A wheel's uniform angular acceleration is constant. Initially its angular velocity is zero. During the first 1.0 s time interval, it rotates through an angle of 90.0°.
(a) Through what angle does it rotate during the next 1.0 s time interval?
(b) Through what angle during the third 1.0 s time interval?
I know similar questions have been posted before, but none are very helpful to me, so if one could give me a setup of how to do these please! I thought that for a it would be 180 degrees and for b) 270 degree, but its wrong. Sorry for re-posting this! :/
theta = (1/2) alpha t^2
pi/2 = (1/2) alpha (1)^2
alpha = pi radians/s^2
total theta after 2 seconds = pi/2(4)
= 2 pi
so during second second it goes 2 pi - (p/2) = 3 pi/2 radians
total theta after 3 seconds = (pi/2)(9)
so during third second = (4.5-2)pi
Thank you Damon :)!
No problem! Let's go through the setup to solve these problems step by step.
We can use the kinematic equation for rotational motion:
θ = ωi * t + (1/2) * α * t^2,
where:
θ is the angular displacement,
ωi is the initial angular velocity,
t is the time interval, and
α is the angular acceleration.
(a) To find the angle during the second time interval, we need to know the angular velocity at the end of the first time interval. Since the problem states that the wheel's uniform angular acceleration is constant, we can assume that the angular acceleration remains constant throughout the entire motion. Given that the initial angular velocity, ωi, is zero, we can calculate the angular displacement during the first time interval using the formula above:
θ1 = ωi * t + (1/2) * α * t^2,
θ1 = 0 * 1.0 s + (1/2) * α * (1.0 s)^2,
θ1 = 0 + (1/2) * α * 1.0 s^2,
θ1 = (1/2) * α * 1.0 s^2,
θ1 = (1/2) * α.
Since the problem states that the wheel rotates through an angle of 90° during the first time interval, we know that θ1 = 90°. Therefore,
(1/2) * α = 90°,
α = 180°/s^2.
Now, we can find the angular displacement during the second time interval by applying the same formula:
θ2 = ωi * t + (1/2) * α * t^2,
θ2 = 0 * 1.0 s + (1/2) * (180°/s^2) * (1.0 s)^2,
θ2 = 0 + (1/2) * (180°/s^2) * 1.0 s^2,
θ2 = (1/2) * (180°/s^2),
θ2 = 90°.
Therefore, the wheel rotates through an angle of 90° during the second 1.0 s time interval.
(b) Similarly, for the third time interval, we can apply the same formula:
θ3 = ωi * t + (1/2) * α * t^2,
θ3 = 0 * 1.0 s + (1/2) * (180°/s^2) * (1.0 s)^2,
θ3 = 0 + (1/2) * (180°/s^2) * 1.0 s^2,
θ3 = (1/2) * (180°/s^2),
θ3 = 90°.
Therefore, the wheel also rotates through an angle of 90° during the third 1.0 s time interval.
So, the correct answers are:
(a) 90°,
(b) 90°.
I hope this helps! If you have any further questions, feel free to ask.