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Where does the center of mass rectangular lamina lie?
The centre of mass of a rectangular lamina lies at the point of intersection of its diagonals.
What is the center of gravity for a lamina?
If the lamina is in two dimensions (i.e. not curled round into a third dimension) then the centre of gravity will be somewhere within the flat shape. The position of the centre of gravity will depend on the distribution of mass across the lamina. If the lamina is curled round into a third dimension then the centre of gravity will be somewhere within the volume enclosed, fully or partially, by the lamina; this may or may not be on the lamina.
What is difference between center of mass and center of gravity?
centre of mass is nothing the mass (volume) situated at centre which is not at all use full for pt of control. but centre of gravity is that pt at which we can hold the total mass or body .
What is the mass moment of inertia of a triangular plate?
The mass moment of inertia of a triangular plate depends on its shape and mass distribution. It can be calculated using the formula for a triangular plate, which involves the mass of the plate and the distances of the plate's vertices from the axis of rotation.
What is the difference between center of mass and center of buoyancy?
Centre of mass is a convinient point where whole mass of the body is supposed to be acting and for bodies of small shape it concides with the geometrical centre. Centre of bouancy is the centre of the immersed portion of the body inside the liquid.
Where does the center of mass rectangular lamina lie?
The centre of mass of a rectangular lamina lies at the point of intersection of its diagonals.
What is the center of gravity for a lamina?
If the lamina is in two dimensions (i.e. not curled round into a third dimension) then the centre of gravity will be somewhere within the flat shape. The position of the centre of gravity will depend on the distribution of mass across the lamina. If the lamina is curled round into a third dimension then the centre of gravity will be somewhere within the volume enclosed, fully or partially, by the lamina; this may or may not be on the lamina.
Can a median divide a triangle in equal area?
Sure. That's true of a median in every isosceles triangle, and every median in an equilateral triangle. In fact it is true for any median of any triangle. The two parts may not be the same shapes but they will have the same area. That is why the point where the three medians meet (centroid) is the centre of mass of a triangular lamina of uniform thickness.
What best describes the centroied of a triangle?
It is the centre of mass of a uniform triangle.
What are medians and what is the point called where they meet?
The median of a triangle is a straight line from a vertex to the midpoint of the opposite side. The three medians of a triangle meet at the centroid. If the triangle is made of uniform material the centroid is the centre of mass of the triangular shape.
How is the center of the gravity determined when an object is at rest or in motion?
If the object is a thin lamina with uniform thickness (e.g. a piece of paper), the the centre of gravity of the object is at its geometrical centre. It can be determined by suspending a load (e.g. pendulum) on an edge of the lamina twice and the point where the plumb lines intersect is the centre of gravity.
What is a centriod of a lamina?
The centroid of a lamina is the point at which it could be balanced if it was suspended. It represents the center of mass of the lamina and is the point where all the mass could be concentrated to achieve balance. The centroid is an important concept in engineering and physics for determining the equilibrium and stability of objects.
Why doesnt the meter stick hang straight when it has its pivot point at the 50 cm mark?
The 50cm mark isn't necessarily the ruler's centre of mass where it will balance. If the ruler is not uniform (i.e. it has a hole in it) its centre of mass will not be exactly in the middle.
How do you find out the area of a composite figure?
There are different formulae for different shapes. Try to break down the composite firgure into components that you can add together (or subtract one from the other). An annulus, for example, is a big circle minus a smaller circle. Areas of squares, triangles, trapeziums, circles, semicircles and the process described anove will answer most high school questions. For more complex figures you may need to look elsewhere. Copy the shape onto a lamina of uniform density. Cut out the shape and find its mass. Also find the mass of a unit square of the lamina. Then area of composite shape = size of unit shape*mass of composite shaped lamina/mass of unit shape.
What is difference between center of mass and center of gravity?
centre of mass is nothing the mass (volume) situated at centre which is not at all use full for pt of control. but centre of gravity is that pt at which we can hold the total mass or body .
What is the mass moment of inertia of a triangular plate?
The mass moment of inertia of a triangular plate depends on its shape and mass distribution. It can be calculated using the formula for a triangular plate, which involves the mass of the plate and the distances of the plate's vertices from the axis of rotation.
How do you square footage?
Square footage is a measure of area. There are formulae for some simple shapes but for more complicated shapes there are essentially two options: you can either use uniform laminae and mass or estimate the area using grids. Uniform Lamina: Copy the shape onto a sheet (lamina) of material with uniform density. Cut the shape out carefully and measure its mass (or weight). Do the same for a unit square of the lamina. Then, because the lamina is of uniform density, the ratio of the two areas is the same as the ratio of the two masses. That is: Area of Shape/Area of Unit Square = Mass of Shape/Mass of Unit Square. Rearranging, and noting that the area of the Unit Square is, by definition, = 1 sq unit Area of Shape = Mass of Shape/Mass of Unit Square. Grid Method: Copy the shape onto a grid, where each grid square has an area of G square units. Count the number of squares that are fully or mostly inside the shape. Call this number W (for whole). Count the number of squares that are approximately half inside the shape and call this number H (for half). Ignore any square that are less than half in the shape. Then a reasonable estimate of the area of the shape is G*[W + H/2] square units. There is some arbitrariness about “mostly inside†and “approximately half†but there is no way around that. You will get more accurate results with finer grids, but they will also require much more effort in terms of counting the grid squares.
