Video: Calculating the Angular Momentum of a Solid Sphere

A spherical boulder with a mass of 20 kg and a radius of 20 cm rolls down the side of a hill, starting from rest. The hill is 15 m high. What is its angular momentum when it is halfway down the hill? What is it angular momentum when it is at the bottom?

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Video Transcript

A spherical boulder with a mass of 20 kilograms and a radius of 20 centimeters rolls down the side of a hill starting from rest. The hill is 15 meters high. What is its angular momentum when it is halfway down the hill? What is it angular momentum when it is at the bottom?

Weโ€™re told the boulder has a mass of 20 kilograms, weโ€™ll call that ๐‘š, and that it has a radius of 20 centimeters, which weโ€™ll refer to as ๐‘Ÿ. Weโ€™re also told that the hill the boulder rolls down has a height of 15 meters, which weโ€™ll call โ„Ž. We want to know in this two-part problem the angular momentum of the boulder when itโ€™s halfway and completely down the hill. Weโ€™ll call these values ๐ฟ sub one-half and ๐ฟ sub ๐‘“ respectively.

Letโ€™s begin our solution by drawing a diagram. In this picture of the ball on a hill, the ball starts out at the top 15 meters above the bottom of the hill. Starting from rest, it rolls downward. And if we take a snapshot when the ball is halfway down the hill, itโ€™s at that moment when we want to solve for the angular momentum on the ball. And likewise, if we take a snapshot when the ball is at the bottom of the hill, itโ€™s at that instead that we want to solve for angular momentum weโ€™ve called ๐ฟ sub ๐‘“.

Letโ€™s start by recalling that energy is conserved within a closed system. This means that the initial energy in our system is equal to the final energy. For our scenario, we can write this, that the initial kinetic plus potential energy is equal to the final kinetic plus potential energy. Since the ball starts out from rest, initially itโ€™s kinetic energy is zero. And further, wherever the ball ends up, wherever we call its position final, we can choose to set that altitude as the value at which height is zero.

Therefore, as we calculate both ๐ฟ sub one-half and ๐ฟ sub ๐‘“, weโ€™ll assume that the final potential energy is zero. Looking at the remaining terms, we can recall that gravitational potential energy equals an objectโ€™s mass times the acceleration due to gravity, ๐‘”, times its height above the zero point. We can enter that information into our equation.

Now what about the final kinetic energy of the ball? This is a case of rolling kinetic energy, which means that the ball doesnโ€™t just translate from one place to another, it also rotates as it translates. So there are two terms to the kinetic energy of this rolling ball. First, we have the linear or translational term. And second, we have the rotational term. And we need to include both terms because the object weโ€™re considering is rolling. So we plug in this expression for KE sub ๐‘“.

So we have our energy balance expression written here. But how does this relate to angular momentum? Recall that angular momentum, ๐ฟ, is equal to an objectโ€™s moment of inertia multiplied by its angular velocity, ๐œ”. This means that if we can solve for ๐ผ and for ๐œ”, weโ€™ll be able to compute ๐ฟ. As far as the moment of inertia, when we look up in a table the ๐ผ value for a uniform sphere, thatโ€™s equal to two-fifths the mass of the sphere multiplied by the square of its radius.

The problem statement tells us ๐‘š and ๐‘Ÿ. So weโ€™ll be able to solve for ๐ผ. That leaves ๐œ”, the angular speed, as an unknown. Letโ€™s look back at our conservation of energy equation. This equation includes ๐œ”. And if we look at the ๐‘ฃ term and recall that linear speed ๐‘ฃ is equal to ๐‘Ÿ times angular speed ๐œ”, we can perform a substitution so that now our energy balance equation has ๐œ” and no linear speed ๐‘ฃ terms.

We now want to algebraically rearrange this equation to solve for ๐œ”. As a first step, letโ€™s substitute in the moment of inertia of the sphere, ๐ผ sub sphere, for capital ๐ผ in our equation. When we do this, we see that the mass ๐‘š of the boulder is common in every term in the equation. So we can cancel it out. If we then multiply the two fractions in the far right term and group the ๐œ” squared by factoring it out, we can add the one-half ๐‘Ÿ squared plus one-fifth ๐‘Ÿ squared to result in seven-10ths ๐‘Ÿ squared.

Now letโ€™s multiply both sides by 10 divided by seven ๐‘Ÿ squared, which cancels out that fraction and the ๐‘Ÿ squared term on the right-hand side. And finally if we take the square root of both sides, then on the-right hand side the square root cancels with the square term and we have an expression for angular speed ๐œ”: ๐œ” equals one divided by ๐‘Ÿ times the square root of 10 over seven times the acceleration due to gravity ๐‘” multiplied by โ„Ž sub ๐‘–.

Now that we have an expression for ๐œ”, letโ€™s bring that expression together with the expression for moment of inertia to calculate the angular momentum ๐ฟ. Since ๐ฟ equals ๐ผ times ๐œ” ๐ฟ equals two-fifths ๐‘š๐‘Ÿ squared multiplied by one over ๐‘Ÿ times the square root of 10-sevenths times ๐‘” timed โ„Ž sub ๐‘–. Notice that in this equation one factor of ๐‘Ÿ cancels out, which simplifies the overall expression. Letโ€™s use this general equation for angular momentum in this problem to solve for ๐ฟ sub one-half, the angular momentum of the boulder when itโ€™s halfway down the hill.

For ๐‘š, we insert the value of 20 kilograms. For ๐‘Ÿ, we convert 20 centimeters to a distance in meters of 0.20. We treat ๐‘”, the acceleration due to gravity, as exactly 9.8 meters per second squared. And for โ„Ž, because the ball is halfway down the 15-meter-tall hill, we use a value of 7.5 meters. When we compute this value, we find that, to two significant figures, it equals 16 kilograms-meters squared per second. Thatโ€™s the angular momentum of the boulder when it is halfway down the hill.

But what about when the boulder is completely at the bottom? In other words, what is ๐ฟ sub ๐‘“? Looking at our equation for ๐ฟ sub one-half, to apply it when the ball is at the bottom of the hill, we only need to change one thing. The height weโ€™ve used of 7.5 meters will change to 15 meters when we compute ๐ฟ sub ๐‘“. All the other values stay the same. When we make this substitution and compute the value for ๐ฟ sub ๐‘“, we find that itโ€™s 23 kilograms-meter squared per second. This is the angular momentum of the boulder when itโ€™s at the bottom of the hill.

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