Moment of Inertia

PHY111LL - Tuesday 10/23/2012

Moment of Inertia

Objectives

The purpose of this lab is to calculate the moment of inertia from the mass and geometric dimensions and also to measure the moment of inertia of a system with a dynamic method(motion). This was done by applying tension to string wound around an axis of rotating shaft from a hanging a mass attached to the string, over a pulley. The mass is allowed to drop vertically, and thus, The linear acceleration of the mass is the same as the linear acceleration of the string, and thus of a point on the axis of the shaft. That linear acceleration of the mass is related to the angular acceleration of the apparatus by a factor of the radius of the rotation.

Data

Table 1

Ra=0.07m, Mh=0.05kg, Rs=0.05135m, H=0.725m, Mr=0.0633kg, Lr=0.342m,

Wing nuts=0.0245kg

Ma(kg) | Δt1(s) | Δt2(s) | Δt3(s) | Δtavg(s) | Imot(kgm2) | Igeo(kgm2) | %error |

Shaft only | 1.45 | 1.65 | 1.48 | 1.53 | 0.00195 | NA | NA |

Shaft and rod | 8.10 | 7.84 | 7.10 | 7.68 | 0.0524 | 0.0530 | 1.18 |

0.1245 | 11.38 | 10.38 | 11.45 | 11.1 | 0.109 | 0.110 | 1.13 |

0.2245 | 12.28 | 13.03 | 12.71 | 12.7 | 0.143 | 0.145 | 1.20 |

0.3245 | 14.05 | 13.89 | 14.14 | 14.0 | 0.175 | 0.177 | 1.26 |

0.4245 | 15.99 | 15.96 | 15.62 | 15.9 | 0.224 | 0.226 | 1.26 |

Sample Calculation

ΔTavg.= ΔT1+ΔT2+ΔT3 = 11.38+10.38+11.45 = 11.1s

3 3

I (motion)= MhRs2(g Ta 22H _ 1)= 0.05kg*(0.05135m)2(9.80m/s2*(11.1s)2-1)=0.109kgm2

2*0.725m

I (geo)= Is+Ir+Ia

I (geo)= MhRs2(g Ta 22H _ 1)+112MrLr2+MaRa2

= 0.05kg*(0.05135m)2(9.80m/s2*(11.1s)2 -1)+112*0.0633kg*(0.342m)2 +0.1245kg*(0.07m)2

2*0.725m

=0.111kgm2

%Error= Igeo-Imot * 100%=0.110kgm2-0.109kgm2 *100%=1.13%

Imot 0.109kgm2...