Stability and centre of mass - class-X

Given,

For a cubical box of side L = 40 cm open at the top, the mass is distributed over 5 faces (bottom + 4 sides). The center of mass of the bottom is at z=0, and the center of mass of the 4 sides is at z=L/2 = 20 cm. Using the weighted average: z_cm = (10 + 420) / 5 = 80 / 5 = 16 cm.

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance  $\dfrac{R}{2}$ from the center.

The moment of inertia depends on the distribution of mass relative to the axis. With only three particles at (a,0,0), (0,a,0), and (0,0,a), the mass distribution is not symmetric enough to determine the Z-axis moment of inertia solely from the X and Y values without knowing the masses or the full body configuration.

Center of mass of a system of particles is at the origin. It follows that:

Centre of mass of rigid bodies may lie either inside or outside the body.

the point where the total weight appears to act for any orientation of the object is centre of gravity .

centre of gravity depends on shape of object and its mass distribution.

The center of mass of a uniform semi-circular plate of radius R is located on the axis of symmetry at a distance of 4R / (3 * pi) from the center of the base.

Masses are m1 = 1 kg at (0,0), m2 = 2 kg at (a,0), m3 = 3 kg at (a,a), and m4 = 4 kg at (0,a) on a square of side a. The X-coordinate of the center of mass is X_cm = (1(0) + 2(a) + 3(a) + 4(0)) / (1+2+3+4) = 5a / 10 = a/2. The Y-coordinate is Y_cm = (1(0) + 2(0) + 3(a) + 4(a)) / 10 = 7a / 10, resulting in coordinates (a/2, 7a/10).

Velocity of center of mass is $=\dfrac{m _1v _1+m _2v _2}{m _1+m _2}$

For a solid cylinder rolling down an inclined plane of height h1, conservation of energy gives mgh1 = (1/2)mv^2 + (1/2)Iomega^2. Since I = (1/2)mR^2 and omega = v/R, mgh1 = (3/4)mv^2, so v^2 = (4/3)gh1. For a sliding body on a frictionless plane of height h2, mgh2 = (1/2)mv^2, so v^2 = 2gh2. Equating the velocities gives (4/3)gh1 = 2gh2, leading to h2 = (2/3)*h1 = (2/3)*2.7 m = 1.8 m.

We know that, $x+cm=\cfrac{4}{\pi R^2}\int^0 _R{-t^2 dt}$

Given$,$

At lowest height

The center of mass of a regular hexagon with identical masses is at the center. If one mass m is doubled to 2m, the system is equivalent to the original system plus an additional mass m at that corner. The shift is (m * r) / (total mass), where r is the distance from the center. Total mass = 7m. Shift = (m * a) / 7m = a/7.

The center of mass is the unique point where the weighted relative position of the distributed mass sums to zero. It is the point at which the entire mass of the body can be considered to be concentrated for the purpose of translational motion analysis.

When a football rolls through the ground, the path followed by center of mass of ball is linear as the center of mass remains always at a fixed height from the ground when the football rolls and moves in a straight line if the ball does not changes its direction of rolling which is assumed in this case.

The center of mass of a uniform semicircular ring of radius R lies on its axis of symmetry at a distance of 2R / pi from the geometric center along the perpendicular bisector.

As the chain falls, the velocity of the falling part increases. The center of mass of the chain is determined by the position of the part still in the air and the part already on the surface, accounting for the loss of potential energy.

If all the particles have positive x  co-ordinates then their COM can't be at the origin. Total KE may not be conserved because of internal forces. Its not given that there is no external force acting on the system, so its momentum may also change.

As we can see that the center of mass will be at $r _c=\dfrac{mr+MR}{m+M}$,

All co-ordinates positive will make the co-ordinate of com positive.
All co-ordinates negative will make the co-ordinates of com negative.
If all the co-ordinates are zero(0), co-ordinate of com is zero(0).
If  co-ordinate of com can be zero,  co-ordinates of some are positive and co-ordinates of some are negative.

The center of mass for a non-uniform rod is found by integrating xlambdadx divided by the total mass. With lambda = A + Bx, the integral of x(A+Bx) from 0 to L yields the numerator, and the integral of (A+Bx) yields the denominator.

The shift in the center of mass is calculated using the principle of negative mass. The mass of the removed square is proportional to its area, and the shift is (m_removed * distance) / (M_total - m_removed).

The internal forces all balance each other so that there is no change in the mass distribution of the body, thus center of mass remains unchanged. Rest all options are correct.

The center of gravity is the point where the total gravitational force (weight) of a body is considered to act. While often coincident with the center of mass, the definition specifically refers to weight.

when passengers stands up on the boat from the sitting situation , the centre of gravity moves upwards and it increases the chances of toppling of boat .

The given statement is true that centre of gravity of rectangular shaped objects is at their geometric centre.

The point through which the total weight of an object appears to act for any orientation of the object is centre of gravity.

The centre of mass of a body can lie within or outside the body.

Using the negative mass method, the center of mass of the remaining part is found by subtracting the mass and moment of the smaller disc from the larger one. The calculation leads to alpha = 1/3.