Good Rotational Motion Report Example

Type of paper: Report

Pages: 10

Words: 2750

Published: 2021/03/28

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Introduction

Purposes of the lab experiment were to analyze rotating effects of a force; to be familiar with torque, moment of inertia and angular acceleration and how these relate to each other; to check whether the angular momentum was conserved when a ring was dropped onto a rotating disk. In the lab, moment of inertia for a solid circular disk and a ring were evaluated. A solid disk and a ring; three photogates connected to the measurement box; a rotational assembly on a heavy “A” shaped base; set of weights; flag; string and pulley assembly; a rotating platform with a square mass attached (used for the leveling the base) were used in this experiment.

Linear acceleration was calculated from the following equation.

a=2x31t31-x21t21(t31-t21)
In the above equation x31, x 21, t31, t21 were measured in this experiment. Angular acceleration was calculated by dividing linear acceleration by the radius of disk.
α=ar

Thus, torque was calculated by following equation which is based on Newton’s Second Law of Inertia.

τ=mr(g-a)
where g is gravitational acceleration. In the above equation, force is m (g-a) and r is radius i.e. lever arm of the torque. Finally, total net torque was determined by evaluating difference between torque exerted by the string mass and the frictional torque in the opposite direction.
τnet=τ-τf=Iα

In the above equation I is moment of inertia.

Initial Data and Preliminary Analysis
Initial Data:
Analysis
a=2x13t13-x12t12t13-t12=20.53m4.873s-0.25m2.732s4.873s-2.732s=0.016ms2
T=mg-a=0.0501kg9.81ms2-0.016ms2=0.49067 N
τ=rspool×F= rspool×T=0.037355m20.49067N=0.009165 Nm
α=ar=0.016ms20.01868m=0.863 rad/s
v3=x3-x1t31=0.63m-0.10m4.873s=0.1088ms
v2=x2-x1t21=0.35m-0.10m2.732s=0.0915ms
t32=t31-t21=4.873s-2.732s=2.141s

