A motorcycle, which has an initial linear speed of 6.6 m/s, decelerates to a speed of 2.1 m/s in 5.0 s. Each wheel has a radius of 0.65 m and is rotating in a counterclockwise (positive) direction. What are a) the constant angular acceleration (in rad/s^2) and b) the angular displacement (in rad) of each wheel?
no
To find the constant angular acceleration and angular displacement of each wheel, we can use the following formulas:
a) Angular acceleration (α) = (final angular velocity - initial angular velocity) / time
b) Angular displacement (θ) = initial angular velocity * time + (1/2) * angular acceleration * time^2
First, let's find the initial angular velocity of each wheel:
ω = linear speed / radius
Given that the initial linear speed is 6.6 m/s and the wheel radius is 0.65 m, we can calculate the initial angular velocity:
ω = 6.6 m/s / 0.65 m
ω ≈ 10.15 rad/s
Now, let's calculate the final angular velocity of each wheel:
ωf = linear speed / radius
Given that the final linear speed is 2.1 m/s and the wheel radius is 0.65 m, we can calculate the final angular velocity:
ωf = 2.1 m/s / 0.65 m
ωf ≈ 3.23 rad/s
Using the given time interval of 5.0 seconds, we can now calculate the angular acceleration (α):
α = (ωf - ω) / t
= (3.23 rad/s - 10.15 rad/s) / 5.0s
= -1.39 rad/s²
Therefore, the constant angular acceleration (α) is approximately -1.39 rad/s².
Now, let's calculate the angular displacement (θ) of each wheel:
θ = ω * t + (1/2) * α * t^2
For the first wheel:
θ1 = 10.15 rad/s * 5.0s + (1/2) * (-1.39 rad/s²) * (5.0s)^2
= 50.75rad - 17.38rad
≈ 33.37 rad
For the second wheel (assuming it's rotating in the opposite direction):
θ2 = -10.15 rad/s * 5.0s + (1/2) * (-1.39 rad/s²) * (5.0s)^2
= -50.75rad - 17.38rad
≈ -68.13 rad
Therefore, the angular displacement (θ) of each wheel is approximately 33.37 rad and -68.13 rad, respectively.
To solve this problem, we will use the equations of linear and rotational motion.
a) To find the constant angular acceleration (α), we need to relate the linear and angular quantities. The linear speed (v) of a point on the wheel is related to the angular speed (ω) by the equation v = rω, where r is the radius of the wheel.
Initially, the linear speed is 6.6 m/s, which corresponds to an angular speed of ω1 = v/r = 6.6 m/s / 0.65 m ≈ 10.15 rad/s.
Finally, the linear speed is 2.1 m/s, which corresponds to an angular speed of ω2 = 2.1 m/s / 0.65 m ≈ 3.23 rad/s.
The change in angular speed (Δω) can be calculated as Δω = ω2 - ω1 = 3.23 rad/s - 10.15 rad/s ≈ -6.92 rad/s.
The time taken (t) is given as 5.0 s.
The constant angular acceleration (α) can be calculated using the formula α = Δω / t.
Substituting the known values, α = -6.92 rad/s / 5.0 s ≈ -1.384 rad/s^2.
Therefore, the constant angular acceleration is approximately -1.384 rad/s^2.
b) To find the angular displacement, we can use the equation of rotational motion, θ = ω1t + (1/2)αt^2.
Substituting the known values, θ = (10.15 rad/s)(5.0 s) + (1/2)(-1.384 rad/s^2)(5.0 s)^2 ≈ 50.75 rad.
Therefore, the angular displacement of each wheel is approximately 50.75 rad.