Question

Figure (8-E17) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. (a) Find the minimum projection-speed ${\nu }_0$ for which the particle reaches the top of the track. (b) Assuming that the projection-speed is $2{\nu }_0$ and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than ${\nu }_0$, where will the block lose contact with the track?

Answer 1

(a) Net force on the particle at A and B,
$\mathrm{F}=mg\sin \mathrm{\theta }$

Work done to reach B from A,
$\mathrm{W}=\mathrm{FS}=mg\sin \mathrm{\theta }l$

Again, work done to reach B to C
$$\begin{gathered} =mgh \\ =mg\mathrm{R}\left(1-\cos \mathrm{\theta }\right) \end{gathered}$$

So, total work done
$$\begin{gathered} =mgl\sin \mathrm{\theta }+mgR\left(1-\cos \mathrm{\theta }\right) \\ =mg\left[l\sin \mathrm{\theta }+R\left(1-\cos \mathrm{\theta }\right)\right] \end{gathered}$$



Now, change in K.E. = Total work done

$$\begin{gathered} \Rightarrow \frac{1}{2}m{\nu }_2^2=mg\left[l\sin \mathrm{\theta }+\mathrm{R}\left(1-\cos \mathrm{\theta }\right)\right] \\ \Rightarrow {\nu }_2=\sqrt{2g\left[\mathrm{R}\left(1-\cos \mathrm{\theta }\right)+l\sin \mathrm{\theta }\right]} \end{gathered}$$

(b) When the block is projected at a speed:



Let the velocity at C be ${\nu }_0$.
Applying energy principle,

$$\begin{gathered} \left(\frac{1}{2}\right)m{\nu }_0^2-\left(\frac{1}{2}\right)m{\left(2{\nu }_0\right)}^2 \\ =-mg\left[l\sin \mathrm{\theta }+\mathrm{R}\left(1-\cos \mathrm{\theta }\right)\right] \\ \Rightarrow {\mathrm{V}}^2=4{\nu }_0^2-2g\left[l\sin g\mathrm{\theta }+\mathrm{R}\left(1-\cos \mathrm{\theta }\right)\right] \\ =4.2g\left[l\sin \mathrm{\theta }+\mathrm{R}\left(1-\cos \mathrm{\theta }\right)\right]- \\ 2g\left[l\sin \mathrm{\theta }+\mathrm{R}\left(1-\cos \mathrm{\theta }\right)\right] \end{gathered}$$

So, force acting on the body,

$\mathrm{N}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=6mg\left[\left(\frac{l}{\mathrm{R}}\right)\sin \mathrm{\theta }+1-\cos \mathrm{\theta }\right]$

(c) Let the loose contact after making an angle θ.

$$\begin{gathered} \frac{m{\nu }^2}{\mathrm{R}}=mg\cos \mathrm{\theta } \\ \Rightarrow {\nu }^2=\mathrm{R}g\cos \mathrm{\theta }...\left(i\right) \\ \mathrm{Again},\frac{1}{2}m{\nu }^2=mg\left(\mathrm{R}-\mathrm{R}\cos \mathrm{\theta }\right) \\ \Rightarrow {\nu }^2=2g\mathrm{R}\left(1-\cos \mathrm{\theta }\right)...\left(ii\right) \end{gathered}$$
$$\begin{gathered} \mathrm{From}\left(i\right)\mathrm{and}\left(ii\right),={\cos }^{-1}\left(\frac{2}{3}\right) \\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{2}{3}\right) \end{gathered}$$