, the total mass of the system is ‘3m’.
In an equilateral triangle, the centroid of the triangle is the point at which three medians of the triangle intersects. In other words, the centroid is the distance that is located at the 2/3 distance from the point which joins to the mid of the opposite side. The centroid of the triangle is denoted as,
${\text{centroid of triangle = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ times length of the median}}$
Where the length of the median for an equilateral triangle is found out by,
Length of Median $= \dfrac{{\sqrt 3 }}{2}a$
where a=length of side,
Here, we know that side of triangle =$l$
Thus the length of median= $= \dfrac{{\sqrt 3 }}{2}l$
Put the value of the median in the formula for the centroid of the triangle, Then,
$\Rightarrow \dfrac{2}{3} \times \dfrac{{\sqrt 3 }}{2}l \\\ = \dfrac{l}{{\sqrt 3 }} \\\$
The centroid of triangle =$r = \dfrac{l}{{\sqrt 3 }}$

Now, we are interested to find out the centripetal force for a particle of mass 3m at a distance $r = \dfrac{l}{{\sqrt 3 }}$ from A.
We know that Centripetal force is given as,
${F_c} = \dfrac{{m{v^2}}}{r}$
We have, $v = r\omega$ where, $v$=velocity of the particle, r=distance of the point from the center of mass, ω=angular frequency. On substituting these values in the centripetal force equation. we get,
${F_c} = \dfrac{{m{r^2}{\omega ^2}}}{r} = mr{\omega ^2}$
Put the values for (r), we get,
${F_c} = (3m)\left( {\dfrac{l}{{\sqrt 3 }}} \right){\omega ^2} = \sqrt 3 ml{\omega ^2}$
This is the horizontal force acting at the hinge.
Therefore, at hinges the horizontal force $= \sqrt 3 ml{\omega ^2}$

Additional information:
A rigid body is a body in which the distance between all pairs of particles does not change during the motion of the body. Hence there is no change in the shape and size of the body during motion under the application of forces. Due to the application of external forces, a rigid body can have either translational or rotational motion.
During translational motion, of a rigid body, all the constituent particles of the body move along parallel straight lines and undergo equal displacement in the same time interval. Hence, the velocity, as well as the acceleration of every particle in the rigid body during translational motion, is the same.

Note: In this type of question, everyone must know the center of mass of regular as well as irregular bodies by using a coordinate system.
The center of mass of a system particles moves as if the entire mass of the system is concentrated at the center of mass.
In the rotational motion of a rigid body, a line of a particle in the body remains fixed and all the other particles describe concentric circles around this fixed line of particles. The fixed-line of particles is called the axis of rotation.