please do parts a-f Class Management 1 Help Homework #7 (Ch 10: Rotation of a Ri

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please do parts a-f

Class Management 1 Help Homework #7 (Ch 10: Rotation of a Rigid Object) Begin Date: 9/162017 12:00:00 AM-Due Date: 11/14/2017 11:59.00 PM End Date 12/20/2017 12:00:00 AM (11%) Problem 7: A uniform wooden meter stick has a mass of m-785 g. A clamp can be attached to the measuring stick at any point P along the stick so that the stuck can rotate freely about point P, which is at a distance d from the zero-end of the stick as shown. Randomized Variables m=785 g ent Status here for ed view Ctheexpertta.com Status Partial Partial & 17% Part (a) Calculate the moment ofinerta mkgm2 of the meter stick if the prot point P 1s at the 50cm mark 17% Part(b) Calculate the moment of inertia in kg m2 of the meter stsek if the prot point p s at the 0-cm mark d-D cm. 2 is at the a sen mark em mak 17% Part te) Calculate the moment of nerta in kgm2 of the meter stek ifthe pivot point p s at the 17% Part (d) Calculate the moment of nerta in kgar of the meter stick f the pivot point 17% Partie ii timr Completed Enter a general expressa nor the momentofinertua of a meter stick4ofmass mun kiloma s prot dabur pot distance d in meters from the zero-cm mark E> Completed r (fi The meter stick is now replaced with a unsform yard stick wnh the same mass of m= 785 i Calculate the moment of inertia 17% Par in kg m2 of the yard stick if the pivor point P is 50 cm frona the end of the yardsick. Grade Summary Potential Subreiscieas 100 Anempts remaining

Explanation / Answer

Icm = m L^2 / 12 = 0.785 x 1^2 / 12

Icm = 0.0654 kg m^2

from parallel axis theorem,

I = Icm + m x^2

(A) x = 50 - 50 = 0

I = I cm = 0.0654 kg m^2

(B) x = 50 - 0 =50 cm = 0.50 m

I = (0.0654) + (0.785 x 0.5^2)

I = 0.262 kg m^2

(C) x = 0.50 - 0.25 = 0.25 m

I = 0.114 kg m^2

(D) x = 0.95 - 0.50 = 0.45

I = 0.224 kg m^2

(E) I = m L^2 / 12 + m (L/2 - d)^2

(F) I = 0.785 (0.914^2)/12 + 0.785 (0.914/2 - 0.50)^2

{ 1 yard = 0.914 m }

I = 0.0561 kg m^2

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