CAIE M2 2004 November — Question 6 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2004
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRotating disc with friction
DifficultyModerate -0.3 This is a straightforward application of circular motion principles where friction provides centripetal force. Part (i) uses F=mω²r with limiting friction μR=μmg, requiring simple algebraic manipulation. Part (ii) uses v=ωr. Both parts are standard textbook exercises with direct formula application and minimal problem-solving insight required, making it slightly easier than average.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

6 A horizontal turntable rotates with constant angular speed \(\omega\) rad s \(^ { - 1 }\) about its centre \(O\). A particle \(P\) of mass 0.08 kg is placed on the turntable. The particle moves with the turntable and no sliding takes place.
  1. It is given that \(\omega = 3\) and that the particle is about to slide on the turntable when \(O P = 0.5 \mathrm {~m}\). Find the coefficient of friction between the particle and the turntable.
  2. Given instead that the particle is about to slide when its speed is \(1.2 \mathrm {~ms} ^ { - 1 }\), find \(\omega\).

AnswerMarks Guidance
(i) For using \(F = m\omega^2 r\)M1 (\(0.36\) N)
For using \(F = \mu R\)M1 (\(0.36 = 0.8\mu\))
\(\mu = 0.45\)A1 3 marks
(ii) (\(\mu\), \(R\) fixed \(\rightarrow\)) \(F = 0.36\)B1 ft
For using \(F = mv^2/r\) or \(F = mr(1.2r)^2\) (\(r \approx 0.5\))M1
\((0.36 = 0.08 \times 1.2^2/r\) or \(0.36 = 0.08 \, r(1.2r)^2)\)
[for \(F = m(v/\omega) \times \omega^2\) treat as for alternative case below.]
For using \(v = \omega r\) (\(r \approx 0.5\))M1
\(r = 0.32\) or unsimplified equation in \(\omega\) onlyA1 ft
Alternative for the above last three marks:
For using \(F = mv\omega\)M2
\(0.08 \times 1.2\omega = 0.36\)A1
\(\omega = 3.75\)A1 5 marks
**(i)** For using $F = m\omega^2 r$ | M1 | ($0.36$ N)
For using $F = \mu R$ | M1 | ($0.36 = 0.8\mu$)
$\mu = 0.45$ | A1 | 3 marks

**(ii)** ($\mu$, $R$ fixed $\rightarrow$) $F = 0.36$ | B1 ft |
For using $F = mv^2/r$ or $F = mr(1.2r)^2$ ($r \approx 0.5$) | M1 |
$(0.36 = 0.08 \times 1.2^2/r$ or $0.36 = 0.08 \, r(1.2r)^2)$ |  |
[for $F = m(v/\omega) \times \omega^2$ treat as for alternative case below.] |  |
For using $v = \omega r$ ($r \approx 0.5$) | M1 |
$r = 0.32$ or unsimplified equation in $\omega$ only | A1 ft |
**Alternative for the above last three marks:** |  |
For using $F = mv\omega$ | M2 |
$0.08 \times 1.2\omega = 0.36$ | A1 |
$\omega = 3.75$ | A1 | 5 marks

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6 A horizontal turntable rotates with constant angular speed $\omega$ rad s $^ { - 1 }$ about its centre $O$. A particle $P$ of mass 0.08 kg is placed on the turntable. The particle moves with the turntable and no sliding takes place.\\
(i) It is given that $\omega = 3$ and that the particle is about to slide on the turntable when $O P = 0.5 \mathrm {~m}$. Find the coefficient of friction between the particle and the turntable.\\
(ii) Given instead that the particle is about to slide when its speed is $1.2 \mathrm {~ms} ^ { - 1 }$, find $\omega$.

\hfill \mbox{\textit{CAIE M2 2004 Q6 [8]}}