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.