A uniform wooden meter stick has a mass of m = 669 g. A clamp can be attached to

Question

A uniform wooden meter stick has a mass of m = 669 g. A clamp can be attached to the measuring stick at any point so that it can freely pivot around point P, which is a distance d from the zero-end of the stick as shown. Randomized Variables m = 669 g Calculate the moment of inertia in kgmiddotm^2 of the meter stick if the pivot point P is at the 50-cm mark. Calculate the moment of inertia in kgmiddotm^2 of the meter stick if the pivot point P is at the 0-cm mark d = 0 cm. Calculate the moment of inertia in kgmiddotm^2 of the meter stick if the pivot point P is at the d = 25 cm mark. Calculate the moment of inertia in kgmiddotm^2 of the meter stick if the pivot point P is at the d = 69 cm mark. Determine a general expression for the moment of inertia of a meter stick I_e of mass m in kilograms pivoted about point P, any distance d in meters from the zero-em mark. The meter stick is now replaced with a uniform yard stick with the same mass of m = 669 g. Calculate the moment of inertia in kgmiddotm^2 of the yard stick if the pivot point P is at the 50-cm mark. I_f =

Explanation / Answer

a) Icm = m*L^1/2

= 0.669*1^2/12

= 0.05575 kg.m^2

b) I = m*L^2/3

= 0.669*1^2/3

= 0.223 kg.m^2

c) IP = Icm + m*d^2

= m*L^2/12 + m*d^2

= 0.669*1^2/12 + 0.669*0.25^2

= 0.0976 kg.m^2

d) IP = Icm + m*d^2

= m*L^2/12 + m*d^2

= 0.669*1^2/12 + 0.669*0.19^2

= 0.08 kg.m^2

e) Ie = Icm + m*(L/2 - d)^2

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

f) we know, 1 yard = 0.9144 m

so,

Ie = m*L^2/12 + m*(L/2 - d)^2

= 0.669*0.9144^2/12 + 0.669*(0.9144/2 - 0.5)^2

= 0.048 kg.m^2

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