You are provided with a retort stand, boss head, clamp, stopwatch, slotted weights, hanger, grooved pulley, thread, measuring tape, and other necessary mate...

(a) Determination of the moment of inertia of a pulley

A mass \(m\) on the hanger unwinds the thread from the pulley (radius \(R\)) and falls through a fixed height \(h = 1.4\ \text{m}\). The apparatus is set up as shown below, with the clamp \(1.5\ \text{m}\) above the floor.

For each value of \(m\) the time of fall \(t\) is recorded twice and averaged, and the derived quantities are computed from

\[ a = \frac{2h}{t^{2}} = \frac{2(1.4)}{t^{2}} = \frac{2.8}{t^{2}},\qquad T = \frac{m}{1000}\,(10 - a),\qquad \alpha = \frac{a}{R},\quad R = 0.08\ \text{m}. \]

xii. Table of readings

S/N	\(m\)/g	\(t_1\)/s	\(t_2\)/s	\(t=\dfrac{t_1+t_2}{2}\)/s	\(t^{2}\)/s\(^2\)	\(a=\dfrac{2.8}{t^{2}}\)/m s\(^{-2}\)	\(T=\dfrac{m}{1000}(10-a)\)/N	\(\alpha=\dfrac{a}{R}\)/rad s\(^{-2}\)
1	50.0	5.00	5.00	5.00	25.000	0.110	0.490	1.380
2	70.0	4.80	4.80	4.80	23.040	0.120	0.690	1.500
3	90.0	4.60	4.60	4.60	21.160	0.130	0.890	1.630
4	110.0	4.40	4.40	4.40	19.360	0.140	1.080	1.750
5	130.0	4.20	4.20	4.20	17.640	0.150	1.280	1.880

where \(h = 1.4\ \text{m} = 140\ \text{cm}\) and \(R = 0.08\ \text{m} = 8\ \text{cm}\).

Worked check of row 1 (\(m = 50.0\) g, \(t = 5.00\) s):

\[ t^{2} = 5.00^{2} = 25.000\ \text{s}^2,\qquad a = \frac{2.8}{25.000} = 0.110\ \text{m s}^{-2}, \] \[ T = \frac{50}{1000}(10 - 0.110) = 0.050 \times 9.890 = 0.490\ \text{N},\qquad \alpha = \frac{0.110}{0.08} = 1.380\ \text{rad s}^{-2}. \]

xiii. Graph of \(\alpha\) against \(T\)

Plotting \(\alpha\) (vertical axis) against \(T\) (horizontal axis) gives a straight line:

xiv. Slope of the graph

Taking two well-separated points on the line of best fit, \((T_1,\alpha_1) = (0.50,\ 1.383)\) and \((T_2,\alpha_2) = (1.30,\ 1.891)\):

\[ s = \frac{\alpha_2 - \alpha_1}{T_2 - T_1} = \frac{1.891 - 1.383}{1.30 - 0.50} = \frac{0.508}{0.800} = 0.635\ \text{rad s}^{-2}\,\text{N}^{-1}. \]

xv. Moment of inertia

Since the driving torque \(TR = I\alpha\), we have \(\alpha = \dfrac{R}{I}\,T\), so the slope \(s = \dfrac{R}{I}\) and

\[ I = \frac{R}{s} = \frac{0.08}{0.635} = 0.126\ \text{kg m}^{2}. \]

xvi. Two precautions

• I avoided parallax error by reading the metre rule and the height marks with the line of sight perpendicular to the scale.
• I ensured the boss head and clamp were firmly tightened so the pulley did not slip or wobble during the fall.

(b)(i) Centripetal force

Centripetal force is the resultant inward force, directed towards the centre of the circular path, that keeps a body moving with constant speed in a circle. Its magnitude is

\[ F = \frac{m v^{2}}{r}. \]

(b)(ii) Speed of a body dropped from a height

An object drops to the ground from a height \(h = 2.0\ \text{m}\). Taking \(g = 10\ \text{m s}^{-2}\) and equating potential energy to kinetic energy:

\[ \tfrac{1}{2} m v^{2} = m g h \;\Rightarrow\; v^{2} = 2 g h = 2 \times 10 \times 2.0 = 40\ \text{m}^2\text{s}^{-2}, \] \[ v = \sqrt{40} = 6.32\ \text{m s}^{-1}. \]

The object strikes the ground with a speed of \(6.32\ \text{m s}^{-1}\).