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A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

Nady [450]

Answer:

The time interval is $t = 3 \ s$

Explanation:

From the question we are told that

The angular acceleration is $\alpha = 4.0 \ rad/s^2$

The time taken is $t = 4.0 \ s$

The angular displacement is $\theta = 80 \ radians$

The angular displacement can be represented by the second equation of motion as shown below

$\theta = w_i t + \frac{1}{2} \alpha t^2$

where $w_i$ is the initial velocity at the start of the 4 second interval

So substituting values

$80 = w_i * 4 + 0.5 * 4.0 * (4^2)$

=> $w_i = 12 \ rad/s$

Now considering this motion starting from the start point (that is rest ) we have

$w__{4.0 }} = w__{0}} + \alpha * t$

Where $w__{0}}$ is the angular velocity at rest which is zero and $w__{4}}$ is the angular velocity after 4.0 second which is calculated as 12 rad/s s