# Lab Report #9 Rotation motion - PHY 122 Labs Name Rotation...

• Lab Report
• 11
• 95% (21) 20 out of 21 people found this document helpful

This preview shows page 1 - 5 out of 11 pages.

PHY 122 Lab’s Name : Rotation Motion Student’s Name: Binhan Tian Group Number: 4 Class Number: 28132 TA’s Name: Bryce Davis Lab Time: Tuesday, April 5, 2016 Lab Report due day: Tuesday, April 12, 2016
Objective The objective for this experiment is to find the relationship between the moment of inertia and the rotation axis or the dimensions for a given object. And to verify that the rotational motion also abided by the law of conservation of energy in a system. Experimental Data Mass of block (M) Mass of hanger(m) Radium of small pulley( ? 𝑠 ) Radium of Medium pulley( ? ? ) 0.443kg 0.02kg 0.0048m 0.0143m a (m) b (m) c (m) 0.03 0.04 0.05 Short axis Small pulley Short axis Medium pulley Long axis Small pulley Long axis Medium pulley Angular acceleration 5.79 16.7 7.71 27.3 Initial angular position 0 0 0 0 Final angular position 88.698 23.003 40.59 17.82 Initial angular velocity 0 0 0 0 Final angular velocity 31.76 27.75 25.31 31.24 Data analysis: Static moment of inertia: 1) Calculate static moment of inertia in case of the short axis. M = 0.443kg , b = 0.04m , c = 0.05m ? 𝑠 = 1 12 ?(? 2 + ? 2 )
? 𝑠 = 1 12 × 0.443 × (0.04 2 + 0.05 2 ) = 1.514 × 10 −4 𝑘𝑔 ∙ 𝑚 2 2) Error propagation for 1 For the error propagation, use the equation below. ∆? 𝑠 = ( ?? 𝑠 ?? ∆?) 2 + ( ?? 𝑠 ?? ∆?) 2 The standard deviation is calculated by the function: 𝜎̅ 𝑥 = 𝜎 𝑥 √? The size of the object was measure by the Vernier caliper, here is the data that measured for five times. Run 1 Run 2 Run 3 Run 4 Run 5 b = 0.041 c = 0.052 b = 0.042 c = 0.051 b = 0.0413 c = 0.0522 b = 0.0412 c = 0.0524 b = 0.0414 c = 0.0511 𝜎 ? = 0.041 + 0.042 + 0.0413 + 0.0412 + 0.0414 5 = 0.04138 𝜎̅ ? = 𝜎 ? √? = 0.04138 √5 = 0.01851 𝜎 ? = 0.052 + 0.051 + 0.0522 + 0.0524 + 0.0511 5 = 0.05174 𝜎̅ ? = 𝜎 ? √? = 0.05174 √5 = 0.023139 So that, ∆b = 0.01851 , ∆c = 0.023139 ∆? 𝑠 = ( ?? 𝑠 ?? ∆?) 2 + ( ?? 𝑠 ?? ∆?) 2 = 1.014 × 10 −4 3) Static moment of inertia case of the long axis
M = 0.443kg , a = 0.03m , b = 0.04m ? 𝑠 = 1 12 ?(? 2 + ? 2 ) ? 𝑠 = 1 12 × 0.443 × (0.03 2 + 0.04 2 ) = 9.229 × 10 −5 𝑘𝑔 ∙ 𝑚 2 4) Error propagation for 2 The size of the object was measure by the Vernier caliper, here is the data that measured for five times.
• • • 