Question 4.6: A sphere attached to a cord of length L = 1 m swings in a ve......
A sphere attached to a cord of length L = 1 m swings in a vertical circle under the influence of gravity. The sphere has a speed of 2 m/s when the cord has an angle θ = 30° with the vertical, as shown in Fig. 4.18. At this instant, find its acceleration in terms of tangential and radial components.
When the cord makes an angle θ to the vertical line, the component of the gravitational acceleration \vec{g} that is tangent to the circular path has a magnitude g sin θ. Thus the magnitude of the tangential acceleration is:
a_{\mathrm{t}}=g\sin\theta=(9.8\,\mathrm{m}/s^{2})(\sin30^{\circ})=4.9\,\mathrm{m}/s^{2}Since the speed of the sphere at this instant is v = 2 m/s and the radius of the circle that the sphere swings about equals the length of the cord, i.e.r = L = 1 m, then the magnitude of the radial acceleration is:
a_{t}={\frac{v^{2}}{r}}={\frac{(2\ \mathrm{m}/s)^{2}}{1\ \mathrm{m}}}=4\ \mathrm{m}/s^{2}Therefore, T − mg cos θ = ma_r, where T is the cord’s tension. From the relation Eq. 4.34
\vec{a}=\vec{a}_{\mathrm{t}}+\vec{a}_{\mathrm{r}} a = \sqrt{a^{2}_{t}+a^{2}_{r}} , and {\mathrm{tan}}{\ \theta}=\frac{a_{\mathrm{t}}}{a_{\mathrm{r}}} (4.34)
we can find the magnitude of a at θ = 30° as follows:
a={\sqrt{\left(4.9\,\mathrm{m}/s^{2}\right)^{2}+(4\,\mathrm{m}/s^{2})^{2}}}=6.32\,\mathrm{\ m}/s^{2}The angle Φ between the vector \vec{a} and the cord will be:
\phi=\tan^{-1}\left({\frac{a_{\mathrm{t}}}{a_{\mathrm{r}}}}\right)=\tan^{-1}\left({\frac{4.9~\mathrm{m/s^{2}}}{4~\mathrm{m/s^{2}}}}\right)=\tan^{-1}\left(1.225\right)=50.77^{\circ}