a) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravit...
Question 1 Report
a) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.
(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:
(i) acceleration;
(ii) total distance covered.
(c ) A body at rest is given an initial uniform acceleration of \(6.0\ \mathrm{ms}^{-2}\) for \(20\mathrm{s}\) after which the acceleration is reduced to \(4.0\ \mathrm{ms}^{-2}\) for the next \(10\mathrm{s}\).
The body maintains the speed attained for \(30\mathrm{s}\).
Draw the velocity-time graph of the motion using the information given above. From the graph, calculate the:
- maximum speed attained during the motion;
- total distance traveled during the first \(30\ \mathrm{s}\);
- average speed during the same time interval as in (ii) above.
Answer Details
(a) Determination of the centre of gravity of the cardboard
- Make three small, well-spaced holes near the edge of the irregular cardboard.
- Clamp the stout pin horizontally in the retort stand. Suspend the cardboard freely from the pin through one hole.
- Hang the simple pendulum from the same pin, with its string very close to the surface of the cardboard.
- Allow the cardboard and pendulum to come to rest. Trace the vertical line indicated by the pendulum string on the cardboard with the pencil.
- Suspend the cardboard successively from each of the other two holes and trace the vertical line in each case.
- The point of intersection of the three vertical lines is the centre of gravity of the cardboard.
Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.
(b) Velocity-time graph
(i) Acceleration
Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,
\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]
(ii) Total distance covered
The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.
(c) Velocity-time graph of the motion
For the first 20 s:
\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]
For the next 10 s:
\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]
The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.
The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(i) Maximum speed
\[\boxed{160\ \text{m s}^{-1}}\]
(ii) Total distance travelled during the first 30 s
This is the area under the graph from 0 s to 30 s:
\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]
\[s=1200+1400=\boxed{2600\ \text{m}}.\]
(iii) Average speed during the first 30 s
\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]
\[\boxed{86.7\ \text{m s}^{-1}}\]
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