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Physics 1

 Group 5 Lab#12: Rotational Dynamics & Moments of Inertia

        Due: 11/27/2017

Purpose The purpose of this lab is to experiment with the dynamical equations of rotation. Using the equations, τ = I α , F = ma , and a = αr to calculate the principal and off-axis moments of inertia of a solid cylinder, a solid sphere, and a cylindrical shell. We can then compare the results with the actual moments of inertia.

Equipment Rotating Platform, Solid Sphere, Solid Cylindrical Disk and Ring, Stopwatch, Meter stick, Vernier Caliper, Hooked Weights, Digital Balance, Digital Timer

Theory Combining the equations mg − T = ma and τ = T r = I shaft α = I shaft a/r, gives us the experimental way of calculating I shaft = m(g − a)r 2 /a, where r is the radius of the cylinder where the wire is wrapped. By putting different objects on top of the shaft and finding the difference in moments of inertia, we can then compare with the theoretical moments of inertia.

Procedure Using the Rotating Platform, we take the radius of the cylinder that the string is wrapped around. The string, hanging over a pulley, is attached to a hanging weight. We find the height by taking the distance from the bottom of the weight to the floor. Now, we release the weight and time how long it takes for the weight to strike the floor. This experiment is repeated with different objects placed on top of the shaft.

Data Radius of Cylinder where string is wrapped, r = 0.019 m Height of hooked weight from the floor, h = 0.73 m Part 0: Platform and Shaft alone Mass of Hooked Weight, m = 0.01 kg

t1 = 0.66 sec

t2 = 0.58 sec

2 aexp = 3.68 m /s

t3 = 0.64 sec

tavg = 0.63 sec

I shaft = 6.00 x 10− 6 kg.m2

Part 1: Platform and Shaft + Solid Disk

Radius of Solid Disk, RD = 0.114 m

Mass of Solid Disk, M D = 1.766 kg

Mass of Hooked Weight, m = 0.052 kg

t1 = 8.43 sec

tavg = 8.58 sec

t2 = 8.64 sec

2 aexp = 0.0198 m /s

t3 = 8.67 sec

I T otD = 9.27 x 10−3 kg.m2

I SolidDiskE xp = I T otD − I Shaft = 9.264 x 10− 3 kg.m2 I SolidDiskT heo = M D R D 2 /2 = 1.148 x 10−2 kg.m2 %Error 1 = 100% x ∣I SolidDiskExp − I SolidDiskT heo∣ /I SolidDiskT heo = 19.3% Part 2: Platform and Shaft + Solid Disk + Cylindrical Ring Inner Radius of Ring, R1 = 0.054 m Outer Radius of Ring, R2 = 0.0662 m Mass of Ring, M R = 1.781 kg

Mass of Hooked Weight, m = 0.052 kg

t1 = 11.13 sec t2 = 10.85 sec t3 = 10.55 sec tavg = 10.84 sec 2 aexp = 0.0124 m /s

I T otR = 1.48 x 10− 2 kg.m2

I RingExp = I T otR − I T otD = 5.53 x 10 −3 kg.m 2 I RingT heo = M R (R 1 2 + R2 2 )/ 2 = 6 .50 x 10−3 kg .m2 %Error 2 = 100% x ∣I RingExp − I RingT heo∣ /I RingT heo = 14.9% Part 3: Platform and Shaft + Solid Disk + Solid Sphere on groove: Radius of Sphere, Rs = 0.017 m Mass of Sphere, M s = 0.0668 kg t1 = 8.95 sec

t2 = 8.54 sec

aexp = 0.0191 m /s 2

Mass of Hooked Weight, m = 0.052 kg

t3 = 8.73 sec

tavg = 8.74 sec

I T otS = 9.61 x 10− 3 kg.m2

I SphereExp = I T otS − I T otD = 3.40 x 10 −4 kg.m 2

d = Shift Amount = 0.0619 m

I SphereT heo = I SphereCM + M sd 2 = 2M sR s 2) /5 + M sd 2 = 2.63 x 10− 4 kg.m2 %Error 3 = 100% x ∣I SphereExp − I SphereT heo∣ /I SphereT heo = 29.3% Extra Credit: Moment of Inertia of Rod Length of Rod, L = 0.606 m Mass of Rod, M = 0.6285 kg

Mass of Hooked Weight, m = 0.052 kg

t1 = 15.11 sec t2 = 14.82 sec t3 = 14.75 sec tavg = 14.89 sec 2 aexp = 6.585 x 10−3 m /s

I T otRod = 0.0279 kg.m 2

I RodExp = I T otRod − I T otD = 0.0187 kg.m 2

I RodT heo = 0.0192 kg.m 2

%Error = 100% x ∣I RodExp − I RodT heo∣ /I RodT heo = 2.83%

Conclusion Given the results of the data and the relatively low margin of error and consideration of human error, I believe it is safe to say that we have proven the dynamical equations of rotation....