Question Video: Refractive Index Relationship to Speed of Light Propagation

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?


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?

We’re told the wheels have a radius of 30.0 centimeters; we’ll call that π‘Ÿ. The bicycle starts from rest and accelerates to a speed of 10.0 meters per second; will call that speed 𝑣 sub 𝑓. And that happens in 10.0 seconds, which we’ll call 𝑑.

We want to know the angular acceleration of the tires, which we’ll refer to as 𝛼. As we start our solution, let’s recall the definition for linear rather than angular acceleration.

Linear acceleration, π‘Ž, is defined as the change in velocity over the change in time. If we move to angular acceleration, 𝛼, the definition is the same as from linear except that instead of linear velocity 𝑣 we have angular velocity πœ”.

Applying the relationship for angular acceleration to our situation, 𝛼 is equal to Ξ”πœ” over Δ𝑑. Or πœ” sub 𝑓, the final angular speed, minus πœ” sub 𝑖, the initial angular speed, divided by the time 𝑑.

We don’t know the initial or final angular speed, but we can recall the relationship between linear speed 𝑣 and angular speed, πœ”. 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‘Ÿ multiplied by the angular speed πœ”.

For example, if we have a rotating wheel of radius π‘Ÿ rotating in an angular speed πœ”, then the linear speed of a point on the circumference of the wheel, which we will call 𝑣, is equal to π‘Ÿ times πœ”. All this means that we can replace πœ” sub 𝑓 and πœ” sub 𝑖 in our equation for angular acceleration.

Since 𝑣 equals π‘Ÿ times πœ”, that means πœ” is equal to 𝑣 divided by π‘Ÿ. So πœ” sub 𝑓 in our final equation becomes 𝑣 sub 𝑓 over π‘Ÿ and πœ” sub 𝑖 becomes 𝑣 sub 𝑖 over π‘Ÿ. We were told in the problem statement that 𝑣 sub 𝑖, the initial speed of the bike, is zero. This means our equation for angular acceleration simplifies to 𝑣 sub 𝑓 divided by π‘Ÿ times 𝑑.

We’ve been given each of these variables in the problem statement and can plug in their values now; 𝑣 sub 𝑓 is 10.0 meters per second; π‘Ÿ, in units of meters, is zero point three zero meters; and time 𝑑 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.

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