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 seconds of braking, we need to find the angular displacement of the wheel.
First, let's calculate the initial angular velocity of the wheel (ωi). Since the wheel is rolling without slipping, the linear speed (v) of a point on the rim can be related to the angular speed (ωi) by the equation:
v = ωi * r
Where r is the radius of the wheel. Given that the radius is 0.7m and the angular speed is 2 rev/s:
v = (2 rev/s) * (0.7m) = 1.4 m/s
Now, we can calculate the acceleration (a) of the wheel during braking. We know that the bicycle stops in 5 seconds, so the final speed (vf) is 0. The formula for acceleration can be written as:
a = (vf - vi) / t
Where vi is the initial velocity and t is the time taken. Substituting the given values:
0 = (0 - 1.4 m/s) / 5 s
Solving for a, we find:
a = -1.4 m/s^2
Since the bicycle is braking, the acceleration has a negative sign.
Next, we can use the constant acceleration kinematic equation to find the angular displacement (θ) during the 5 seconds of braking:
θ = ωi * t + 0.5 * a * t^2
Substituting the values:
θ = (2 rev/s) * 5 s + 0.5 * (-1.4 m/s^2) * (5 s)^2
Simplifying:
θ = 10 rev + -17.5 rev
θ = -7.5 rev
The negative sign indicates that the wheel has rotated in the opposite direction during braking.
Therefore, during the 5 seconds of braking, the wheel turned -7.5 revolutions.