CAIE FP2 2010 June — Question 4 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRadial and transverse acceleration
DifficultyStandard +0.8 This is a Further Maths mechanics question requiring understanding of no-slip conditions, angular kinematics with time-varying angular velocity, and decomposition of acceleration into radial and tangential components. While the individual concepts are standard (ω₁r₁ = ω₂r₂, a_radial = ω²r, a_tangential = αr), students must correctly differentiate to find angular acceleration, apply Pythagoras for magnitude, and use trigonometry for the angle—a multi-step problem requiring careful coordination of several techniques beyond typical A-level.
Spec6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration

\includegraphics{figure_4} Two coplanar discs, of radii \(0.5\) m and \(0.3\) m, rotate about their centres \(A\) and \(B\) respectively, where \(AB = 0.8\) m. At time \(t\) seconds the angular speed of the larger disc is \(\frac{1}{2}t\) rad s\(^{-1}\) (see diagram). There is no slipping at the point of contact. For the instant when \(t = 2\), find
  1. the angular speed of the smaller disc, [2]
  2. the magnitude of the acceleration of a point \(P\) on the circumference of the larger disc, and the angle between the direction of this acceleration and \(PA\). [7]

Question 4:

(ii) ---
4 (i)
AnswerMarks
(ii)Equate tangential speeds to find ω : 0⋅5ω = 0⋅3ω , ω = 5/3 [rad s -1 ] M1 A1
B A B B
Find tangential acceleration, r d 2θ/dt 2 : 0⋅5 × ½ = 0⋅25 M1 A1
Find radial acceleration, r (dθ/dt) 2 : 0⋅5 × 1 2 = 0⋅5 B1
Combine to give mag. of acceln: √(0⋅25 2 + 0⋅5 2 ) M1
= √5/4 or 0⋅559 [m s -2 ] A1
Find angle made with PA (A.E.F.): tan -1 (0⋅25/0⋅5) M1
AnswerMarks Guidance
= 0⋅464 rad or 26⋅6° A12
7[9]
Page 5Mark Scheme: Teachers’ version Syllabus
GCE A LEVEL – May/June 20109231 22
Question 4:
--- 4 (i)
(ii) ---
4 (i)
(ii) | Equate tangential speeds to find ω : 0⋅5ω = 0⋅3ω , ω = 5/3 [rad s -1 ] M1 A1
B A B B
Find tangential acceleration, r d 2θ/dt 2 : 0⋅5 × ½ = 0⋅25 M1 A1
Find radial acceleration, r (dθ/dt) 2 : 0⋅5 × 1 2 = 0⋅5 B1
Combine to give mag. of acceln: √(0⋅25 2 + 0⋅5 2 ) M1
= √5/4 or 0⋅559 [m s -2 ] A1
Find angle made with PA (A.E.F.): tan -1 (0⋅25/0⋅5) M1
= 0⋅464 rad or 26⋅6° A1 | 2
7 | [9]
Page 5 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – May/June 2010 | 9231 | 22
\includegraphics{figure_4}

Two coplanar discs, of radii $0.5$ m and $0.3$ m, rotate about their centres $A$ and $B$ respectively, where $AB = 0.8$ m. At time $t$ seconds the angular speed of the larger disc is $\frac{1}{2}t$ rad s$^{-1}$ (see diagram). There is no slipping at the point of contact. For the instant when $t = 2$, find

\begin{enumerate}[label=(\roman*)]
\item the angular speed of the smaller disc, [2]
\item the magnitude of the acceleration of a point $P$ on the circumference of the larger disc, and the angle between the direction of this acceleration and $PA$. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2010 Q4 [9]}}