i. Answer question 1 and one other question which will be assigned to you by the supervisor and examiner.
ii. Mark are given mainly for a clear record of observation actually made and for the use made of them. Candidates are therefore expected to record on their scripts all their observations as soon as they are made and to plan the presentations of the records so that it will not be necessary to recopy them.
iii. Details of question paper should not be reproduced; neither will the derivation of the theory of the experiment is required. However, knowledge of the basic principles will be essential in order to be able to comment on the results where necessary. Any major precautions taken must be stated.
iv. Each question carries 30 marks.
v. Only scientific calculator can be used.
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## QUESTION 1A
**(a)** The first general equation of motion developed was Newton's second law of motion. However, another familiar equation according to the Euler's laws of motion are similar to Newton's laws, but they are applied specifically to the motion of rigid bodies. The Newton–Euler equations combine the forces and torques acting on a rigid body into a single equation.
Newton's second law for rotation takes a similar form to the translational case:
$$F = \frac{dP}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} = ma \tag{1}$$
Since m is a constant in Newtonian mechanics.
Newton's second law applies to point-like particles, and to all points in a rigid body. They also apply to each point in a mass continuum. Again, by equating the torque acting on the body to the rate of change of its angular momentum L. Analogous to mass times acceleration, the moment of inertia tensor (I) depends on the distribution of mass about the axis of rotation, and the angular acceleration is the rate of change of angular velocity, to be in the form of:
$$\tau = \frac{dL}{dt} \quad \text{or} \quad \tau = I\alpha \tag{2}$$
These equations apply to point like particles, or at each point of a rigid body. Likewise, for a number of particles.
In general, we assume the equation of motion for one particle to be:
$$\tau = I\alpha^k \tag{3}$$
Where I and k are constants parameters.
Some experimental values to verify equation (3) are obtained in table 1.
**Table 1**
| τ / Nm | 3.7 | 20.7 | 79.1 | 76.9 | 106.4 | 162.2 | 254.7 | 278.0 | 319.9 | 331.9 |
|---|---|---|---|---|---|---|---|---|---|---|
| α / rads⁻² | 1.3 | 3.2 | 9.5 | 10.3 | 14.4 | 17.9 | 21.1 | 25.5 | 29.6 | 31.4 |
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### QUESTION 1A — Parts (i) to (v): SOLUTIONS
#### (i) Transform equation (3) into a suitable straight-line graph form to determine I and k
Starting from:
$$\tau = I\alpha^k$$
Take logarithm (base 10) of both sides:
$$\log \tau = \log I + k \log \alpha$$
This is in the form **Y = mX + C**, where:
- Y = log τ
- X = log α
- Gradient (slope) = **k**
- Y-intercept = **log I** → so **I = 10^(intercept)**
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#### (ii) Composite Table for plotting log τ against log α
| τ (Nm) | α (rads⁻²) | log τ | log α |
|---|---|---|---|
| 3.7 | 1.3 | 0.568 | 0.114 |
| 20.7 | 3.2 | 1.316 | 0.505 |
| 79.1 | 9.5 | 1.898 | 0.978 |
| 76.9 | 10.3 | 1.886 | 1.013 |
| 106.4 | 14.4 | 2.027 | 1.158 |
| 162.2 | 17.9 | 2.210 | 1.253 |
| 254.7 | 21.1 | 2.406 | 1.324 |
| 278.0 | 25.5 | 2.444 | 1.407 |
| 319.9 | 29.6 | 2.505 | 1.471 |
| 331.9 | 31.4 | 2.521 | 1.497 |
> Plot **log τ** (Y-axis) against **log α** (X-axis) to get a straight line.
