A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s², what is its angular velocity at t = 2.50 s? (b) Through what angle has the wheel turned between t = 0 and t = 2.50 s?

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Welcome back. Everyone in this problem. A wind turbine has a steady angular acceleration of 0.1 radiance per second squared and an initial angular speed of 0.5 radiance per second find its angular speed after four seconds. And the angle the turbine will have turned through between T equals zero and T equals four seconds. A says the angular velocity is 1.5 radiance per second and the angle is six radiance B says they are 0.9 radiance per second and six radiance respectively. C says they are 1.5 radiance per 2nd and 2.8 radiance respectively. And B says they are 0.9 radiance per 2nd and 2.8 radiance respectively. Now let's start by trying to find the angular speed after four seconds. What information are we given so far? Well, here we are given the angular acceleration alpha of 0.1 radiance per second squared and our initial angular speed omega knot of 0.5 radiance per second. So how can we relate both of those to figure out the angular speed after four seconds? Well, recall, OK. This is for the first one. So let me put this in red here. Recall that for our speed under uniform acceleration, the final speed omega equals our initial speed omega knot plus or angular acceleration alpha multiplied by the time. Now, for our problem, we have all of this information and we know that our time is four seconds. So our final angular speed is going to be equal to 0.5 radius per second plus our acceleration, 0.1 radiance per second squared multiplied by four seconds. I know when we go ahead and solve here, that's going to be 0.5 radiance per second plus 0.4 radiance per second, which equals 0.90 radiance per second. So that is going to be our angular speed after four seconds. Next. What's the angle the turbine will have when turned between T equals zero and T equals for a second. Well, to calculate how much our anger the has been turned between time intervals, we can use the formula that says our anger theta is equal to Omega knot multiplied by T plus half of our angular acceleration alpha multiplied by T squared. Again, we have all of this information so we can go ahead and solve. OK. Because we know Omega knott is 0.5 radiance per second. Our change in time is from 0 to 4 seconds. So T is four seconds. OK? Or angular acceleration is 0.1 radiance per second squared. And again, our time is four seconds. So that's four seconds squared. Now, we can solve for our angle. Now, when we do that, ok, then we should get at our angle equals 2.8 radiance. Thus, after four seconds, our wind turbine will have an angular speed of 0.9 radiance per second. And it will turn through an angle of 2.8 radiance. If we look back at our answer choices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s², what is its angular velocity at t = 2.50 s? (b) Through what angle has the wheel turned between t = 0 and t = 2.50 s?

Verified video answer for a similar problem:

Key Concepts

Angular Velocity

Angular Acceleration

Angular Displacement

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s2, what is its angular velocity at t = 2.50 s? (b) Through what angle has the wheel turned between t = 0 and t = 2.50 s?

Verified video answer for a similar problem:

Key Concepts

Angular Velocity

Angular Acceleration

Angular Displacement

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s², what is its angular velocity at t = 2.50 s? (b) Through what angle has the wheel turned between t = 0 and t = 2.50 s?