I've been stuck on this problem for a while. Im not sure where to even begin.
A bike wheel is rotating with an angular velocity of 3.1 rev/s. If the coefficient of kinetic friction between the wheel and brakes is .87, then find the revolutions the wheel makes in coming to rest. The moment of inertia of the wheel is .24 kg/m and its radius is .32m.
Im especially confused on the part where the wheel comes to rest. Without a set distance or time, Im not sure where to start.
To solve this problem, you can use the concept of rotational motion and the relationship between torque, angular acceleration, and moment of inertia.
First, let's identify the known values given in the problem statement:
- Angular velocity, ω = 3.1 revolutions per second
- Coefficient of kinetic friction, μ = 0.87
- Moment of inertia, I = 0.24 kg/m
- Radius, r = 0.32 m
The objective is to find the number of revolutions the wheel makes in coming to rest.
Now, let's break down the problem into steps:
Step 1: Convert angular velocity to radians per second.
Since most formulas in physics use radians as the unit of angle, we need to convert the angular velocity from revolutions per second to radians per second.
1 revolution = 2π radians, so the angular velocity in radians per second is:
ω = 3.1 rev/s * (2π radians/rev) = 19.48 radians/s
Step 2: Calculate the angular acceleration.
The equation relating angular acceleration (α), torque (τ), and moment of inertia (I) is:
τ = I * α
In this case, the torque acting on the wheel is due to the friction between the wheel and the brakes. The formula for torque is:
τ = r * F, where F is the frictional force.
Substituting the coefficient of kinetic friction (μ) for the frictional force, we have:
τ = r * μ * N
The normal force (N) can be calculated as the weight (mg), where m is the mass and g is the acceleration due to gravity.
Assuming the bike wheel is not slipping, there is no linear acceleration, and the frictional torque provides the angular acceleration. Hence:
τ = I * α becomes r * μ * N = I * α
Step 3: Solve for the angular acceleration.
Rearrange the equation from the previous step to solve for α:
α = (r * μ * N) / I
Substituting the expression for N, we get:
α = (r * μ * m * g) / I
Step 4: Calculate the mass.
To find the mass, we use the relationship between moment of inertia and mass:
I = m * r^2
Rearranging the equation, we have:
m = I / r^2
Substituting the given values, we obtain:
m = 0.24 kg/m / (0.32 m)^2 = 2.34375 kg
Step 5: Calculate the angular acceleration.
Substituting the mass into the equation from Step 3, we obtain:
α = (0.32 m * 0.87 * 2.34375 kg * 9.8 m/s^2) / (0.24 kg/m) = 38.0375 rad/s^2
Step 6: Calculate the time taken to stop.
The formula that relates final angular velocity (ω_f), initial angular velocity (ω_i), and angular acceleration (α) is:
ω_f = ω_i + α * t
In this case, the final angular velocity is 0, because the wheel comes to rest. The initial angular velocity is 19.48 rad/s, as discussed in Step 1. Substituting these values, we can solve for the time (t) it takes to stop.
0 = 19.48 rad/s + (38.0375 rad/s^2) * t
Solving this equation for t gives:
t = -19.48 rad/s / (38.0375 rad/s^2) ≈ -0.512 seconds
Note that the negative sign indicates the direction of the angular velocity change.
Step 7: Calculate the number of revolutions.
The number of revolutions made by the wheel can be determined by multiplying the final angular velocity (0 rad/s) by the time it takes to stop (t) and converting it to revolutions:
Revolutions = 0 * t * (1 rev / 2π radians) ≈ 0 revolutions
Therefore, the wheel does not make any complete revolutions in coming to rest.
In conclusion, the bike wheel does not complete any revolutions before coming to rest.