Density rhoxy y gcm2 radius a cmSolutionA lamina is a 2di

Solution

A lamina is a 2-dimensional object. In other words, it is a flat object whose thickness is we can ignore. If a body has a line of symmetry, the centre of mass will lie on this line. We know that the centre of mass lies somewhere on the axis of symmetry of the semi-circle, although, where on the axis of symmetry, we do not know. We therefore divide the semi-circle into many tiny strips, each with thickness x. We know that the centre of mass lies on the black vertical line, since this is a line of symmetry of the object. If we label this line the x-axis and introduce a y-axis as follows, this will allow is to use calculus:

Each strip is a distance of x from the y-axis. The mass of each strip is its volume × density. The volume of each strip is just its area, since it is a lamina and so has no depth. The area of each strip is x × 2y . Since the body is uniform, which means that the density is constant. It doesn\'t matter what the density is, lets call it . So the mass of each strip is 2yx. Now the total area of the semicircle is ½ r² . So the total mass is ½ r² . Let O be the mid- point of the horizontal line between –a and a.

Now, we know that (distance of centre of mass from O) × (weight of body) = the sum of: (the mass of each particle) × (the distance of each particle from O) Therefore: (centre of mass) × ½ r 2 = 2y x x . The sum is from x = 0 to x = r . (centre of mass) × ½ r 2 = 2y x dx . The integral is with respect to x, so we must replace the y by a function of x.Now, we can deduce that y/r = x/r, so y = x.

Therefore: (Centre of mass) × ½ r² = 2x² dx (Centre of mass) × ½ r² = (2r³ )/3. So the centre of mass is a distance of 4r/3 from O, on the axis of symmetry.