What Forces Affect the Rotation of a Submerged Disc?

[JordanB87]

1. A 20-lbf disc with diameter 18" and thickness of 3" is held static while completely submerged in water. Upon release of a lock, the disc experiences a torque from a torsional spring that causes rotation about its center of mass along the x/y axis (think coin toss, not wheel). If the spring is wound to 180-degrees with a constant of 10/π [lb/rad], what is the initial acceleration of the disc?
How fast is the disc moving after a quarter rotation?Attempt at a solution:
Spring force must overcome moment of disc and drag from water

20lbf = (20/32.2) slugs = 0.62 slugs
ι_disc=(1/4)Mr² = (1/4)*0.62+(9²) = 12.58

τ _total = 180(π/180)(10/π) = 10 in-lbs

Initially: no velocity means no drag force. Thus...

τ _total= ι_discα
α = τ _total/ι_disc = 10/12.58 = 0.79 rad/s²

Final Velocity at π/2 rotations
ρ = 0.036 [lb/in³]
C_D = 1.17 for disc
A = πr²
F_drag = C_D(ρ/2)V²A= 1.17*(.036/2)*(0)² = 0
velocity = π*diameter/2 <-- simplified as max velocity (at edge)
r_drag = r/2 = 4.5in

τ _total= ι_discα - F_dragr_drag

Am I on the right track?
Do I need to represent the drag force more qualitatively so that we can integrate to get the equation in terms of ω?

I am a bit lost at this point. Thanks for your time!

 

[anathema]

Hi there,

I would approach this problem by first identifying all the forces acting on the disc and then using Newton's laws of motion to determine the initial acceleration and final velocity.

So, let's start by listing the forces acting on the disc:

1. Weight of the disc (20 lbf)
2. Buoyant force from water (equal to weight of displaced water)
3. Torsional force from the spring
4. Drag force from the water

Now, let's use Newton's second law of motion (F=ma) to determine the initial acceleration and final velocity.

Initial acceleration:
Net force = Torsional force - Weight - Buoyant force
Net force = 10 in-lbs - 20 lbf + 20 lbf = 10 in-lbs
Net force = 10 in-lbs = mass * acceleration
Acceleration = 10 in-lbs / mass = 10 in-lbs / (20 lbf / 32.2 ft/s^2) = 1.61 ft/s^2

Final velocity:
To determine the final velocity, we need to consider the angular motion of the disc. We can use the equation ω = ω0 + αt, where ω0 is the initial angular velocity (which is zero in this case), α is the angular acceleration (which we calculated above), and t is the time.

Since we are looking for the final velocity after π/2 rotations, we can use the equation θ = ω0t + 0.5*α*t^2 to determine the time taken for π/2 rotations.

θ = π/2, ω0 = 0, α = 0.79 rad/s^2
π/2 = 0*t + 0.5*0.79*t^2
t = √(π/0.79) = 2.81 s

Now, we can use the equation ω = ω0 + αt to determine the final angular velocity:

ω = 0 + 0.79*2.81 = 2.22 rad/s

To convert this into linear velocity, we can use the equation v = ω*r, where r is the radius of the disc (9 inches):

v = 2.22*9 = 19.98 in/s

So, the final velocity of the disc after π/2