A bicycle wheel of radius .7m is rolling without slipping on a horizontal surface with an angular speed of 2 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5 s. Through how many revolutions did the wheel turn during the 5 s of braking?

I tried it but got the wrong answer and I'm not sure what to do. Please help!

To determine the number of revolutions the wheel turned during the 5 s of braking, we need to calculate the angular displacement of the wheel.

First, let's find the initial angular velocity (ω_initial) of the wheel. We know that the angular speed (ω) is given by the formula:

ω = 2πf,

where f is the frequency in hertz. Since 1 revolution is equal to 1 cycle, the frequency here is 2 rev/s. Substituting the value into the formula, we get:

ω = 2π * 2 = 4π rad/s.

Next, we need to find the final angular velocity (ω_final) of the wheel when it comes to rest. We know that the angular acceleration (α) can be calculated using the equation:

α = (ω_final - ω_initial) / t,

where t is the time taken for the wheel to stop. Substituting the values given, we have:

α = (0 - 4π) / 5 = -4π/5 rad/s^2.

Since the wheel comes to rest, the final angular velocity is 0 rad/s. So, we can rearrange the formula to solve for ω_final:

0 = ω_initial + α * t,

0 = 4π + (-4π/5) * 5,

0 = 4π - 4π,

0 = 0.

As we can see, the final angular velocity is indeed 0 rad/s.

Now we can calculate the angular displacement (θ) of the wheel using the formula:

θ = ω_initial * t + 0.5 * α * t^2.

Substituting the given values, we get:

θ = 4π * 5 + 0.5 * (-4π/5) * (5^2),

θ = 20π - 4π,

θ = 16π rad.

Since 1 revolution is equal to 2π rad, we can convert the angular displacement to revolutions:

θ (in revolutions) = (16π rad) / (2π rad/rev),

θ (in revolutions) = 8 revolutions.

Therefore, the wheel turned through 8 revolutions during the 5 s of braking.