Video: Motion of Rolling Objects

In this video we learn how rolling objects combine translational and rotational motion, as well as how to calculate the kinetic energy of a rolling object.

10:56

Video Transcript

In this video, we’re going to learn about the motion of rolling objects. We’ll see how these objects move and we’ll see how to look out for the energy of a rolling object.

To get started, imagine that it’s in between periods at a professional ice hockey game. And as the winner of a raffle, you’ve been chosen to compete in a game for a prize. The game involves going out to center ice and then hitting a puck as hard as you can at a target at one end of the rink. The twist is that the target is equipped with an electronic sensor that measures impulse. And in order to win a prize, you not only need to hit the target, but it also needs to read out above a certain minimum impulse level. With a crowd of thousands eagerly looking on, you have just one shot at winning the prize.

Assuming that you can hit the puck with the same speed towards the target in either case, would you rather hit it and cause it to slide like a puck normally moves or would you turn it on its edge and then hit it so that it rolls along without slipping to the target? If you can give the puck the same translational speed in each case, which way should you choose? To understand which way is best, it will be helpful to know something about the motion of rolling objects.

When an object rolls without slipping, it combines rotational and translational motion. It turns out that there’s a helpful relationship between these two types of motion on a rolling object. If we call the linear speed of the object’s centre of mass 𝑣 sub cm, then that speed is equal to the angular speed of a rolling object multiplied by its radius 𝑟. This equation may bring to mind a similar relationship. When we picked out a point on the tangent of a rotating object and drew in an arrow representing its speed calling it 𝑣 sub 𝑡, we found a similar relationship that the tangential speed of the edge of a rotating disk is equal to its radius times its angular speed.

These two relationships taken together tell us that the tangential speed of a rolling object is equal to the speed of its centre of mass. This equality is only true when our object is rolling without slipping. But under that condition, we have a helpful correspondence between these two linear speeds, the angular speed, and the radius of our object. When we consider the energy of a rolling object, we can say that the kinetic energy of a rolling object brings together both rotational and translational energy. That is, the total kinetic energy is equal to the sum of translational plus rotational kinetic energy. These two terms are connected to one another by the fact that the speed of the centre of mass of our rolling object is equal to 𝑟 times 𝜔.

Whenever you encounter a rolling motion scenario, be sure to include these two types of energy in the accounting of overall kinetic energy. To get a bit of a clearer sense for rolling motion, let’s consider the two types of motion that go into rolling motion separately. First, let’s consider an object which moves from left to right without rotating at all. That is, this is purely translational motion. If we were told that this translation happened at a speed 𝑣, then if someone asked us, what is the energy of this object in motion? thinking in terms of linear kinetic energy, we would say that it’s equal to one-half the object’s mass times its speed squared.

But now, let’s consider an object that moves totally differently. This object rather than translating only rotates. Its centre of mass keeps a fixed point in space and it simply turns about that point. If it turns with an angular speed 𝜔 and again someone asked us for the energy of that object, we could reply that it’s equal to one-half the object’s moment of inertia times that angular speed squared. With an object that does both, that rotates and translates, we have rolling motion. And if this rolling object has a linear speed 𝑣 and an angular speed 𝜔, then it actually has more energy than either the translating or rotating object by itself. So we can think of rolling motion as an especially energetic type of movement.

Thinking back to our game at the hockey arena, this helps us understand why we’d rather have a rolling puck than a sliding one. Let’s get some practice now working with rolling motion through an example.

A hollow marble is rolling across the floor at a speed of 7.0 metres per second when it starts up a plane inclined at 30 degrees to the horizontal. How far along the plane does the marble travel before coming to a rest? How much time elapses while the marble moves up the plane?

We can call the distance along the plane that the marble travels capital 𝐿 and the elapsed time for that travel 𝑡. We’ll start on our solution by drawing a diagram. We’re told that our marble starts out at the base of an incline of an angle—we’ve called 𝜃—of 30 degrees, moving with a linear speed—we’ve called 𝑣—of 7.0 metres per second. Thanks to this initial speed, the marble travels some distance up the incline—we’ve called 𝐿—until it reaches a speed of zero. We can move towards solving for 𝐿 by recognizing this as an energy conservation scenario. In a closed and isolated system, the original energy of this system or initial energy is equal to the final energy of that system.

