Turning effects of forces - class-XI

Yes, Centre of gravity of a freely suspended body always lies vertically below the point of suspension. The center of gravity of an object is the point you can suspend the object from without there being any rotation because of the force of gravity, no matter how the object is oriented. If you suspend an object from any point, let it go and allow it to come to rest, the center of gravity will lie along a vertical line that passes through the point of suspension. Unless you've been exceedingly careful in balancing the object, the center of gravity will generally lie below the suspension point.

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 bodies centre of gravity lies at the centre of the body.Hence we know that there will be a centre for a uniform ring lamina.Hence this centre of the ring will be centre of gravity.

The position of the centre of gravity of an object affects its stability. The lower the centre of gravity is, the more stable the object. The higher it is the more likely the object is to topple over if it is pushed.
In general, increasing the size of the base (that is area) of support increases stability.

Centre of gravity of the circle lies at its geometric centre because the resultant weight of the circle acts at its centre.

Centre of gravity of an object is a specific point such that the body behaves as if the entire gravitational force is acting only at that point. It is unique for an object and can't be more than 1.

Weight always point towards the center of the earth that is perpendicular to the surface of the earth that is towards the table and $perpendicular$ to surface of the table.

Centre of gravity of an object is a specific point such that the body behaves as if the entire gravitational force is acting only at that point. It is unique for an object and can't be more than 1.

Centre of mass is a point on a physical body which behaves as if the whole mass is concentrated over there. Whenever a force acts on a body it appears to be acting on the centre of mass. Whenever gravitational force is acting on a body it appears to be acting on a specific point on the body which is known as the centre of gravity.
 It usually coincides with the centre of mass. Earth being a larger planet than the moon, the gravitational force is more intense on earth as compared to the moon.
 The location of the centre of mass does not vary with the intensity of the force acting on the body. Similarly, centre of gravity does not vary with the intensity of the gravitational force acting on it.  Hence the centre of gravity on earth and moon will be the same

Center of gravity is a point on which behaves as if the whole weight of the body appears to be concentrated. So, it depends upon its mass, weight and geometry (i.e. shape)

It is a point on which behaves as if the whole weight of the body appears to be concentrated.

The center of gravity is defined as the ratio of sum of its moments to the total weight of the object.

Since, short person has lower center of gravity hence it is difficult to knocked them out

Centre of gravity of an object is a specific point such that the body behaves as if the entire gravitational force is acting only at that point.

It is a point on which behaves as if the whole weight of the body appears to be concentrated.

Center of gravity is the point at which the entire weight of a body may be considered as concentrated so that if supported at this point the body would remain in equilibrium.

In case of a ring, the centre of gravity lies at the centre of the ring. Hence option c is the incorrect option

Option A is the definition of the centre of gravity. B is a contradiction. As far as C ad D are concerned, the force of gravity is equal at all points provided the object is small as compared to the radius of the planet. 

Centre of mass and of gravity coincide for objects which are small as compared to the radius of the earth.

The mass of the body is scattered equally in all space around the centre of mass. It is closer to the heavier regions of the object and hence need not be at the geometric centre. 

A is the proper definition. In some bodies like hollow bodies, frames and rings, the center of gravity is at the geometric centre where the matter of the object is not present.

The stability of objects depends on the base, C.G and shape of the body. For a body to be the stable,

1. The base of the body need to be broad.
2. The Centre of gravity should be as low as possible.
3. The vertical line through the center of gravity should fall within the base.

The center of mass x_cm is calculated by the integral of (x * lambda * dx) / integral of (lambda * dx) from 0 to L. With lambda = 2 + x and L = 3, the numerator is integral from 0 to 3 of (2x + x^2) dx = [x^2 + x^3/3] from 0 to 3 = 9 + 9 = 18. The denominator is integral from 0 to 3 of (2 + x) dx = [2x + x^2/2] from 0 to 3 = 6 + 4.5 = 10.5. Thus, x_cm = 18 / 10.5 = 180 / 105 = 12 / 7.

By energy conservation of the suspended block we can say 

When rod falls due to earth magnetic field an emf is induced 

The apparent weight is given by N = m(g + a) for upward acceleration. If N = 2mg, then m(g + a) = 2mg, which simplifies to g + a = 2g, so a = g. The elevator is moving upward with an acceleration of g.

