# Video: Refractive Index Relationship to Speed of Light Propagation

Ed Burdette

A boy rides his bicycle which has wheels with a radius of 30.0 cm. If the boy on the bicycle accelerates from rest to a speed of 10.0 m/s in 10.0 s, what is the angular acceleration of the tires?

03:42

### Video Transcript

A boy rides his bicycle, which has wheels with a radius of 30.0 centimeters. If the boy on the bicycle accelerates from rest to a speed of 10.0 meters per second in 10.0 seconds, what is the angular acceleration of the tires?

Weβre told the wheels have a radius of 30.0 centimeters; weβll call that π. The bicycle starts from rest and accelerates to a speed of 10.0 meters per second; will call that speed π£ sub π. And that happens in 10.0 seconds, which weβll call π‘.

We want to know the angular acceleration of the tires, which weβll refer to as πΌ. As we start our solution, letβs recall the definition for linear rather than angular acceleration.

Linear acceleration, π, is defined as the change in velocity over the change in time. If we move to angular acceleration, πΌ, the definition is the same as from linear except that instead of linear velocity π£ we have angular velocity π.

Applying the relationship for angular acceleration to our situation, πΌ is equal to Ξπ over Ξπ‘. Or π sub π, the final angular speed, minus π sub π, the initial angular speed, divided by the time π‘.

We donβt know the initial or final angular speed, but we can recall the relationship between linear speed π£ and angular speed, π. For a circularly rotating object, the linear speed at a point on an object is equal to the distance from the axis of rotation to that point π multiplied by the angular speed π.

For example, if we have a rotating wheel of radius π rotating in an angular speed π, then the linear speed of a point on the circumference of the wheel, which we will call π£, is equal to π times π. All this means that we can replace π sub π and π sub π in our equation for angular acceleration.

Since π£ equals π times π, that means π is equal to π£ divided by π. So π sub π in our final equation becomes π£ sub π over π and π sub π becomes π£ sub π over π. We were told in the problem statement that π£ sub π, the initial speed of the bike, is zero. This means our equation for angular acceleration simplifies to π£ sub π divided by π times π‘.

Weβve been given each of these variables in the problem statement and can plug in their values now; π£ sub π is 10.0 meters per second; π, in units of meters, is zero point three zero meters; and time π‘ is 10.0 seconds.

When we calculate this fraction, we find a value to three significant figures of three point three three radians per second squared. This is the angular acceleration of the tires on the bicycle.