Challenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple sophisticated techniques: calculating moment of inertia for a composite lamina using parallel axis theorem, then applying energy conservation to rotational motion with a non-uniform mass distribution. The 12-mark allocation and multi-stage reasoning (finding MI, then using it in energy analysis) place it well above average difficulty, though the methods are standard for FM students who have studied the topic.
\(AB\) is a diameter of a uniform circular disc \(D\) of mass \(9m\), radius \(3a\) and centre \(O\). A lamina is formed by removing a circular disc, with centre \(O\) and radius \(a\), from \(D\). Show that the moment of inertia of the lamina, about a fixed horizontal axis \(l\) through \(A\) and perpendicular to the plane of the lamina, is \(112ma^2\). [5]
A particle of mass \(3m\) is now attached to the lamina at \(B\). The system is free to rotate about the axis \(l\). The system is held with \(B\) vertically above \(A\) and is then slightly displaced and released from rest. The greatest speed of \(B\) in the subsequent motion is \(k\sqrt{(ga)}\). Find the value of \(k\), correct to 3 significant figures. [7]
Find MI of disc D about O: ID = ½ (9m)(3a) = (81/2)ma B1
Find MI of removed disc D′ about O: ID′ = ½ ma 2 B1
Find MI of circular lamina about O: IO = ID – ID′ [= 40 ma 2 ] M1
Find MI of circular lamina about A: IA = IO +8m(3a) 2 or ½(243 – 19)ma 2
= 112ma 2 A.G. M1 A1
Find MI of lamina plus particle about A: IA ′ = IA + (3m)(6a) 2 = 220ma 2 B1
State or imply that speed is max when AB
vertical M1
Use energy when AB vertical (or at general
point): ½ IA ′ω 2 = 8mg × 6a + 3mg × 12a
or 2 × 11mg × 42a/11 [= 84mga] M1 A1
Substitute for I A ′ and find max ang. speed ω
2 2 2
(or ω ): ω = 84mga/110ma = 42g/55a A1
Answer
Marks
Equate 6aω to k√(ga) to find k: k = √(36 × 42/55) = 5⋅24 M1 A1
5
7
[12]
Question 4:
4 | 2 2
Find MI of disc D about O: ID = ½ (9m)(3a) = (81/2)ma B1
Find MI of removed disc D′ about O: ID′ = ½ ma 2 B1
Find MI of circular lamina about O: IO = ID – ID′ [= 40 ma 2 ] M1
Find MI of circular lamina about A: IA = IO +8m(3a) 2 or ½(243 – 19)ma 2
= 112ma 2 A.G. M1 A1
Find MI of lamina plus particle about A: IA ′ = IA + (3m)(6a) 2 = 220ma 2 B1
State or imply that speed is max when AB
vertical M1
Use energy when AB vertical (or at general
point): ½ IA ′ω 2 = 8mg × 6a + 3mg × 12a
or 2 × 11mg × 42a/11 [= 84mga] M1 A1
Substitute for I A ′ and find max ang. speed ω
2 2 2
(or ω ): ω = 84mga/110ma = 42g/55a A1
Equate 6aω to k√(ga) to find k: k = √(36 × 42/55) = 5⋅24 M1 A1 | 5
7 | [12]
$AB$ is a diameter of a uniform circular disc $D$ of mass $9m$, radius $3a$ and centre $O$. A lamina is formed by removing a circular disc, with centre $O$ and radius $a$, from $D$. Show that the moment of inertia of the lamina, about a fixed horizontal axis $l$ through $A$ and perpendicular to the plane of the lamina, is $112ma^2$. [5]
A particle of mass $3m$ is now attached to the lamina at $B$. The system is free to rotate about the axis $l$. The system is held with $B$ vertically above $A$ and is then slightly displaced and released from rest. The greatest speed of $B$ in the subsequent motion is $k\sqrt{(ga)}$. Find the value of $k$, correct to 3 significant figures. [7]
\hfill \mbox{\textit{CAIE FP2 2012 Q4 [12]}}