PLEASE ANSWER PARTS E AND F AND USE THE LETTERS GIVED ON THE KEY PAD IN THE PICT

Question

PLEASE ANSWER PARTS E AND F AND USE THE LETTERS GIVED ON THE KEY PAD IN THE PICTURE TO ANSWER THE QUESTION. I HAVE THE REST OF THE ANSWERS

LAST PERSON WAS INCORRECT ON PART E THEY SAID IT WAS 1/3 ML^2+md^2 THERE IS NOT A L OR CAPITAL M AS MY CHOICES SO I TRIED 1/3md^2+md^2 and the was incorrect

I Help Practice Problem 9: Rotational Motion Begin Date: 1122015 12:00:00 AMDue Date: 11/122015 11:59:00 PM End Date: 11/12/2015 11:59:00 PM (8%) Problem 6: A uniform wooden meter stick has a mass of m-697 g. A clamp can be attached to the measuring stick at any point so that it can freely pivot around poimt P, which is a distance d from the zero-end of the stick as shown. Randomized Variables m= 697 g 0 1 2 304 567 89 10 11 12 13 14 15 16 17 18 Otheexpertta.com 17% Part (a) Calculate the moment of merta m kgm2 of the meter stick ifthe pivot point P is at the 50-cm mark. 17% Part (b) Calculate the moment of merta m kgm2 of the meter stick ifthe prot point P is at the 0-cm mark d-0 cm q) 17% Part (c) Calculate the moment of merta m kg-m2 of the meter stick if the prot pont Pis at the d-25 cm mark. tm aterial 17% Part (d) Calculate the moment of mertia in kgm2 of the meter stick if the pivot point P is at the d= 79 cm mark. 17% Part (e) Determne a genera expression for the moment of nerta of a meter stick le of mass m m kilograms proted about pont P anv distance d in meters from the zero-cm mark. Grade Su Potential Submissions 496 9600 Attempts remaining: 2 40 per attempt) detailed view 496 Submit Hint I give up Hints: 490 deduction per hint. Hints Feedback: 590 deduction per feedbad:. 17% Part (f) The meter stick is now replaced with a uniform yard stick with the same mass of m = 69 g . Calculate the moment of merta m kg m2 of the yard stick: if the pivot point P is at the 50-cm mark.

Explanation / Answer

A) I = (1/12)*M*L^2 = (1/12)*697*1^2 = 58.1 kg-m^2

B) I = (1/3)*M*L^2 = (1/3)*697*1^2 = 232.33 kg-m^2

C) I = Icm + M*d^2 = 58.1+(697*0.25*0.25) = 101.6625 kg-m^2

D) I = 58.1+(697*0.29*0.29) = 116.7177kg-m^2

E) I = Icm +M*d^2

here Icm = (1/12)*M*L^2

F) 1 yard = 0.9144 m

Icm = (1/12)*697*0.9144*0.9144 = 48.56 kg-m^2

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