Question

anyone please help me to solve this question​

Answer 1

Answer:

(a)

(1) Sphere will reach the bottom first.

(2) Ring will reach the bottom last.

(b) M.O.I parallel to diameter = (5/4)x(M)x(R^2)

Explanation:

(WORDS OF CAUTION: If you want a mathematical explanation let me know I'll provide it.)

(a) In simple words Moment of inertia tells us about resistance to change in state of body i.e. how difficult it is to move or stop a body from its prior state.

Now that we know M.O.I tells us about resistance in motion, we can conclude that whichever body has least M.O.I will reach the bottom first, as it will experience lesser resistance in motion.

M.O.I of solid sphere = (2/5)x(M)x(R^2)

M.O.I of disc = (1/2)x(M)x(R^2)

M.O.I of ring = (M)x(R^2)

Hence sphere will reach first and ring will reach last at the bottom of incline plane.

(b) This part involves use of Parallel axes theorem and perpendicular axes theorem.

According to perpendicular axes theorem,

M.O.I (X-Xaxis) + M.O.I (Y-Yaxis) = M.O.I (Z-Zaxis)

According to parallel axes theorem,

M.O.I (through center) + { (M)x(R^2) } = M.O.I (parallel axis)

In the question we are provided with M.O.I (Z-Z axes) = (1/2)MR^2, which is perpendicular to plane of paper.

Hence we need M.O.I about X-X or Y-Y axes, as these are the only axis which are parallel to diameter of disc.

Therefore using Perpendicular axes theorem,

I(X-X) +I(Y-Y) = I(Z-Z)

Due to symmetry I(X-X) =I(Y-Y)

(You can choose any axis among X or Y as both are parallel to diameter.)

Therefore, 2 I(X-X) = I(Z-Z)

2 I(X-X) = (1/2) MR^2

I(X-X)= (1/4) MR^2

Now, to find M.O.I about an axis along the edge parallel to diameter we will use parallel axes theorem.

I(X-X) + MR^2 = I(axes parallel to diameter passing through edge)

(1/4)MR^2 + MR^2 = I(....)

(5/4) MR^2= I(....)