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 5s. Through how many revolutions did the wheel turn during the 5s of braking?
displacament=wi*time-1/2*(2/5 rev/s^2)*time^2
solve for displacement (in revs).
To find out how many revolutions the wheel turned during the 5 seconds of braking, we need to calculate the angular displacement of the wheel during this time.
Let's start by finding the initial angular displacement of the wheel. The initial angular speed is given as 2 rev/s, which means the wheel completes 2 full revolutions in 1 second. Therefore, the initial angular displacement is:
Initial angular displacement = Initial angular speed * Time
= 2 rev/s * 0.5 s (since the braking period is 1/2 of a second)
= 1 revolution
Now, let's find the final angular displacement. The wheel stops rotating after 5 seconds of braking, so the final angular displacement is 0 revolutions.
Finally, we can find the total angular displacement during the 5 seconds of braking by subtracting the initial angular displacement from the final angular displacement:
Total angular displacement = Final angular displacement - Initial angular displacement
= 0 revolutions - 1 revolution
= -1 revolution
However, since we are interested in the total number of revolutions, we should take the absolute value of the angular displacement:
Total number of revolutions during braking = |Total angular displacement|
= |-1 revolution|
= 1 revolution
Therefore, the wheel turned 1 revolution during the 5 seconds of braking.
To find the number of revolutions the wheel turned during the 5s of braking, we need to calculate the angular displacement of the wheel.
The formula to calculate angular displacement is:
θ = ωi * t + (1/2) * α * t^2
Where:
- θ is the angular displacement
- ωi is the initial angular velocity
- t is the time interval
- α is the angular acceleration
In this case, the wheel is rolling without slipping, so its angular acceleration is zero. Therefore, the equation simplifies to:
θ = ωi * t
Given that the wheel's initial angular speed (ωi) is 2 rev/s and the braking time (t) is 5s, we can substitute these values into the equation and solve for θ:
θ = 2 rev/s * 5s
θ = 10 revolutions
Therefore, the wheel turned 10 revolutions during the 5s of braking.