Wheel A of radius ra = 6.1 cm is coupled by belt B to wheel C of radius rc = 33.3 cm. Wheel A increases its angular speed from rest at time t = 0 s at a uniform rate of 7.1 rad/s2. At what time will wheel C reach a rotational speed of 119.3 rev/min, assuming the belt does not slip?
To solve this problem, we can use the rotational kinematic equation:
ωf = ωi + αt
where
ωf is the final angular velocity,
ωi is the initial angular velocity,
α is the angular acceleration, and
t is the time.
First, let's convert the final angular velocity of wheel C from revolutions per minute to radians per second:
ωf_C = 119.3 rev/min = 119.3 * 2π rad/60 s = 12.54 rad/s
Next, we need to find the initial angular velocity of wheel A. Since it starts from rest, ωi_A = 0 rad/s.
The problem states that the wheel A increases its angular speed at a uniform rate of 7.1 rad/s^2, so α_A = 7.1 rad/s^2.
We know that wheel C is larger than wheel A, so their angular velocities are related by the belt B. The ratio between their angular velocities is equal to the ratio between their radii:
ω_C / ω_A = rc / ra
Substituting the given values:
12.54 rad/s / ω_A = 33.3 cm / 6.1 cm
Simplifying, we find:
ω_A = (12.54 rad/s)(6.1 cm) / 33.3 cm ≈ 2.30 rad/s
Now we can use the kinematic equation to solve for the time t:
ωf_C = ωi_C + α_C t
Substituting the known values:
12.54 rad/s = 2.30 rad/s + 7.1 rad/s^2 * t
Simplifying, we have:
10.24 rad/s = 7.1 rad/s^2 * t
Dividing both sides by 7.1 rad/s^2:
t = 10.24 rad/s / 7.1 rad/s^2
t ≈ 1.44 s
Therefore, wheel C will reach a rotational speed of 119.3 rev/min at approximately 1.44 seconds.