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 ms\(^{-2}\) for 20s after which the acceleration is reduced to 4.0 ms\(^{-2}\) for the next 10s.
The body maintains the speed attained for 30s.
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 s;
- average speed during the same time interval as in (ii) above.
Explanation
(a) PROCEDURE
Make at least 3 well-spaced pin holes round the edge of the cardboard. Clamp the pin horizontally and suspend the cardboard on it through one of the pin-holes such that the cardboard can swing freely
Hang the simple pendulum on the same pin and let its string be very close to the cardboard.
When the whole system is at rest (or in equilibrium) trace the plumbline on the cardboard. Repeat the procedure for each of the two other pin holes.
CONCLUSION
The point at which the (three) traced lines intersect is the centre of gravity of the cardboard.
PRECAUTIONS
- Repeat procedure
- Pin rigidly and firmly held by retort stand and clamp
- Allow the simple pendulum to rest before tracing the shadow of
the plumbline on the cardboard.
- The string to be close to the cardboard
b) Diagram:
Both axes correctly labelled
Any correct shape of graph showing acceleration segment
- Acceleration = gradient of AB
- Total distance covered = area under the graph
(c) Diagram
- At least one axis labelled
- correct shape of graph
Let V\(_1\) = maximum velocity after 20 sec.
V\(_2\) = maximum velocity after 30 sec.
(i) Then \(\frac{V_1}{20}\) = 6
∴ V\(_1\) = 120 ms\(^{-1}\)
Also V\(\frac{V_2 - V_1}{10}\) = 4
\(\frac{V_2 - 120}{10}\) = 4
V\(_2\) = 160 ms\(^{-1}\)
Maximum speed = V\(_2\) = 160ms\(^{-1}\)
(ii) Total distance covered = Area of Δ + Area of trapezium after 1st 30 seconds
= ( ½ x 20 x 120) + ½ (120 + 160) x 10
= 1200 + 1400
= 2600m
(iii) Average speed = \(\frac{\text{Total distance}}{\text{ Total time}}\)
= \(\frac{ 2600}{30}\)
= 86.67 ms\(^{-1}\)