Challenging +1.8 This is a multi-part Further Maths mechanics question requiring parallel axis theorem, energy conservation, and small angle approximations for SHM. While the individual techniques are standard for FM students, the combination of rotational dynamics concepts and the need to set up differential equations makes this significantly harder than typical A-level questions. The 14 marks and three distinct parts requiring different approaches place it well above average difficulty.
A uniform circular disc has diameter \(AB\), mass \(2m\) and radius \(a\). A particle of mass \(m\) is attached to the disc at \(B\). The disc is able to rotate about a smooth fixed horizontal axis through \(A\). The axis is tangential to the disc. Show that the moment of inertia of the system about the axis is \(\frac{5}{2}ma^2\). [4]
The disc is held with \(AB\) horizontal and released. Find the angular speed of the system when \(B\) is directly below \(A\). [5]
The disc is slightly displaced from the position of equilibrium in which \(B\) is below \(A\). At time \(t\) the angle between \(AB\) and the vertical is \(\theta\). Write down the equation of motion, and find the approximate period of small oscillations about the equilibrium position. [5]
Find MI of disc about axis at A by par. axes thm: Idisc = ¼ 2ma + 2ma [= 5ma /2] M1 A1
2 2
Find MI of particle about axis at A: Im = m(2a) [= 4ma ] B1
Combine to find MI of system : I = 13ma 2 /2 A.G. A1
Use conservation of energy (lose A1 for one error): ½IΩ2 = 2mg × a + mg × 2a M1 A2
Substitute for I to find angular speed Ω: Ω = √(16g/13a) A.E.F. M1 A1
State eqn of motion (A.E.F.): I d 2θ /dt 2 = – 4mga sin θ M1 A1
Approximate sin θ by θ (implied by use of SHM): I d 2θ /dt 2 = – 4mga θ M1
Find approx. period T from SHM formula: T = 2π /√(8mga/13ma 2 )
Answer
Marks
= 2π √(13a/8g) A.E.F. M1 A1
4
5
Answer
Marks
5
[14]
Question 5:
5 | 2 2 2
Find MI of disc about axis at A by par. axes thm: Idisc = ¼ 2ma + 2ma [= 5ma /2] M1 A1
2 2
Find MI of particle about axis at A: Im = m(2a) [= 4ma ] B1
Combine to find MI of system : I = 13ma 2 /2 A.G. A1
Use conservation of energy (lose A1 for one error): ½IΩ2 = 2mg × a + mg × 2a M1 A2
Substitute for I to find angular speed Ω: Ω = √(16g/13a) A.E.F. M1 A1
State eqn of motion (A.E.F.): I d 2θ /dt 2 = – 4mga sin θ M1 A1
Approximate sin θ by θ (implied by use of SHM): I d 2θ /dt 2 = – 4mga θ M1
Find approx. period T from SHM formula: T = 2π /√(8mga/13ma 2 )
= 2π √(13a/8g) A.E.F. M1 A1 | 4
5
5 | [14]
A uniform circular disc has diameter $AB$, mass $2m$ and radius $a$. A particle of mass $m$ is attached to the disc at $B$. The disc is able to rotate about a smooth fixed horizontal axis through $A$. The axis is tangential to the disc. Show that the moment of inertia of the system about the axis is $\frac{5}{2}ma^2$. [4]
The disc is held with $AB$ horizontal and released. Find the angular speed of the system when $B$ is directly below $A$. [5]
The disc is slightly displaced from the position of equilibrium in which $B$ is below $A$. At time $t$ the angle between $AB$ and the vertical is $\theta$. Write down the equation of motion, and find the approximate period of small oscillations about the equilibrium position. [5]
\hfill \mbox{\textit{CAIE FP2 2010 Q5 [14]}}