Base EQUATIONS

Solving for Values

α (Angular Acceleration)

Solution 1: Use angular velocity equation

ω = ω₀ + αt

ω-ω₀ = αt

α = ω-ω₀/t-t₀

α= -0.296 rad/sec²

Solution 2: Use slope of angular velocity graph

Since ω =αt + ω₀, (similar to y = mx+b), the slope of the graph represents α

The slope of our best fit line suggests a value of α = -0.296 rad/sec^2, which is the same as solution 1

Solution 3: Find best fit line of acceleration graph

Assuming the acceleration stays constant, the best fit line of the acceleration graph should be a horizontal line, the y value of this line would represent the acceleration.

α = 0*t - 0.2963

α= -0.2963 rad/sec² -> which is nearly equal to the other solutions.

Since all 3 solutions produce a similar value, I am quite certain the angular acceleration is correct. 

Thus α = -0.296 rad/sec^2

I (Inertia)

Solution: Physical Pendulum

In class we learned that:

T = 2π root L/g

However, if it is a physical pendulum with a rigid body, (ex: a rigid stick), then

T = 2π root 𝐼/MgL

(T/2π)² = 𝐼/MgL

T²MgL/4π² = 𝐼stick

Theorem of Parallel axis states 𝐼stick = 𝐼spinner + md²

T²MgL/4π² = 𝐼spinner + mL²

(T²MgL/4π²) = (m)(L²) + (𝐼spinner)

y     =   m   x    +    b 

Thus if we plot y as T²MgL/4π² and x as L², the y-intercept would be 𝐼spinner.

To plot this graph, we need to find the mass of the objects + two sets/points of data.

MASS:

(Note: the tissue was for COVID safety reasons.)

mass of stick = (mass of stick+tissue) - (mass of tissue) = 0.024kg - 0.002kg = 0.022 kg

m (mass of spinner) = (mass of spinner+tissue) - (mass of tissue) = 0.058kg - 0.004kg = 0.054kg

mass of band = (mass of band+tissue) - (mass of tissue) = 0.006kg - 0.004kg = 0.002kg

M (mass of total) = mass of spinner + mass of stick + mass of band = 0.054 + 0.022 + 0.002 = 0.078 kg

POINT 1:

Time t0 = 6.721 s -> x = 21.62cm

Time t1 = 7.654 s -> x = 21.62 cm 

T (period) = t1 - t0 = 7.654 - 6.721 = 0.933 seconds

L (length) = 21.59 cm = 0.2159 m

x = L² = (0.2159)² =  0.04661281

y = T²MgL/4π² = (0.933)² * 0.078*9.8*0.2159/(4*π²) = 0.003638956562

Thus Point 1 = (0.04661281,0.003638956562)

POINT 2:

Time t0 = 4.626 -> 18.78 cm

Time t1 = 5.426 -> 18.81 cm

T (period) = t1-t0 = 5.426 - 4.626 = 0.800s

L (length) = 10.35cm = 0.1035m

x = L² = 0.1035² = 0.01071225

y = T²MgL/4π² = (0.800)² * 0.078*9.8*0.1035/(4*π²) = 0.001282570556

Thus, Point 2 = (0.01071225,0.001282570556)

The graph shows the y-intercept to be 5.79*(10)⁻⁴ 

Thus the inertia of the fidget spinner = 5.79*(10)⁻⁴ kg*m²

Photos of Mass

L (Angular Momentum)

L = 𝐼ω 

Using the inertia found above, we can simply multiply it by the angular velocity to find the angular momentum.

Since we are multiplying angular velocity by a constant, the degree of the equation is preserved and thus the angular momentum equation will be linear as well.

ΔL (Impulse)

Now that we have calculated the angular momentum, we can simply calculate the impulse by finding the change in angular momentum.

ΔL = Lf - Li

ΔL = -Li

ΔL = -0.004 kg*m²/s

τ(friction) (Frictional Torque) 

Since ΔL = τ*Δt and the only torque acting upon the fidget spinner (once the finger is not in contact) is friction, then

τ(friction) = ΔL/Δt = -0.004/(37.917-3.2) = -1.15*(10)⁻⁴ N*m

E (Energy)

E = Kt + Kr + Ug

Since the fidget spinner only rotates and doesn't move across the table, we can assume Kt = 0

Since the fidget spinner is rotating on an axis parallel to the floor, we can set the base height to be the height of the fidget spinner, resulting in Ug being equal to 0 too.

Thus, E = Kr

Kr = 0.5𝐼ω²

[Plug in 𝐼 and then rotational kinetic energy can be graphed by simply modifying the ω graph]

[Should be a degree two polynomial]