Centre of Mass of Continuous Bodies

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a thin steel ring and cut it in half: a semicircular wire. Where is its centre of mass? Most students point somewhere on the wire, near the top. But is that really true? Keep this puzzle in mind; we will solve it in this lesson. So, the centre of mass of continuous bodies, like rods, rings, discs and cones. These come up often in JEE problems, so let's begin.

0:29From a sum to an integral

So, recall the last lesson. For particles, r c m, the r with a small arrow on top and c m below, is the position vector of the centre of mass. It equals the sum of m i times r i, over the sum of m i, where the small i counts the particles. The big Greek sigma means add them all up. Okay? But a rod, or a disc, has no separate particles; the mass is spread out continuously. So what replaces this sum? An integral! Yes, exactly. Cut the body into tiny pieces, each of mass d m. Then r c m equals the integral of r d m, over the integral of d m. The tall S is the integral sign. Right? And the bottom, the integral of d m, adds up all the tiny masses: capital M, the total mass of the body. Clear?

1:24Writing dm: choose the right density

Now, the real skill is writing d m. For a rod or a wire, use lambda, the linear density. So d m equals lambda d x, where d x is a tiny length. With me? For a plate or a thin shell, use small sigma, the surface density: d m equals sigma d A, with d A a tiny area. Okay? Now you tell me. For a solid ball, which density? The volume density, rho! Correct! So d m equals rho d V, with d V a tiny volume. Got it?

1:57Standard results for uniform bodies

Now, the standard results. These are for uniform bodies only, okay? The figure on the right shows four of them; the dot marks each centre of mass, and capital O is the centre. For a uniform rod, the centre of mass is at its midpoint. For a rectangular or circular plate, a ring or a sphere, it is at the geometric centre. Right? Next, an arc of radius R that subtends an angle two alpha at the centre, where alpha is the Greek letter you see. Its centre of mass is at R sine alpha, over alpha, from the centre, on the bisector of that angle. With me? Now, our semicircular wire is an arc too. What is alpha for it? Pi over two! Yes, exactly. Two alpha is pi, so alpha is half of pi. Sine of pi over two is one, so we get R over half pi: two R over pi. That is less than R, so the point is inside the curve, in empty space, not on the wire at all! Look at the first shape in the figure. Puzzle solved! Okay? For a sector of a disc, a pizza slice of angle two alpha: two R sine alpha, over three alpha, from the centre. Clear? Put alpha equal to pi over two again, and you get the semicircular disc: four R over three pi. Right?

3:20Solid or hollow? Read carefully

Now, a solid hemisphere of radius R, the third shape in the figure: three R over eight from the flat face, okay? And a hollow hemisphere, a thin shell? Also three R over eight, right? No, no, careful! The hollow one is at R over two. So always read the question: solid or hollow? Got it? Same story for a cone of height h. Solid cone, the last shape in the figure: h over four from the base. Notice, from the base, not from the tip! With me? Hollow cone, only the curved surface: h over three from the base. Okay? And a triangular plate: at the centroid, h over three from any side, along the height to that side.

4:06Topper's trick: symmetry and pieces

Right? Now, two tricks that toppers use. First: for a uniform body, the centre of mass lies on its axis of symmetry. So you only find one distance, along that axis. Clear? Second: for a body made of pieces, treat each piece as a particle sitting at its own centre of mass. Then use the particle formula. Got it?

4:29Example 1: non-uniform rod with lambda = lambda0 x

Alright, Example one. There is a rod of length L along the x axis, from x equals zero to x equals L, okay? Its linear density is lambda equals lambda zero times x; lambda zero, with a small zero below, is a constant. Find x c m, the x coordinate of the centre of mass. Pause the video and try it first. Done? Okay, let's check. It is a rod, so the midpoint, L over two, right? No, no, wait! The midpoint is only for a uniform rod. Here lambda grows with x, so the rod gets heavier as x increases. So we integrate, with d m equals lambda d x, that is, lambda zero x d x. Okay? On top, the integral of x d m: x times lambda zero x d x. At the bottom, the integral of d m. The zero and the L on the integral sign are the limits, from one end of the rod to the other. Right? The top gives lambda zero L cubed over three; the bottom, lambda zero L squared over two. Lambda zero cancels: two L over three. That's it! Quick sense check: two L over three is greater than L over two, as the greater than sign shows. Makes sense: the heavier end pulls the centre of mass towards it. Clear?

5:56Example 2: centre of mass of a D-shaped wire

Now a harder one, in the JEE pattern. Remember our semicircular wire? We close it with its diameter, so the uniform wire looks like the capital letter D, with radius R, okay? Find the distance of the centre of mass from the diameter. Pause the video and try it. Done? Okay. Some students write two R over pi again, straight from the standard result. Is that correct? No! The straight piece is part of the wire too. So use our trick: two pieces, each like a particle. The arc has length pi R, so its mass is lambda pi R, at two R over pi from the diameter. With me? The diameter has length two R, so its mass is two lambda R, okay? Its centre of mass is its midpoint, on the diameter itself, so its distance is zero. Right? Now the particle formula. y c m, the height above the diameter, equals lambda pi R times two R over pi, over lambda pi R plus two lambda R; the dot means times. The pi on top cancels, and lambda R cancels from top and bottom. We get two R, over pi plus two, about zero point three nine R; the wavy equals sign means approximately. See? The straight piece pulled the centre of mass down, closer to the diameter.

7:22Quick revision: integral formula, dm, standard results

Got it? So, quick revision. For a continuous body, r c m is the integral of r d m over the integral of d m, with d m written using lambda, sigma or rho. For uniform bodies, use the standard results, and check: solid or hollow? Base or tip? And for a composite body, treat each piece as a particle at its own centre of mass. Now go and try the practice problems on this page. All the best!