The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio 5 : 3 then its velocity is
Text Solution
Verified by ExpertsA
In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant.
Tension at lowest point \(\Gamma_{me} = - \frac{mv^{2}}{r} + mg\)
Tension at highest point \(\Gamma_{rr} = -\frac{mv^{2}}{r} - mg\)
\(\frac{T_{\max}}{T_{\min}} = \frac{r v^2 + r mg}{r v^2 - r mg} = \frac{5}{3}\)
by solving we get, \(v = \sqrt{2 g y}\) \(\sqrt{4 \times 9.8 \times 2.5}\) \(= \sqrt{98} \text{ m/s}\)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems