Ball a, of mass m4 , is connected to ball b, of mass mB, by a massless rod of le

Question

Ball a, of mass m4 , is connected to ball b, of mass mB, by a massless rod of length L. (Figure 1) The two vertical dashed lines in the figure, one through each ball, represent two different axes of rotation, axes a and b. These axes are parallel to each other and perpendicular to the rod. The moment of inertia of the two-mass system about axis a is Ia , and the moment of inertia of the system about axis b is Ib . It is observed that the ratio of I. to Ib is equal to 3: Ia/Ib = 3 Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass Find the ratio of the masses of the two balls. Find da, the distance from ball A to the system's center of mass.

Explanation / Answer

Moment of Inertia about axis a will be due to mass b and as mass a passes through the axis therefore it will not contribute to the moment of inertia therefore

Ia = mb*(L)^2

Similarly Moment of Inertia about axis b will be due to mass a and as mass b passes through the axis therefore it will not contribute to the moment of inertia therefore

Ib = ma*(L)^2

Now Ia/Ib = mb*(L)^2/ma*(L)^2 = 3

therefore ma/mb = 1/3 (Answer)

Centre of mass is given by
x = [m1(r1)^2 + m2(r2)^2] / [m1+m2]

Now taking centre of the rod as the orogin
m1 = ma
r1 = -L/2
m2 = mb
r2 = L/2

therefore

x = [ma*(-L/2) + mb*(L/2)]/[ma + mb]
and on putting mb = 3ma from Part A

we get

x = L/4
therefore centre of mass is located at a distance of L/4 from the origin
Now
da = distance from ball a to the centre of mass
da = L/2 + L/4

da = 3L/4 (Answer)

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