question_answer

A stone is tied to a string of length \[l\] and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed \[u\]. The magnitude of the change in velocity as it reaches a position where the string is horizontal {g being acceleration due to gravity) is:  [AIPMT (S) 2004]

A)       \[\sqrt{2({{u}^{2}}-gl)}\]

B) \[\sqrt{{{u}^{2}}-gl}\]

C) \[u-\sqrt{{{u}^{2}}-2gl}\]

D) \[\sqrt{2gl}\]

Correct Answer: A

Solution :

Key Idea: When scone reaches a position where string is horizontal, it attains the energy partially as kinetic and partially as potential. When stone is at its lowest position, it has only kinetic energy, given by

\[K=\frac{1}{2}m{{u}^{2}}\]

At the horizontal position, it has energy

\[E=U+K=\frac{1}{2}mu{{'}^{2}}+mgl\]

According to conservation of energy,

\[K=E\]

\[\therefore \]      \[\frac{1}{2}m{{u}^{2}}=\frac{1}{2}mu{{'}^{2}}+mgl\]

or         \[\frac{1}{2}mu{{'}^{2}}=\frac{1}{2}m{{u}^{2}}-mgl\]

or         \[u{{'}^{2}}{{u}^{2}}-2gl\]

or         \[u'=\sqrt{{{u}^{2}}-2gl}\]                               ...(i)

So, the magnitude of change in velocity

\[|\Delta \overset{\to }{\mathop{u}}\,|=|\overset{\to }{\mathop{u}}\,|=\sqrt{u{{'}^{2}}+{{u}^{2}}+2u'u\cos {{90}^{0}}}\]

\[|\Delta \overset{\to }{\mathop{u}}\,|=\sqrt{u{{'}^{2}}+{{u}^{2}}}\]

\[=\sqrt{2(u{{'}^{2}}-gL)}\]                  [from Eq. (1)]