A thin uniform rod AB has mass kM and length 2a. The end A of the rod is rigidly attached to the surface of a uniform hollow sphere with centre O, mass kM and radius 2a. The end B of the rod is rigidly attached to the circumference of a uniform ring with centre C, mass M and radius a. The points C, B, A, O lie in a straight line. The horizontal axis L passes through the mid-point of the rod and is perpendicular to the rod and in the plane of the ring (see diagram). The object consisting of the rod, the ring and the hollow sphere can rotate freely about L. (i) Show that the moment of inertia of the object about L is $rac{1}{3}(8k + 3)Ma^2$.

A-Level CIE Further Maths 14.1 Work, Energy & Power: A thin uniform rod AB has mass k

A thin uniform rod AB has mass kM and length 2a - CIE - A-Level Further Maths - Question 5 - 2019 - Paper 1

Question 5

A thin uniform rod AB has mass kM and length 2a. The end A of the rod is rigidly attached to the surface of a uniform hollow sphere with centre O, mass kM and radius... show full transcript

Worked Solution & Example Answer:A thin uniform rod AB has mass kM and length 2a - CIE - A-Level Further Maths - Question 5 - 2019 - Paper 1

Step 1

Moment of Inertia of the Rod About Axis L

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Answer

The moment of inertia of the rod about the axis L, which is at a distance of $rac{2a}{2} = a$ from end A, can be calculated using the parallel axis theorem:

$I_{rod} = I_{CM} + Md^2$

where $I_{CM} = \frac{1}{3} kM (2a)^2$ and $d = a$ .

So, we calculate:

$I_{rod} = \frac{1}{3} kM (2a)^2 + kM a^2 = \frac{4}{3} kMa^2 + kMa^2 = \frac{7}{3} kMa^2$ .

Step 2

Moment of Inertia of the Hollow Sphere About Axis L

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Answer

The moment of inertia of the hollow sphere about its own center is given by:

$I_{sphere} = \frac{2}{3} kM (2a)^2$

Using the parallel axis theorem again to shift to axis L:

$I_{sphere} = \frac{2}{3} kM (2a)^2 + kM\cdot(2a)^2 = \frac{2}{3} kM (4a^2) + kM (4a^2) = \frac{8}{3} kMa^2 + 4kMa^2 = \frac{8}{3} kMa^2 + \frac{12}{3} kMa^2 = \frac{20}{3} kMa^2$ .

Step 3

Moment of Inertia of the Ring About Axis L

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Answer

The moment of inertia of the ring about its own center is:

$I_{ring} = Ma^2$

Here, the distance from the center of the ring to axis L is 0, which means:

$I_{ring} = Ma^2$ .

Step 4

Total Moment of Inertia About Axis L

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Answer

Adding all contributions:

$I_{total} = I_{rod} + I_{sphere} + I_{ring}$

Substituting the values:

$I_{total} = \frac{7}{3} kMa^2 + \frac{20}{3} kMa^2 + Ma^2 = \frac{7}{3} kMa^2 + \frac{20}{3} kMa^2 + \frac{3}{3} Ma^2$

$= \frac{(7k + 20k + 3)Ma^2}{3} = \frac{(8k + 3)Ma^2}{3}$ .

Thus, we have shown that the moment of inertia of the object about L is indeed:

$\frac{1}{3}(8k + 3)Ma^2$ .

A thin uniform rod AB has mass kM and length 2a - CIE - A-Level Further Maths - Question 5 - 2019 - Paper 1

Worked Solution & Example Answer:A thin uniform rod AB has mass kM and length 2a - CIE - A-Level Further Maths - Question 5 - 2019 - Paper 1

Moment of Inertia of the Rod About Axis L

Moment of Inertia of the Hollow Sphere About Axis L

Moment of Inertia of the Ring About Axis L

Total Moment of Inertia About Axis L

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