Uncertainties,

∆v3,x=∂v∂x=∆xt31=0.0005m4.873s=0.000103ms
∆v3,t=∂v∂t=∆tx31t312=0.001s0.53m4.873s2=0.0000223ms
∆v2,x=∂v∂x=∆xt21=0.0005m2.732s=0.000183ms
∆v2,t=∂v∂t=∆tx21t212=0.001s0.25m2.732s2=0.0000335ms
∆v3=∆v3,x2+∆v3,t2=0.000103ms2+0.0000223ms2=0.0001054ms
∆v2=∆v2,x2+∆v2,t2=0.000183ms2+0.0000335ms2=0.000186ms
∆t32=2∆t=2*0.001s=0.001414s
∆av3=∂a∂v=2∆v3t32=20.0001050ms2.141s=0.000098ms2
∆av2=∂a∂v=2∆v2t32=20.000186ms2.141s=0.000174ms2
∆at32=∂a∂t=2v3-v2∆t32t322=20.1088ms-0.0915ms0.001414s2.141s2=0.0000107ms2
∆a=∆av32+∆av22+∆at322=0.000098ms22+0.000174ms22+0.0000107ms22=0.0002ms2
∆αa=∂α∂a=∆arspool=0.0002ms20.01868m=0.01rads2
∆αr=∂α∂r=a∆rrspool2=0.016ms20.00026m0.01868m2=0.012rads2
∆α=∆αa2+∆αr2=0.01rads2+0.012rads2=0.0156 rad/s2
∆τa=∂τ∂a=mg-a∆r=0.0501kg9.81ms2-0.016ms20.00026m=0.000128N
∆τr=∂τ∂r=mrspool∆a=0.0501kg0.01868m0.0002ms2=0.000000187N
∆τ=∆τa2+∆τr2=0.000128N2+0.000000187N2=0.000128N
Id0=12MR2=121.431 kg0.11375m2=0.00926kg*m2
∆Id,M0=∂ ∆Id0∂M∆M= 12R2∆M=120.11375m20.0001kg=0.000000647kgm2
∆Id,R0=∂∆Id0∂R∆R= MR∆R=1.431kg0.11375m0.00055m=0.00009 kgm2
∆Id= ∆IM2+∆IR2=0.000000647 kgm22+0.00009 kgm22=0.00009 kgm2
Ird= Id 0+Ir0
Ir0 =12Mr Ri2+R02=(12)(1.431kg)(0.0537m2+0.06362m2))=0.00496 kg m^2
Ird= Id 0+Ir0 =0.00926kg m2+0.00496 kg m2=0.0142 kg m2
∆Ird, Id = ∂Ird ∂Id0∆Id0= ∆Id0=0.00009 kg m2
∆Ird, Mr =∂Ird ∂Mr∆Mr=12Ri2+Ro2∙∆Mr=(12)(0.0537m)2+0.06362m20.0001kg=0.00000035 kg m2
∆Ird, Ri =∂Ird ∂Ri∆Ri = MrRi∆Ri=1.431kg0.0537m0.000029m=0.0000022 kg m2
∆Ird, Ro =∂Ird ∂RO∆RO = MrRO∆RO =1.431kg0.06362m0.000037m=0.0000034 kg m2
∆Ird = ∆Ird, Id 2+ ∆Ird, Mr 2+∆Ird, Ri 2+ ∆Ird, Ro 2=0.00009 kg m22+0.00000035 kg m22+0.0000022 kg m22+0.0000034 kg m22=0.00009 kg m2
ω1=wRp*tb1=0.012570 m0.10497 m*0.039s=3.07 rad/s
∆ω1,w=∂ω1∂w∆w= 1Rp∙tb1∆w=0.0001m0.10497m*0.039s=0.0244rad/s
∆ω1, rp=∂ω1∂Rp∆Rp= -wRp2∙tb1∆Rp=0.01257m0.000017m0.10497m2(0.039s)=0.0005 rad/s
∆ω tb1=∂ω1∂tb1∆tb1=-wRp∙tb12∆tb1=0.01257m0.001414s0.10497m0.039s2=0.111 rad/s
∆ω1= ∆ω1,w 2+∆ω1,Rp 2+∆ω1,tb1 2=0.0244rad2+0.0005 rad2+0.111 rad2=0.114 rad
ω2=wRp*tb2=0.01257m(0.10497m)(0.0643s)=1.86rads
∆ω2,w=∂ω2∂w∆w= 1Rp*tb2∆w=0.0001m(0.10497m)(0.0643s)=0.0148 rad/s
∆ω2,Rp=∂ω1∂Rp∆Rp= -wRp2∙tb2∆Rp=0.01257m0.000017m0.10497m20.0643s=0.0003 rad/s
∆ω tb2=∂ω2∂tb2∆tb2=-wRp∙tb22∆tb2=0.01257m0.001414s0.10497m0.0643s2=0.04 rad/s
∆ω2= ∆ω2, w2+∆ω2, Rp 2+∆ω 2,tb22=0.0148rads2+0.0003rads2+0.04rads2=0.0436rads
L1= Id0ω1=0.00926 kgm23.07rads=0.0284 kgm2s
∆L1 w1=∂L1∂ω1∆ω1=Id0∆ω1=0.00926 kgm20.0244rads=0.000226 kg m/s2
∆L1, Id= ∂L1∂Id0∆Id0=ω1∆Id0=3.07rads0.00009 kgms2=0.0002763 kg m/s2
∆L1= ∆L1,w22+∆L1,Ird 2=(0.000226 kgrads2)^2 +0.0002763 kgrads22=0.00036rad/s2
L2= Ird ω2=0.014316kg m21.86rads=0.027 kg rad/s2
∆L2,w2 = ∂L2∂ω2∆ω2= Ird ∆ω2=0.01431 kg m20.0148rads=0.0002 kg rad/s2
∆L2,Ird =∂L2∂Id ∆Ird =ω2∆Ird =1.86rads0.00009 kg m2=0.0001674 kg rad/s2
∆L2= ∆L2,w22+∆L2,Ird 2=0.0002kg m22+0.0001674 kgrads22=0.00026 kg rad/s2

Final Result

Id0=0.00926±0.00009kgm2;∆Id0Id0=0.00979.7%
Id=0.00926±0.00009kgm2;∆IdId=0.00979.7%
τf=0.009±0.000128kgm2;∆τfτf=0.014(1.4%)
L1=0.0284 ±0.00036 kg∙m2s;∆L1L1=0.0126 (1.26%)
L2=0.027 ±0.00026 kg∙m2s;∆L2L2=0.009 (0.9%)

Conclusion

The experiment was carried out safely. Experimental results obtained were as desired and percentage of error was low. It was also found out that angular momentum was conserved in this experiment. Probable sources of errors may be the frictional force on torque that affected the final angular momentum of the ring disk system. Nonetheless, objectives of the experiment were fully met. So, it can be said that the lab experiment was successful.

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WePapers. (2021, March, 28) Good Rotational Motion Report Example. Retrieved May 14, 2021, from https://www.wepapers.com/samples/good-rotational-motion-report-example/
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Good Rotational Motion Report Example. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/good-rotational-motion-report-example/. Published Mar 28, 2021. Accessed May 14, 2021.
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