# Video: Calculating the Angular Momentum of a Wheel

A mountain bike is traveling at 10 m/s when it jumps, and the wheels lose contact with the ground. The mass of the bike’s front wheel is 750 g and the wheel’s radius is 35 cm. What is the angular momentum of the spinning wheel in the air at the moment that the bike leaves the ground?

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### Video Transcript

A mountain bike is traveling at 10 meters per second when it jumps, and the wheels lose contact with the ground. The mass of the bike’s front wheel is 750 grams. And the wheel’s radius is 35 centimeters. What is the angular momentum of the spinning wheel in the air at the moment that the bike leaves the ground?

We’re told in this statement that the bike is traveling at 10 meters per second, which we’ll call 𝑣. We’re also told the mass of the bike’s front wheel is 750 grams, which we’ll name 𝑚. And we’ve learned that the wheel’s radius is 35 centimeters, which we’ll call 𝑟.

We want to know the angular momentum of the wheel just after the bike leaves the ground. We’ll call that capital 𝐿. Let’s draw a sketch of our situation. The mountain bike is moving along at a speed, 𝑣, of 10 meters per second. The rider then jumps off the ground with the bike. And while the bike is in the air, the wheels continue to spin.

To solve for the angular momentum of the front wheel of the bike, let’s recall the relationship that defines that term. The angular momentum of an object 𝐿 is equal to its moment of inertia, 𝐼, times its angular velocity, 𝜔. So 𝐿 equals 𝐼 𝜔. We can solve for these two terms in order.

First, the moment of inertia 𝐼, if we treat the front wheel of the bicycle as a hoop of uniform radius and mass throughout the hoop’s circumference, then when we look up the moment of inertia of an object of that type in a table, we find that it’s equal to the mass of the hoop times its radius squared. So we can rewrite 𝐿 as 𝑚𝑟 squared 𝜔.

Now let’s recall the relationship between linear and rotational velocity. Linear velocity, 𝑣, is equal to 𝑟 times the angular velocity, 𝜔. This means that 𝜔 is equal to linear velocity, 𝑣, divided by 𝑟. We can make this substitution into our equation for 𝐿. In doing so, we see that a factor of 𝑟 cancels out. And we’re left with an equation that reads angular momentum 𝐿 equals 𝑚 times 𝑟 times 𝑣.

Each of these three values is given in our problem statement. So we can plug them into this equation now. As we do so, we’ll be careful to use SI units of meters, kilograms, and meters per second. The mass 𝑚, in units of kilograms, is 0.750. The radius 𝑟, in units of meters, is 0.35. And the bike’s speed, 𝑣, is already given in units of meters per second.

When we multiply these numbers together, we find a result of 2.6 kilograms-meter squared per second. To two significant figures, this is the angular momentum of the front wheel of the bike just after the jump starts.