The largest distance between the origin and center of mass would be 'r' when there is only one particle.
If there are more than one particle, the center of mass would be inside the circle of radius 'r' centered at the origin.
Hence option B is correct.

The position vector of COM of the.three particles will be given by
$\vec{r} _{COM}, =, \displaystyle \frac {m _1\vec{r} _1, +, m _2\vec{r} _2, +, m _3\vec{r} _3}{m _1, +, m _2, +, m _3}$
Substituting the values, we get
$\vec{r} _{COM}, =, \displaystyle \frac {(1) (\hat{i}, +, 4\hat{j}, +, \hat{k}), +, (2) (\hat{i}, +, \hat{j}, +, \hat{k}), +, (3) (2\hat{i}, -, \hat{j}, -, 2\hat{k})}{1+2+3}, =, \displaystyle \frac {1}{2}, (3\hat{i}, +, \hat{j}, -, \hat{k}), m$.
Hence, the position vector of their center of mass is $, \displaystyle \frac {1}{2}, (3\hat{i}, +, \hat{j}, -, \hat{k}), m$.

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 bodies centre of gravity lies at the centre of the body. Hence this will be at the centre of the circular lamina.

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. Hence for a regular shaped bodies it will have at the centre of that body. Hence for a rectangle it is nothing but at the point of intersection of diagonals.

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.

Initial K.E. = 490 J. At height h, K.E. = 245 J. By conservation of energy, initial K.E. = final K.E. + potential energy at height h. 490 = 245 + mgh. 245 = 2 * 9.8 * h. h = 245 / 19.6 = 12.5 m.

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, $d=\dfrac{R _{E}}{2}$
$\therefore W^{\prime}=mg\left(1-\dfrac{{R _{E}}/{2}}{R _{E}}\right)=\dfrac{mg}{2}=\dfrac{W}{2}=\dfrac{250N}{2}=125N$

Centre of mass is a point on a physical body which behaves as if the whole mass is concentrated over there. Whenever a force acts on a body it appears to be acting on the centre of mass. 

Racing car can corner rapidly without turning due to low centre of gravity. The lower the centre of gravity is more stable is the object The higher it is more likely  the object is to topple over if it is pushed.Racing cars have really low centre of gravity so that they can corner rapidly withoit turning.

For a freely falling body starting from rest, the distance covered in time t is s = (1/2)gt^2. Using g = 10 m/s^2 and t = 5 s, s = 0.5 * 10 * 25 = 125 meters.

The height a person can jump is inversely proportional to the acceleration due to gravity g. g = G * M / R^2. Since M = density * volume = rho * (4/3) * pi * R^3, g is proportional to rho * R. If rho' = rho/4 and R' = R/3, then g' = g * (1/4) * (1/3) = g/12. The jump height h' = h * (g/g') = 1.5 * 12 = 18 m.

$mr\omega^{2} = \dfrac {GMm}{r^{2}}$
$\Rightarrow r\omega^{2} = \dfrac {GM}{r^{2}}$
$\Rightarrow r^{3} = \dfrac {GM}{\omega^{2}} = \dfrac {GM}{R^{2}} . \dfrac {R^{2}}{\omega^{2}}$
$\Rightarrow r^{3} = g \dfrac {R^{2}}{\omega^{2}}$
Hence (D) is correct.

A hydrometer is an instrument used to measure the specific gravity (or relative density) of liquids based on the concept of buoyancy.

Force due to liquid (stokes law):

If a charged particle moves through a region without deflection, the net force must be zero. The Lorentz force is F = q(E + v x B). If E = 0 and B = 0, the force is zero. While other configurations exist (like E and B being parallel to v), the most fundamental condition for no deflection regardless of velocity is that both fields are zero.

We know that angular momentum of spin $\displaystyle =I\omega $
Bythe conservation of angular momentum
$\displaystyle \frac { 2 }{ 5 } M{ R }^{ 2 }.\frac { 2\pi  }{ T } =\frac { 2 }{ 5 } M{ \left( \frac { R }{ 4 }  \right)  }^{ 2 }.\frac { 2\pi  }{ T' } $
$\displaystyle T'=\frac { T }{ 16 } =\frac { 24 }{ 16 } =1.5h$

This relates to the effective weight of a body in a rotating frame or a non-inertial frame. The term m(v_e + v)^2 / R represents the centrifugal force component acting on the body, which reduces the effective gravitational weight.

When the passengers stand, the center of gravity rises. If it rises much it becomes unstable and may topple down.