Question

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time interval, the wheel rotates through 70.4 rad. what is the angular speed of the wheel at the end of the 3.80-s interval

Answer 1

Answer:8.65

Step-by-step explanation:

70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)²

ω₀ = (70.4 + 37.544) / 3.80

ω₀ = 28.406 rad/s

Using equation 2:

ω² = (28.406)² + 2(-5.2)70.4

ω = 8.65 rad/s

Answer 2

We can answer this using the rotational version of the kinematic equations:
θ = θ₀ + ω₀t + ½αt² -----> 1

ω² = ω₀² + 2αθ -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)²

ω₀ = (70.4 + 37.544) / 3.80

ω₀ = 28.406 rad/s


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4


ω = 8.65 rad/s