175 gram Frisbee. 24cm diameter. Half the mass is equally spread in the disc and the other half of the mass is equally spread in the rim. If Moses Rifkin throws a Frisbee at 1000rpm with a 1/8-turn of the wrist, what is…
- The total moment of inertia?
Step 1: Convert all numbers + organize information
Disk: radius = .12m – containing half mass of .0875kg
Hoop: radius = .12m – containing the other half of mass of .0875kg
Moment of Inertia disk = 1/2MR^2
Moment of Inertia hoop = MR^2
1000rpm = 104rad/sec
Step 2: Solve for I
In other words, I = Moment of Inertia disk + Moment of Inertia hoop
Total Moment of inertia = 1/2MR^2 + MR^2
= .5(.0875)(.12)^2+(.0875)(.12)^2
I = .00189 kg m 2
- The magnitude of the torque?
Step 1: Use Torque equation of τ=Iα (however I first want to derive this formula). I want to explain the connection between torque and angular acceleration.
τ = torque
I= moment of inertia
α= angular acceleration
Imagine a mass moving in a circle away from distance r with a tangential force
We know that F=ma
We can also say that a=rα
So, F=mrα
τ about the center of rotation from F: τ =Fr
Therefore τ=mr^2 α
Finally, τ=I α
Step 2: Since we already know I, solve for α
I will want to use Wf^2=Wi^2+2 α θ. This equation is parallel to the kinematics equations for linear motion. It is really cool how we can relate rotational motion and linear motion.
Wf= Angular momentum final
Wi= Angular momentum initial
Θ = angular displacement (think unit circle)
Wf=104rad/sec
Wi= 0
Θ = π/4
α = (Wf^2)/2 θ
α = 104^2 / (2(π/4))
α = 6885.6 rad/sec^2
Step 3: Plug it all back in
τ=Iα
τ = (.00189)( 6885.6)
τ=13J
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