CAIE M2 2006 November — Question 2 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2006
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRotating disc with friction
DifficultyModerate -0.3 This is a straightforward circular motion problem requiring standard application of F=mrω² for centripetal force and resolving forces vertically/horizontally. Part (i) involves routine calculations (vertical: N=mg, horizontal: F=mrω²), and part (ii) simply applies the friction condition F≤μN. The problem is slightly easier than average as it's a direct application of formulas with no conceptual subtlety or multi-step reasoning required.
Spec3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

2 A horizontal turntable rotates with constant angular speed \(3 \mathrm { rad } \mathrm { s } ^ { - 1 }\). A particle of mass 0.06 kg is placed on the turntable at a point 0.25 m from its centre. The coefficient of friction between the particle and the turntable is \(\mu\). As the turntable rotates, the particle moves with the turntable and no sliding takes place.
  1. Find the vertical and horizontal components of the contact force exerted on the particle by the turntable.
  2. Show that \(\mu \geqslant 0.225\).

Question 2:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(R = 0.6\)NB1
\(F = 0.06 \times 3^2 \times 0.25\)M1 For using Newton's second law and \(a = r\omega^2\)
\(F = 0.135\)A1 (3)
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mu \geq 0.225\)B1 (1) From using \(\mu \geq F/R\)
## Question 2:

### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 0.6$N | B1 | |
| $F = 0.06 \times 3^2 \times 0.25$ | M1 | For using Newton's second law and $a = r\omega^2$ |
| $F = 0.135$ | A1 (3) | |

### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu \geq 0.225$ | B1 (1) | From using $\mu \geq F/R$ |

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2 A horizontal turntable rotates with constant angular speed $3 \mathrm { rad } \mathrm { s } ^ { - 1 }$. A particle of mass 0.06 kg is placed on the turntable at a point 0.25 m from its centre. The coefficient of friction between the particle and the turntable is $\mu$. As the turntable rotates, the particle moves with the turntable and no sliding takes place.\\
(i) Find the vertical and horizontal components of the contact force exerted on the particle by the turntable.\\
(ii) Show that $\mu \geqslant 0.225$.

\hfill \mbox{\textit{CAIE M2 2006 Q2 [4]}}