If we consider the point in time when our marble is at the base of the incline to be our initial point and the point where it’s reached its highest ascent the final time, then we can say that the initial kinetic plus potential energy of our marble is equal to its final kinetic plus potential energy. When we consider the final point, we’ve specified this as the point where the marble has a speed of zero. And therefore, its kinetic energy will also be zero. Looking at the initial energy of our marble, if we set the height of the base of the incline to be zero, that is our reference point for gravitational potential energy, then at that initial point the marble has zero height and therefore zero potential energy due to gravity. Since there are no other contributors to potential energy for the marble, this whole term is equal to zero. We find then that the marble’s initial kinetic energy is equal to its final potential energy due to gravity.

We can recall the mathematical relationships for these two types of energy. And for the kinetic type, we want to recall both rotational and translational kinetic energy forms. Recalling these expressions for kinetic and potential energy, respectively, we’re ready to expand on our energy balance equation. On the initial energy side of our equation, we have one-half the marble’s mass times its speed squared plus one-half its moment of inertia times its angular speed squared. On the right side, we have 𝑚 times 𝑔 times the elevation of the marble when its speed is zero. That elevation above our reference level at the base of the incline is equal to the whole length we’re trying to solve for 𝐿 that the marble moves up the incline multiplied by the sin of the angle 𝜃.

Looking at this energy balance equation, we can expand on the rotational kinetic energy term because we’re told the general shape of our marble that is a hollow sphere, which means, looking up this value from a table, that its moment of inertia is two-thirds its mass times its radius squared. In addition, looking at the angular speed 𝜔, we recall that there is a relationship between that angular speed and the linear speed 𝑣 for a rolling object. Specifically, 𝑣 the linear speed is equal to 𝑟 the radius of the object times its angular speed, which means that we can replace 𝜔 with 𝑣 over 𝑟 in this equation.

Writing out a new energy balance equation with this expanded expression for rotational kinetic energy, we see that the mass value 𝑚 is common to all three terms involved and therefore can be cancelled. And we also see that the factors of 𝑟, the radius of our hollow sphere, cancel one another out in the term for rotational kinetic energy. This leaves us with an expression that says 𝑣 squared times one-half plus one-third is equal to 𝑔𝐿 times the sin of 𝜃, where 𝑔 the acceleration due to gravity we’ll treat as exactly 9.8 metres per second squared. The fraction multiplying 𝑣 squared simplifies to five-sixths. And since we want to solve for the length 𝐿, we’ll rearrange this equation to isolate that term on one side. When we do, we see 𝐿 is equal to five 𝑣 squared over six 𝑔 sin 𝜃. Since we know 𝜃, 𝑣, and 𝑔, we’re ready to plug in and solve for 𝐿. When we do and enter these values on our calculator, we find that to two significant figures 𝐿 is 8.3 metres. That’s how far up the ramp the marble rolls before it comes to a stop.

Next, we wanna solve for 𝑡, the time value for how long it took the marble to move this distance. To do that, we can realize that as the marble ascends this incline, it’s under a constant acceleration. And therefore, the kinematic equations of motion apply to describe its motion. In particular, the relationship that says that the distance travelled is equal to the average of the initial and final speed multiplied by time will be useful to us in solving for 𝑡.

We can write that 𝐿 is equal to 𝑣 𝑓 plus 𝑣 the initial speed of the marble at the base of the incline all divided by two multiplied by the time 𝑡. We’ve specified the final speed of the marble to be zero. So 𝐿 is equal to 𝑣 over two times 𝑡 or 𝑡 is equal to two times 𝐿 over 𝑣. Plugging in for the values of 𝐿 and 𝑣, we find that to two significant figures 𝑡 is 2.4 seconds. That’s how long the marble would roll up the incline before coming to a stop. Let’s summarize what we’ve learned about motion of rolling objects.

In this video, we’ve seen that objects that roll without slipping combine rotational and translational motion. Keep in mind that if an object did start to slip, it would no longer be rotating, but rather just translating. And we’ve also seen that the kinetic energy of rolling objects includes energy from both types of motion. Written as an equation, we can say that the overall or total kinetic energy of a rolling object is equal to its translational kinetic energy plus its rotational kinetic energy. And finally, we’ve seen that angular speed 𝜔 and linear speed 𝑣 connect through the relationship 𝑣 is equal to 𝑟, the radius of the rolling object, times 𝜔.

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