Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?

U=0,t=T,

'A ball starts from rest and accelerates at 0.505 m/s2 while moving down an inclined plane 9.95 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 16.0 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane

To answer this question, we need to understand that the average angular speed (ω_avg) is equal to the final angular speed (ω_final) divided by 2.

We can use the formula for angular acceleration (α) to relate the final angular speed to the initial angular speed (ω_initial), time (t), and angular acceleration:

ω_final = ω_initial + αt

Since the wheel starts from rest (ω_initial = 0), the equation simplifies to:

ω_final = αt

We can now substitute the relationship between ω_avg and ω_final:

ω_avg = ω_final / 2

Rewriting the equation:

ω_final = 2ω_avg

Substituting this back into the previous equation:

2ω_avg = αt

Solving for t:

t = (2ω_avg) / α

So, the wheel reaches an angular speed equal to its average angular speed after a time of T = (2ω_avg) / α.

To solve this problem, we need to find the time at which the angular speed of the wheel is equal to its average angular speed for the given interval.

First, let's define the problem variables:
- Initial angular speed: ω0 (assume it is zero since it starts from rest)
- Final angular speed: ωf (to be determined)
- Average angular speed: ωavg (to be determined)
- Time interval: T

The average angular speed can be calculated by dividing the change in angular displacement by the time interval:
ωavg = Δθ / T

To find the change in angular displacement, we need to use the equation of motion for angular acceleration:

θ = ω0 * t + (1/2) * α * t^2

where:
- θ is the angular displacement
- ω0 is the initial angular speed
- α is the angular acceleration
- t is the time

Since the wheel starts from rest (ω0=0), the equation simplifies to:
θ = (1/2) * α * t^2

To find the final angular speed (ωf) at the given time T, we use the equation for angular speed:
ωf = ω0 + α * t

Since ω0 is zero, we get:
ωf = α * t

Now, let's substitute this equation for ωf into the equation for average angular speed:
ωavg = Δθ / T

Since Δθ = (1/2) * α * t^2, the equation becomes:
ωavg = (1/2) * α * t^2 / T

Simplifying this equation, we can find the time at which the wheel reaches an angular speed equal to its average angular speed:
t = sqrt(2 * ωavg * T / α)

Therefore, the wheel reaches an angular speed equal to its average angular speed at time t = sqrt(2 * ωavg * T / α).