A 46.5-cm diameter disk rotates with a constant angular acceleration of 2.4 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.
(Hint: Remember the counter-clockwise rule for positive angular velocity and acceleration.)
(a) Find the angular speed of the wheel at t = 2.30 s.
___ rad/s
(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
linear velocity ___ m/s
tangential acceleration __ m/s2
c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
(Hint: Your answer must be between 0 and 360 degrees.)
____°
I will be happy to critique your thinking.
I don't know where to start off.
To solve this problem, we can use the equations of rotational motion.
(a) For part (a), we need to find the angular speed of the wheel at t = 2.30 s. We know that the angular acceleration, α, is given as 2.4 rad/s^2. The initial angular speed, ω(0), is 0 since the disk starts from rest.
We can use the equation: ω(t) = ω(0) + αt.
Plugging in the values, we have:
ω(t) = 0 + 2.4 rad/s^2 * 2.30 s
ω(t) = 5.52 rad/s
So, the angular speed of the wheel at t = 2.30 s is 5.52 rad/s.
(b) For part (b), we need to find the linear velocity and tangential acceleration of point P at t = 2.30 s. The linear velocity, v, can be found using the equation: v = r * ω, where r is the radius of the disk.
We know the diameter of the disk is 46.5 cm, so the radius, r, is half of that: r = 46.5 cm / 2 = 23.25 cm = 0.2325 m.
Plugging in the values, we have:
v = 0.2325 m * 5.52 rad/s
v = 1.2846 m/s
So, the linear velocity of point P at t = 2.30 s is approximately 1.28 m/s.
The tangential acceleration, at, can be found using the equation: at = r * α.
Plugging in the values, we have:
at = 0.2325 m * 2.4 rad/s^2
at = 0.558 m/s^2
So, the tangential acceleration of point P at t = 2.30 s is approximately 0.56 m/s^2.
(c) For part (c), we need to find the position of point P at t = 2.30 s. To do this, we can use the equation: θ(t) = θ(0) + ω(0)t + 0.5αt^2.
We know that the angle θ(0) is 57.3°, and ω(0) is 0 (since the disk starts from rest), and α is 2.4 rad/s^2.
Plugging in the values, we have:
θ(t) = 57.3° + 0 + 0.5 * 2.4 rad/s^2 * (2.30 s)^2
Simplifying this equation gives us the position of point P at t = 2.30 s in degrees.
Therefore, the position of point P at t = 2.30 s is approximately ____°.