| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Angular kinematics – constant angular acceleration/deceleration |
| Difficulty | Standard +0.8 This M4 question requires conservation of angular momentum with moment of inertia calculations for a square lamina (non-standard shape), followed by rotational kinematics. The first part demands recall of the perpendicular axis theorem or parallel axis theorem and correct application to a square, which is more conceptually demanding than standard circular motion problems. The multi-step nature and requirement to work with unfamiliar moments of inertia places this moderately above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| [working/answer] | [mark e.g. M1] | [guidance notes] |
| [working/answer] | [mark e.g. A1] | [guidance notes] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = \frac{1}{3}(4.5)(0.3^2 + 0.3^2)\) \((= 0.27)\) | B1 | |
| \(I_2 = I_1 + m(0.3^2 + 0.3^2)\) \((= 0.27 + 0.18m)\) | B1 | |
| \(I_2 \times 1.5 = I_1 \times 2.2\) | M1 | Using angular momentum |
| \(I_2 = 0.396\) | ||
| Mass is \(0.7\) kg | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(36 = \frac{1}{2}(1.5 + 0)t\) | M1 | Using \(\theta = \frac{1}{2}(\omega_0 + \omega_1)t\) |
| Time is \(48\) s | A1 [2] |
## Question 1:
[working/answer] | [mark e.g. M1] | [guidance notes]
[working/answer] | [mark e.g. A1] | [guidance notes]
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# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = \frac{1}{3}(4.5)(0.3^2 + 0.3^2)$ $(= 0.27)$ | B1 | |
| $I_2 = I_1 + m(0.3^2 + 0.3^2)$ $(= 0.27 + 0.18m)$ | B1 | |
| $I_2 \times 1.5 = I_1 \times 2.2$ | M1 | Using angular momentum |
| $I_2 = 0.396$ | | |
| Mass is $0.7$ kg | A1 [4] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $36 = \frac{1}{2}(1.5 + 0)t$ | M1 | Using $\theta = \frac{1}{2}(\omega_0 + \omega_1)t$ | Or other complete method $(a = -0.03125)$ |
| Time is $48$ s | A1 [2] | |
---1 A uniform square lamina, of mass 4.5 kg and side 0.6 m , is rotating about a fixed vertical axis which is perpendicular to the lamina and passes through its centre. A stationary particle becomes attached to the lamina at one of its corners, and this causes the angular speed of the lamina to change instantaneously from $2.2 \mathrm { rad } \mathrm { s } ^ { - 1 }$ to $1.5 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(i) Find the mass of the particle.
The lamina then slows down with constant angular deceleration. It turns through 36 radians as its angular speed reduces from $1.5 \mathrm { rad } \mathrm { s } ^ { - 1 }$ to zero.\\
(ii) Find the time taken for the lamina to come to rest.
\hfill \mbox{\textit{OCR M4 2012 Q1 [6]}}