An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD. The lamina has mass m, and AB = DC = 6a, BC = AD = 3a. The rod has mass M and length 3a. The end P of the rod is attached to the mid-point of AB. The rod is perpendicular to AB and in the plane of the lamina (see diagram). Show that the moment of inertia of the object about a smooth horizontal axis I_x through O, the centroid of the lamina, is displacement in small angle approximation. Show that the moment of inertia of the object about the mid-point of PQ and perpendicular to the plane of the lamina, is $I_y = rac{1}{3}(m + M)a^2$. Find expressions for the periods of small oscillations of the object about the axes I_x and I_y, and verify that these periods are equal when m = M.

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An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD - CIE - A-Level Further Maths - Question 10 - 2015 - Paper 1

Question 10

An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD. The lamina has mass m, and AB = DC = 6a, BC = AD = 3a. The rod has mass ... show full transcript

Worked Solution & Example Answer:An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD - CIE - A-Level Further Maths - Question 10 - 2015 - Paper 1

Step 1

Find the moment of inertia of the lamina about Q:

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Answer

To find the moment of inertia of the lamina about point Q, we use the formula: $I_{lamina} = rac{1}{3}m(6a)^2 + rac{1}{3}m(3a)^2$ Calculating this yields: $I_{lamina} = rac{1}{3}m(36a^2 + 9a^2) = rac{15}{3}ma^2 = 5ma^2.$

Step 2

State or find the moment of inertia of rod about Q:

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Answer

The moment of inertia of the rod about point Q (which is at the mid-point of AB) is given by: $I_{rod} = I_{cm} + Md^2,$ where ( d = 3a ) is the distance from the center of mass of the rod to point Q. Therefore: $I_{rod} = rac{1}{3}(M(3a)^2) + M(3a)^2 = 3Ma^2.$

Step 3

Sum to find the moment of inertia of the object about Q:

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Answer

Now, we sum the moments of inertia: $I_{total} = I_{lamina} + I_{rod} = 5ma^2 + 3Ma^2 = (5m + 3M)a^2.$

Step 4

Find the moment of inertia about the mid-point of PQ:

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Answer

To find the moment of inertia about the mid-point of PQ, we calculate: $I_y = rac{1}{3}(m + M)(3a)^2 = rac{1}{3}(m + M)9a^2 = 3(m + M)a^2.$

Step 5

Find expressions for the periods of small oscillations about axes I_x and I_y:

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Answer

The period of oscillation for a body about an axis is given by: $T = 2 ext{ extit{pi}} imes ext{ extit{sqrt}} rac{I}{mg}$ . Thus, using the previously found moments of inertia:

For axis I_x: $T_x = 2 ext{ extit{pi}} imes ext{ extit{sqrt}} rac{I_x}{m g} = 2 ext{ extit{pi}} imes ext{ extit{sqrt}} rac{(5m + 3M)a^2}{mg}$
For axis I_y: $T_y = 2 ext{ extit{pi}} imes ext{ extit{sqrt}} rac{I_y}{m g} = 2 ext{ extit{pi}} imes ext{ extit{sqrt}} rac{3(m + M)a^2}{mg}$

Setting $T_x = T_y$ leads to: $rac{(5m + 3M)}{m} = rac{3(m + M)}{m}$ when m = M.

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An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD - CIE - A-Level Further Maths - Question 10 - 2015 - Paper 1

Worked Solution & Example Answer:An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD - CIE - A-Level Further Maths - Question 10 - 2015 - Paper 1

Find the moment of inertia of the lamina about Q:

State or find the moment of inertia of rod about Q:

Sum to find the moment of inertia of the object about Q:

Find the moment of inertia about the mid-point of PQ:

Find expressions for the periods of small oscillations about axes I_x and I_y:

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