Class Management Help rotational dynamics part 1 Begin Date: 10/25/2017 1:00:00
ID: 1782288 • Letter: C
Question
Class Management Help rotational dynamics part 1 Begin Date: 10/25/2017 1:00:00 PM Due Date: 10/31/2017 11:59:00 PM End Date: 11/14/2017 11:59:00 PM (17%) Problem 2: Suppose we want to calculate the moment of inertia of a 59.5 kg skater, relative to a vertical axis through their center of mass 50% Part (a) First calculate the moment of inertia (in kg-m2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m > 50% Part (b) Now calculate the moment of inertia of the skater (in kg-m2) with their arms extended by assuming that each arm is 5% if the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.85 m long rods extending straight out from their body being rotated at the ends Grade Summary Deductions Potential Late Work I= 2.001 3% 97% 75% 72.75% cos0 789HOME sin() cotan atan)acotan(.O cosh0tanh0 Late Potential asin acos Submissions Attempts remaining: 6 (3%per attempt) detailed view tanh)cotanh( END Degrees Radians BACKSPACE CLEAR 3% Submit Feedback I give up! Hints: 1 for a 0% deduction. Hints remaining: 0 Feedback: 1% deduction per feedback. What is the moment of inertia of a thin rod? Make sure you look up the moment of inertia relative to the right locationExplanation / Answer
given
Part a )
mass m = 59.5 kg ,
r = 0.11 m
the moment of inertia is " I "
I = 1/2 m r2
= 0.5 X 59.5 X 0.112
I = 0.359975 kg m2
b )
L = 0.85 m
mass of the arms marms = 0.05 X 59.5
= 2.975 kg
mass of the remaining body mbody = 59.5 - 2.975 - 2.975
= 53.55 kg
I = Io + 2 I1
= 0.5 mbodyr2 + 2 X ( 1/3 marms L2 )
= 0.5 X 53.55 X 0.112 + 2 X 0.33 X 2.975 X 0.852
= 0.3239 + 1.4186
I = 1.7425 kg m2