Tag: turning effects of forces

Centre of gravity means a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the centre of mass. For regular shaped bodies centre of gravity lies in the centre of the particular body. Hence for triangular lamina centre lies at the centroid which is the intersection of the three lines drawn from the vertex to the midpoint of the opposite side. Hence centre of gravity lies at the intersection of the three medians.

let us assume body of mass m and divide it into many small particles. The centre of mass means it is the mean value of the all the small particles . it would more clear by assuming the body in 3-D coordinate system and calculate its mean of all small particles with the co ordinates in three dimension . where as the centre of gravity is also same but it is the mean of its weight. it is the point were total weight acts

let us assume body of mass m and divide it into many small particles. The centre of mass means it is the mean value of the all the small particles . it would more clear by assuming the body in 3-D coordinate system and calculate its mean of all small particles with the co ordinates in three dimension . where as the centre of gravity is also same but it is the mean of its weight. it is the point were total weight acts

Center of gravity need not always lie inside object. Suppose take a ring. Its center of mass lies at its center. Hence it is not inside the ring but it is outside the body of the ring ie. at its center. All the other statements are correct.

Resultant torque due to gravity force vanishes around the centre of gravity because perpendicular distance between the gravitational force and the point about that toque is calculated becomes very very small or almost zero.

For an object to undergo rotational motion, a net torque must be exerted on the object. If all the forces are acting at its centre of gravity, net torque acting on the object is zero and so the object will not undergo any rotational motion.

given $\dfrac { { m } _{ 1 } }{ { m } _{ 2 } }$=$\dfrac{n}{1}$
Each mass will have the acceleration=$a=\dfrac { { (m } _{ 1 }-{ m } _{ 2 })g }{ { m } _{ 1 }+{ m } _{ 2 } } $
However ${m} _{1}$ which is heavier will have the will have acceleration ${a} _{1}$ vertically down while the lighter mass ${m} _{2}$ will have acceleration ${a} _{2}$ vertically up -${a} _{2}$=${a} _{1}$
${a} _{c.o.m}$=$\dfrac{{m} _{1}\times{a} _{1}+{m} _{2}\times{a} _{2}}{{m} _{1}+{m} _{2}}$
so ${a} _{c.o.m}$=$\dfrac { { (m } _{ 1 }-{ m } _{ 2 }){a} }{ { m } _{ 1 }+{ m } _{ 2 } } $=$\dfrac{{m} _{1}-{m} _{2}}{{m} _{1}+{m} _{2}}$$\times$$\dfrac { { (m } _{ 1 }-{ m } _{ 2 })g }{ { m } _{ 1 }+{ m } _{ 2 } } $=$\dfrac { { ({ { m } _{ 1 }-{ m } _{ 2 }) }^{ 2 }g } }{ { { (m } _{ 1 }+{ m } _{ 2 }) }^{ 2 } }$
Since $\dfrac{{m} _{1}}{{m} _{2}}$ =n, diving by ${m} _{2}$ and simplifying
${a} _{c.o.m}$=$\dfrac { { (n-1) }^{ 2 } }{ { (n+1) }^{ 2 } }$g

The metacentre remains directly above the center of buoyancy regardless of the tilt of a floating body, for example that of the ship. (Center of gravity is the point in a body about which all parts of the body balance each other).

Let m be mass of a body.
$\therefore$ Weight of the body on the surface of the earth is W=mg=250N
Acceleration due to gravity at a depth d below the surface of the earth is $g^{\prime}=g\left(1-\dfrac{d}{R _{E}}\right)$
Weight of the body at depth d is
$W^{\prime}=mg^{\prime}=mg\left(1-\dfrac{d}{R _{E}}\right)$
Here, $d=\dfrac{R _{E}}{2}$
$\therefore W^{\prime}=mg\left(1-\dfrac{{R _{E}}/{2}}{R _{E}}\right)=\dfrac{mg}{2}=\dfrac{W}{2}=\dfrac{250N}{2}=125N$