You are provided with a grooved inclined plane, a solid sphere, a stopwatch, and other necessary apparatus. Place the pile of paper towels at the tail end o...

Answer Details

Test of Practical Knowledge: solid sphere rolling down a grooved inclined plane

For each distance \(D\) the sphere is released from rest at the marked point on the groove, and the time it takes to roll down to the paper-towel stop is measured. Two readings \(t_1\) and \(t_2\) are taken and averaged to give the mean time \(t\). Then \(W=\dfrac{D}{t}\) is evaluated and \(V=2W\) is calculated.

Table of readings

\(D\) (cm)	\(t_1\) (s)	\(t_2\) (s)	mean \(t\) (s)	\(W=\dfrac{D}{t}\) (cm s\(^{-1}\))	\(V=2W\) (cm s\(^{-1}\))
140.0	4.50	5.00	4.750	29.474	58.948
120.0	4.20	4.40	4.300	27.907	55.814
100.0	4.00	3.80	3.900	25.641	51.282
80.0	3.80	3.50	3.650	21.918	43.836
60.0	3.20	3.40	3.300	18.182	36.364

Graph of \(V\) against \(t\)

The five points are plotted with \(V\) (cm s\(^{-1}\)) on the vertical axis and mean \(t\) (s) on the horizontal axis, and the best straight line is drawn through them.

Slope of the graph

Taking two well-separated points on the best-fit line, \((t=2.3\,\text{s},\,V=32\,\text{cm s}^{-1})\) and \((t=5.6\,\text{s},\,V=70\,\text{cm s}^{-1})\):

\[ s=\frac{\Delta V}{\Delta t}=\frac{70-32}{5.6-2.3}=\frac{38}{3.3}=11.51\ \text{cm s}^{-2} \]

Significance of \(s\): the slope \(s\) represents the acceleration of the sphere as it rolls down the inclined plane (i.e. the acceleration due to gravity along the incline).

Two precautions:

• I avoided parallax error when taking readings on the metre rule by viewing the scale directly from above (perpendicular to the rule).
• I took repeated timings for each distance and averaged them, releasing the sphere gently from rest each time, to reduce random errors.

(b)(i) Velocity ratio of an inclined plane

\[ \text{V.R.}=\frac{\text{length of the inclined plane}}{\text{vertical height}}=\frac{1}{\sin\theta} \]

where \(\theta\) is the angle of inclination of the plane to the horizontal.

(b)(ii) Force on the object perpendicular to the plane

The question asks for the force perpendicular (normal) to the inclined surface. This is the component of the weight at right angles to the plane:

\[ F=mg\cos\theta=5\times10\times\cos30^{\circ}=50\times0.866=\textbf{43.3 N} \]

(The remaining component \(mg\sin\theta=25\,\text{N}\) acts along the plane, not perpendicular to it.)