You are provided with a retort stand, boss head, clamp, stopwatch, slotted weights, hanger, grooved pulley, thread, measuring tape, and other necessary mate...
You are provided with a retort stand, boss head, clamp, stopwatch, slotted weights, hanger, grooved pulley, thread, measuring tape, and other necessary materials.
i. Measure and record the radius \(R\) of the pulley.
ii. Setup the apparatus as illustrated in the diagram above, such that the clamp is 1.5 m above the floor.
iii. Tie one end of the thread to the pulley.
iv. Tie the other end of the thread to the hanger.
v. Slot a mass \(m = 50\ \text{g}\) on the hanger.
vi. Wind the thread around the groove of the pulley until the base of the hanger is at a height \(h = 1.4\ \text{m}\) above the floor. Maintain this height \(h\) for every other value of \(m\) through out the experiment.
vii. Release the mass to unwind the thread.
viii. Determine and record the time \(t\) taken by the mass \(m\) to reach the floor.
ix. Evaluate \(t^{2}\)
x. Also evaluate a = \(\frac{2h}{t^{2}}\), T = \(\frac{m}{1000}(10 - a)\) and \(\propto = \frac{a}{R}\)
xi. Repeat the procedure for four other values of \(m = 70\ \text{g}, 90\ \text{g}, 110\ \text{g}\) and \(130\ \text{g}\)
xii. Tabulate your readings.
xiii. Plot a graph with \(\propto\) on the vertical axis and T on the horizontal axis.
xiv. Determine the slope s, of the graph.
xv. Evaluate \(I = \frac{R}{s}\).
xvi. State two precautions taken to obtain accurate results.
(b)i. Define centripetal force
ii. An object drops to the ground from a height of 2.0 m. Calculate the speed with which it strikes the ground. [g=10 ms\(^{-2}\)]
(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.
Apparatus: falling mass unwinding a thread from a grooved pulley clamped 1.5 m above the floor; the hanger base starts at h = 1.4 m.
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):
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 \(6.32\ \text{m s}^{-1}\).
(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.
Apparatus: falling mass unwinding a thread from a grooved pulley clamped 1.5 m above the floor; the hanger base starts at h = 1.4 m.
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):
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 \(6.32\ \text{m s}^{-1}\).
