A uniform disk with mass 9.6 kg and radius 2.7 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 31.5N is applied to the rim of the disk. The force direction makes an angle of 35 degree with the tangent to the rim. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 5.7 revolution? The unit of the tangential velocity is m/s.
To find the magnitude of the tangential velocity v of a point on the rim of the disk after it has turned through 5.7 revolutions, we can use the concept of rotational dynamics.
The first step is to find the angular acceleration of the disk. We can use the torque equation:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Since the disk is pivoted at its center, we can use the following formulas for the moment of inertia:
I = (1/2)mr^2
where m is the mass of the disk and r is its radius.
Substituting the given values:
I = (1/2)(9.6 kg)(2.7 m)^2
I ≈ 104.0448 kg·m²
Next, we need to calculate the torque. The torque can be found using the equation:
τ = Frsinθ
where F is the force applied, r is the radius, and θ is the angle between the force vector and the tangent to the rim.
Substituting the given values:
τ = (31.5 N)(2.7 m)sin(35°)
τ ≈ 52.8802 N·m
Now, we can solve for the angular acceleration:
α = τ/I ≈ 52.8802 N·m / 104.0448 kg·m²
α ≈ 0.508 rad/s²
To find the angular velocity (ω) after the disk has turned through 5.7 revolutions, we can use the formula:
θ = ωt + (1/2)αt²
Since the disk starts from rest, the initial angular velocity (ω₀) is zero. We can rearrange the equation and solve for ω:
5.7 revolutions = (1/2)(0.508 rad/s²)t²
(2)(5.7 revolutions) = 0.508 rad/s²)t²
t² ≈ (2)(5.7 revolutions) / 0.508 rad/s²
t² ≈ 22.4409 s²
t ≈ √(22.4409 s²)
t ≈ 4.742 s
Finally, we can find the angular velocity (ω) using the equation:
ω = ω₀ + αt
Since ω₀ = 0, we have:
ω = αt ≈ (0.508 rad/s²)(4.742 s)
ω ≈ 2.410 rad/s
The tangential velocity (v) of a point on the rim of the disk can be found by multiplying the angular velocity (ω) by the radius (r):
v = ωr ≈ (2.410 rad/s)(2.7 m)
v ≈ 6.5087 m/s
Therefore, the magnitude of the tangential velocity of a point on the rim of the disk after it has turned through 5.7 revolutions is approximately 6.5087 m/s.