A dumbbell is made by attaching two spheres of equal size to the end of a thin r

Question

A dumbbell is made by attaching two spheres of equal size to the end of a thin rod as shown in the figure. The mass of the rod is m, the mass of the left sphere is also m and the mass of the right sphere is 2m. (All objects are solid. The origin of the coordinate system is at the geometrical center of the dumbbell. The x-axis is labeled as î and the y-axis as ?.) Use the following values:

m = 1.20 kg R = 10.0 cm L = 30.0 cm

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a. What is xcm the x-coordinate of the center of mass?

A. 0 cm B. 6.25 cm C. 12.5 cm D. 18.8 cm E. 25.0 cm

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b. What is the moment of inertia of the dumbbell about the y-axis?

A. 0.0186 kg m2B. 0.0234 kg m2C. 0.169 kg m2D. 0.248 kg m2E. 0.496 kg m2

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c. Consider the moment of inertia of the dumbbell about an axis parallel to the y-axis that passes through the center of mass of the dumbbell. How does this moment of inertia compare to the moment of inertia about the y-axis?

Explanation / Answer

here,

m = 1.2 kg , R = 0.1 m

L = 0.3 m

a)

the coordinate of the centre of mass , x= ( - m * (L/2 + R) + 2 * m * (L/2+ R) /( m + m + 2m)

x = ( l/8 + R/4) = 0.0625 m = 6.25 cm

the x coordinate of the centre is B. 6.25 cm

b)

the moment of inertia of the dumbbell about the y-axis , I = 2 * m ( L/2 + R)^2 + m * (L/2 + R)^2 + m * L^2/12

I = 3 * 1.2 * ( 0.3/2 + 0.1)^2 + 1.2 * 0.3^2 /12

I = 0.234 kg.m^2

the correct option is B) 0.234 kg.m^2

c)

the moment of inertia of the dumbbell about an axis parallel to the y-axis that passes through the center of mass of the dumbbell , I' = I + 4 * m * x^2

so, moment of inertia compare to the moment of inertia about the y-axis is G. 4mx^2 larger

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