A stationary bicycle is raised off the ground, and its front wheel (m = 1.3 kg) is rotating at an angular velocity of 12.3 rad/s (see the drawing). The front brake is then applied for 3.0 s, and the wheel slows down to 3.3 rad/s. Assume that all the mass of the wheel is concentrated in the rim, t
and the question is....
To solve this problem, we can use the equation for rotational kinetic energy:
Kinetic energy (KE) = 1/2 * I * ω^2
where I is the moment of inertia of the wheel and ω is the angular velocity.
The moment of inertia for a solid disc can be calculated as:
I = (1/2) * m * r^2
where m is the mass of the wheel and r is the radius of the wheel.
First, let's find the initial moment of inertia of the wheel:
I_initial = (1/2) * m * r^2
Given that the mass of the wheel is 1.3 kg and we know the radius of the wheel. However, the radius is not mentioned in your question, so we will assume it is given or can be calculated separately.
Now, let's calculate the initial kinetic energy of the wheel:
KE_initial = 1/2 * I_initial * ω_initial^2
where ω_initial is the initial angular velocity of the wheel (12.3 rad/s).
Next, we need to find the final kinetic energy of the wheel:
KE_final = 1/2 * I_final * ω_final^2
where ω_final is the final angular velocity of the wheel (3.3 rad/s).
To find the final moment of inertia, we can use the equation for conservation of angular momentum:
I_initial * ω_initial = I_final * ω_final
Rearranging the equation we get:
I_final = (I_initial * ω_initial) / ω_final
Now, we can substitute the values we have and calculate the final moment of inertia.
Finally, we substitute the values of I_final and ω_final into the equation for KE_final and solve for the final kinetic energy.
By finding the difference between the initial and final kinetic energy, we can determine the work done by the brake on the wheel during the braking process.