---
#### (iii) Determine I and k from graph
Using two well-separated points from the best-fit line — taking extreme points from the table:
**Point 1:** (0.114, 0.568)
**Point 2:** (1.497, 2.521)
**Gradient = k:**
$$k = \frac{2.521 - 0.568}{1.497 - 0.114} = \frac{1.953}{1.383} \approx \boxed{1.41}$$
**Y-intercept (c = log I):**
Using point 1:
$$c = Y - kX = 0.568 - 1.41 \times 0.114 = 0.568 - 0.161 = 0.407$$
$$\log I = 0.407 \Rightarrow I = 10^{0.407} \approx \boxed{2.55 \text{ kg·m}^2}$$
---
#### (iv) Value of α when τ = 187.6
From the linearised equation:
$$\log \tau = \log I + k\log\alpha$$
$$\log(187.6) = 0.407 + 1.41 \times \log\alpha$$
$$2.273 = 0.407 + 1.41\log\alpha$$
$$\log\alpha = \frac{2.273 - 0.407}{1.41} = \frac{1.866}{1.41} = 1.323$$
$$\alpha = 10^{1.323} \approx \boxed{21.0 \text{ rads}^{-2}}$$
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#### (v) Values of I and k when τ = 187.63, and find α
This is essentially the same equation as (iv). With τ = 187.63:
$$\log(187.63) = 2.273$$
Using the same values of I = 2.55 and k = 1.41 (from graph):
$$\log\alpha = \frac{2.273 - 0.407}{1.41} = 1.323$$
$$\boxed{\alpha \approx 21.0 \text{ rads}^{-2}}$$
> The constants remain: **k ≈ 1.41**, **I ≈ 2.55 kg·m²**
---
### QUESTION 1B — Percentage error in μ
**Given:**
$$\mu = w^2 - \sqrt{z}$$
$$w = 10 \pm 0.3, \quad z = 1 \pm 0.02$$
**Step 1: Calculate μ (nominal value)**
$$\mu = (10)^2 - \sqrt{1} = 100 - 1 = 99$$
**Step 2: Find the absolute error in μ**
Using error propagation:
For $w^2$:
$$\Delta(w^2) = 2w \cdot \Delta w = 2 \times 10 \times 0.3 = 6$$
For $\sqrt{z} = z^{1/2}$:
$$\Delta(\sqrt{z}) = \frac{1}{2\sqrt{z}} \cdot \Delta z = \frac{1}{2 \times 1} \times 0.02 = 0.01$$
Total absolute error (errors add for both + and −):
$$\Delta\mu = \Delta(w^2) + \Delta(\sqrt{z}) = 6 + 0.01 = 6.01$$
**Step 3: Percentage error**
$$\% \text{ error} = \frac{\Delta\mu}{\mu} \times 100 = \frac{6.01}{99} \times 100 \approx \boxed{6.07\%}$$
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## QUESTION 2A — Refraction through a Glass Prism
**Procedure Summary (as instructed):**
Place glass prism **abcd** on drawing paper. Set i = 30°. Use pins p1, p2 on incident side; locate p3, p4 on emergent side to coincide with images of p1, p2. Remove prism, draw ray path. Measure **θ** (angle of refraction), **r** (refracted ray), and perpendicular distance **d**. Repeat for i = 35°, 40°, 45°, 50°, 60°.
**Graph:** Plot **cos r** against **sin θ**.
**Relationship:** d cos r = t sin θ
This rearranges to:
$$\cos r = \frac{t}{d} \sin\theta$$
So the graph of cos r vs sin θ gives a straight line through the origin with:
$$\text{Slope } s = \frac{t}{d}$$
Therefore:
$$\boxed{d = \frac{t}{s}}$$
**Sources of Error:**
1. Parallax error in pin alignment when sighting through the glass.
2. Pins not vertical, leading to incorrect ray tracing.
## QUESTION 3A — Electric Circuit (Figure 2)
**Setup:** V = 1.5V, R = 2Ω. Switch closed. Record I and V for five values of V (increased). Form composite table. Plot **log I** against **log V**.
**Given relation:**
$$I = \frac{V}{R + \lambda}$$
**Linearising:** Take logs:
$$\log I = \log V - \log(R + \lambda)$$
This is in the form Y = X + C, where the slope = 1 and the intercept = −log(R + λ).
From the graph:
- **Slope S = 1** (confirms the linear relationship)
- **Y-intercept = −log(R + λ)**
To find **λ**:
$$-\text{intercept} = \log(R + \lambda)$$
$$R + \lambda = 10^{-\text{intercept}}$$
$$\lambda = 10^{-\text{intercept}} - R$$
**Precautions:**
1. Allow the circuit to stabilise before recording readings to avoid transient fluctuations.
2. Ensure connections are firm and tight to minimise contact resistance errors.