A cyclist is riding such that the wheels of the bicycle have a rotation rate of 3.0 revolutions per second. If the cyclist brakes such that the rotation rate of the wheels decrease at a rate of 0.3 revolutions per second squared, how long does it take for the cyclist to come to a complete stop?
Let’s start by highlighting some of the important information we’re given. We’re told that, initially, the bicycle is moving so that the wheels rotate at a rate of 3.0 revolutions per second. So instead of being given a linear speed, we’re given a rotational speed.
We’re also told that once the brakes are applied, the rotation rate of the wheels is decreasing at a rate of 0.3 revolutions per second squared. We want to solve for the time value — we’ll represent it as lowercase 𝑡 — that it takes for the cyclist to come to a complete stop.
Let’s begin by drawing a diagram of the wheel of the bicycle. We were told that the bicycle wheel is initially rotating with a rotational rate of 3.0 revolutions per second.
We’ll represent that rate using the Greek letter lowercase 𝜔. And since this rotational rate is the initial rotational rate of the wheel, which changes over time, let’s add the subscript 𝑖 to 𝜔 to represent the fact that this is an initial rotational rate.
We were told that once the brakes are applied to this rotating wheel, the wheel experiences a deceleration, and that deceleration happens at a rate of 0.3 revolutions per second squared. We’ll represent that angular acceleration using the Greek letter 𝛼.
We know that once the bike has come to a stop, the rotational rate of the wheel will be zero, so let’s assign a name for that; we’ll call that 𝜔 sub 𝑓, where that final rotational speed is exactly zero revolutions per second.
You may recall that, in situations where acceleration is constant, we’ve referred to a set of equations called the kinematic equations. These equations describe translational or linear motion, but it turns out there’s a rotational analogue set of kinematic equations. And just as with the translational kinematic equations, we can write the rotational versions below.
Look over these equations for a moment and hopefully they’ll seem a bit familiar. In place of the final and initial speeds, 𝑣 sub 𝑓 and 𝑣 sub 𝑖, we have the final and initial rotational speeds, 𝜔 sub 𝑓 and 𝜔 sub 𝑖. In place of the translational acceleration 𝑎, we have the rotational acceleration 𝛼. And instead of 𝑑 for displacement, we use the angular displacement 𝜃.
So we’ve essentially substituted in rotational values for translational ones, but otherwise the form of the equations is the same as we’ve seen before. So as we’ve done before, let’s choose one of the kinematic equations to help us solve this problem.
The very first rotational kinematic equation, 𝜔 sub 𝑓 equals 𝜔 sub 𝑖 plus 𝛼 times 𝑡, applies to our situation. As we look at each term in this equation, we see that, for each one, we’re either given that term in the problem statement or in the case of 𝑡 asked to solve for it.
So let’s rearrange this rotational kinematic equation to solve for the time 𝑡. First we subtract 𝜔 sub 𝑖 from both sides, which results in that term canceling out on the right side of our equation. We then divide both sides of the equation by the angular acceleration 𝛼, which results in that term canceling from the right side of the equation.
We can now rewrite this equation with 𝑡 by itself on one side: 𝑡 equals 𝜔𝑓, the final rotational speed, minus 𝜔𝑖, the initial rotational speed, divided by 𝛼, the rotational acceleration.
Each of these three terms is given to us in the problem statement, so we now insert those into this relationship. When we enter these values into our calculator, we find a time of 10 seconds. That’s how long it would take this bicycle wheel to come to a complete stop.