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.)
_____°
12
To solve this problem, we first need to find the angular displacement of the disk at time t = 2.30 s. We can use the formula:
θ = θ_0 + ω_0 * t + 0.5 * α * t^2
where θ is the angular displacement, θ_0 is the initial angular displacement, ω_0 is the initial angular velocity, α is the angular acceleration, and t is the time.
In this case, θ_0 = 57.3° (since the line from the center of the disk to point P makes an angle of 57.3° with the positive x-axis at t = 0), ω_0 = 0 rad/s (since the disk starts from rest), α = 2.4 rad/s^2, and t = 2.30 s.
Plugging these values into the formula, we get:
θ = 57.3° + 0 + 0.5 * 2.4 rad/s^2 * (2.30 s)^2
Let's calculate the value of θ first:
θ = 57.3° + 0 + 0.5 * 2.4 rad/s^2 * 5.29 s^2
θ = 57.3° + 0 + 6.084 rad
θ ≈ 63.384 rad
Now let's answer the questions:
(a) The angular speed of the wheel at t = 2.30 s is given by the formula:
ω = ω_0 + α * t
Plugging in the values ω_0 = 0 rad/s and α = 2.4 rad/s^2, and t = 2.30 s, we get:
ω = 0 + 2.4 rad/s^2 * 2.30 s
ω ≈ 5.52 rad/s
So, the angular speed of the wheel at t = 2.30 s is approximately 5.52 rad/s.
(b) The linear velocity of point P on the rim of the disk can be found using the formula:
v = r * ω
where v is the linear velocity, r is the radius of the disk, and ω is the angular speed.
The radius of the disk is half the diameter, so r = 46.5 cm / 2 = 23.25 cm = 0.2325 m.
Plugging in the values r = 0.2325 m and ω ≈ 5.52 rad/s, we get:
v = 0.2325 m * 5.52 rad/s
v ≈ 1.283 m/s
So, the linear velocity of point P at t = 2.30 s is approximately 1.283 m/s.
The tangential acceleration of point P can be found using the formula:
a_t = r * α
where a_t is the tangential acceleration and α is the angular acceleration.
Plugging in the values r = 0.2325 m and α = 2.4 rad/s^2, we get:
a_t = 0.2325 m * 2.4 rad/s^2
a_t ≈ 0.558 m/s^2
So, the tangential acceleration of point P at t = 2.30 s is approximately 0.558 m/s^2.
(c) To find the position of point P at t = 2.30 s, we can use the formula:
θ = θ_0 + ω_0 * t + 0.5 * α * t^2
Plugging in the values θ_0 = 57.3°, ω_0 = 0 rad/s, α = 2.4 rad/s^2, and t = 2.30 s, we get:
θ = 57.3° + 0 + 0.5 * 2.4 rad/s^2 * (2.30 s)^2
θ = 57.3° + 0 + 6.084 rad
θ ≈ 63.384 rad
To convert this angle from radians to degrees, we can use the conversion factor:
1 rad ≈ 57.3°
So, the position of point P at t = 2.30 s is approximately 63.384 * 57.3° ≈ 3,627.67°.
Therefore, the position of point P at t = 2.30 s is approximately 3,627.67°, with respect to the positive x-axis.