You are provided with two meter rules and other necessary apparatus. Place one of the rules on a knife edge and determine its centre of gravity C. Mark this...

Answer Details

(a) Torsional oscillation of a suspended metre rule

The metre rule is balanced on the knife edge and its centre of gravity is located at the 49.5 cm mark, which is marked with chalk. The mass printed on the reverse of the rule is M_R = 135 g. The 100 g mass is fixed at C, the rule is suspended by two parallel threads of length \(h\) attached at the 10 cm marks, and the separation of the threads is kept constant at d = 80 cm = 0.80 m. The rule is twisted through a small angle about the vertical axis through C and released; the time \(t\) for 20 complete oscillations is taken with a stopwatch. The period is \(T=\dfrac{t}{20}\) and \(T^{2}\) is evaluated. The procedure is repeated for \(h = 40, 50, 60, 70\) and \(80\) cm.

Table of readings

\(h\) (cm)	\(d\) (m)	\(t\) (s)	\(T=\dfrac{t}{20}\) (s)	\(T^{2}\) (s\(^2\))
40	0.80	28.0	1.40	1.9600
50	0.80	29.0	1.45	2.1025
60	0.80	31.0	1.55	2.4025
70	0.80	34.0	1.70	2.8900
80	0.80	37.0	1.85	3.4225

Graph of \(T^{2}\) against \(h\)

Slope of the graph

Two points are taken on the line of best fit: \((h_1, T^2_1) = (40\ \text{cm}, 1.81\ \text{s}^2)\) and \((h_2, T^2_2) = (80\ \text{cm}, 3.30\ \text{s}^2)\).

\[ S=\frac{\Delta T^{2}}{\Delta h}=\frac{(3.30-1.81)\ \text{s}^2}{(80-40)\ \text{cm}}=\frac{1.49\ \text{s}^2}{40\ \text{cm}} \] \[ S = 0.0371\ \text{s}^2\,\text{cm}^{-1} \]

Evaluation of \(k\)

Given \(k=\dfrac{S}{Q}\) where \(Q=\dfrac{2}{25\,d^{2}}\), then

\[ k=\frac{S}{Q}=\frac{S\times 25\,d^{2}}{2} \]

With \(S = 0.0371\ \text{s}^2\,\text{cm}^{-1}\) and \(d = 0.80\ \text{m}\) so that \(d^{2}=0.64\ \text{m}^2\):

\[ k=\frac{0.0371\times 25\times 0.64}{2} \] \[ k=\frac{0.5936}{2}=0.297\ \text{s}^2\,\text{cm}^{-1}\,\text{m}^{2} \]

Two precautions

• The two suspension threads were kept exactly equal in length, vertical and parallel so that the rule hung horizontally and oscillated smoothly in a horizontal plane.
• Only a small angular twist was given, and the stopwatch was read at eye level to avoid parallax error while timing 20 complete oscillations.

(b)(i) Couple

A couple is a pair of two forces that are equal in magnitude, parallel and opposite in direction, but whose lines of action do not pass through the same point. A couple produces a turning (rotational) effect only, with no resultant translational force. Its moment (torque) is:

\[ \tau = F\times d \]

where \(F\) is the magnitude of one of the forces and \(d\) is the perpendicular distance between their lines of action.

(b)(ii) Two practical applications of a couple

• Turning a tap or a water valve on and off with the fingers.
• Turning a spanner or a screwdriver, and turning the steering wheel of a vehicle with both hands.