Energy of Rolling; Rolling on an Incline

Transcript

0:00Introduction

Hello everyone! Here is a race. Four bodies sit at the top of one slope: a ring, a disc, a solid sphere and a hollow sphere. We let them go together, and they roll down without slipping. Who reaches the bottom first? The heaviest? Many students guess the heaviest. Hold that guess. Today: the energy of rolling, and rolling on an incline. It comes up often in JEE, so let's begin.

0:29Kinetic energy of a rolling body

So, a rolling body moves forward and spins at once. Its kinetic energy K equals half M v squared, plus half I c m omega squared. M is the mass, v the speed of the centre, and I c m, with the small c m below, the moment of inertia about the axis through the centre of mass. Omega is the angular speed. Okay? Now, I want everything in terms of v alone. Which two facts from the earlier concepts will help us here? Very good! First, I c m equals M k squared, small k being the radius of gyration. Second, rolling without slipping: v equals omega times capital R, the radius, so omega equals v over R, right? Put both in, and take half M v squared common. K equals half M v squared, times one plus k squared over R squared. The one is translation; k squared over R squared is the extra from rotation. Clear? Now a topper's trick. K rot over K tr, with small rot and tr below for rotational and translational, is simply k squared over R squared. Only the shape is left, okay? And the rotational fraction of the total is k squared over R squared, over one plus k squared over R squared. With me?

1:50Four shapes, four shares

Now the four standard shapes. Each line gives k squared over R squared, the ratio K rot to K tr, with the colon meaning is to, and the rotational fraction. Ring or hollow cylinder: all the mass is at the rim, so k squared over R squared is one. Ratio one is to one: half rotational. Right? Next, the disc, or solid cylinder: k squared over R squared is one half, so the ratio is one is to two. Quick question: what fraction of the total is rotational? One half? Oh no, careful! One is to two means one part rotational, two parts translational: three parts in all. So the fraction is one third, not one half. Ratio and fraction are different things. Got it? Solid sphere: two fifths. So the ratio is two is to five, and the fraction is two sevenths. With me? Hollow sphere: two thirds. Ratio two is to three, fraction two fifths. Write these four lines down, okay?

2:57Rolling down an incline

Now, the slope. In the figure, a body rolls down an incline of angle theta. Weight M g acts down, the normal force N is perpendicular to the slope, friction f acts up the slope, and the arrow marked a is the acceleration. On the right are formulas for a, f, and mu min, the least friction coefficient; friction comes in the next concept. With me? Now, there is friction here. So some energy is lost to friction as the body rolls down, right? What do you think? No, no, careful! This is static friction. It acts at the contact point, which does not move in rolling without slipping. So friction does no work, and mechanical energy is conserved in pure rolling. Clear? So, from rest, the body comes down a height h; g is the acceleration due to gravity. M g h, the loss in potential energy, equals half M v squared, times one plus k squared over R squared. Right? Cancel M and solve: v equals the square root of two g h, over one plus k squared over R squared. The long tick sign is the square root, okay? Now the acceleration along the slope: a equals g sine theta, over one plus k squared over R squared. The bottom is more than one, so a rolling body speeds up more slowly than g sine theta. With me? Finally, the time t to roll a length L from rest: the square root of two L, times one plus k squared over R squared, over g sine theta. Careful: this L is the length of the slope, not angular momentum. Got it?

4:35So who wins the race?

So, the shape enters only through k squared over R squared. Smaller k squared over R squared, larger acceleration, sooner at the bottom; the double arrow means implies. Now, back to our race. Who wins? The solid sphere, with zero point four! Yes, exactly. Then the disc, zero point five, then the hollow sphere, about zero point six seven, and last, the ring, with one. Puzzle solved! But wait. What if the ring is very heavy, and the disc is very light? Then the heavy ring wins, right? No, no! These formulas have no M, and no R by itself. Mass and radius do not matter, only the shape. So the disc still wins; the heaviest-first guess was the trap. Clear? One last case: a smooth incline. No friction, so the spin cannot change. The body just slides with acceleration g sine theta, beats every rolling body to the bottom, and keeps the omega it started with. Got it?

5:38Example 1: total kinetic energy of a rolling solid sphere

Alright, example one. Here, a two kilogram solid sphere rolls at five metres per second; the small minus one on the s means per second. Find its total kinetic energy. Pause the video, and try it first. Done? Okay, let's check. Solid sphere, so k squared over R squared is two fifths. Now, half M v squared is half times two times twenty five: twenty five joules. So the answer is twenty five joules, right? Careful! That is only the translational part; the sphere also spins. So multiply by one plus two fifths, that is one point four. Twenty five times one point four: thirty five joules, capital J for joules. Okay? So, twenty five joules translational, and ten rotational. Check with the trick: ten over twenty five is two fifths, exactly k squared over R squared for a solid sphere. Perfect!

6:39Example 2: identify the body, then its incline acceleration

Now a harder one, JEE level. Here, a body rolls down an incline, with rotational kinetic energy forty percent of its translational kinetic energy. Identify it, and find its acceleration on a thirty seven degree incline, the small circle meaning degrees, with g as ten metres per second squared. Pause and try. Done? Okay, let's check. Forty percent is two fifths, and in our table the hollow sphere had a rotational fraction of two fifths. So it is a hollow sphere, right? No, no, wait! Forty percent of the translational energy, not of the total. So K rot over K tr is zero point four, which by our trick is k squared over R squared. Two fifths: a solid sphere! Read every word, okay? Now the acceleration. So, a equals g sine thirty seven degrees, over one plus zero point four. Sine thirty seven degrees is taken as zero point six, so the top is ten times zero point six. Right? Six over one point four: about four point two nine metres per second squared. The wavy equals sign means approximately, and the minus two on the s means per second squared. Clear?

7:57Quick revision: rolling KE, incline acceleration, the race

So, quick revision. Rolling: K is half M v squared, times one plus k squared over R squared, and K rot over K tr is k squared over R squared. Incline from rest: the acceleration is g sine theta over one plus k squared over R squared; static friction does no work. Smaller k squared over R squared wins: solid sphere, disc, hollow sphere, ring. Now try the practice problems on this page. All the best!