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ADNAN MUHAMMED

Grade 12,

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A body is allowed to slide down a frictionless track freely under gravity. The track end in a semicircular shaped part of a diameter D. The height (minimum) from which the body must fall so that it completes the circle, is:

A body is allowed to slide down a frictionless track freely under gravity. The track end in a semicircular shaped part of a diameter D. The height (minimum) from which the body must fall so that it completes the circle, is:

Grade:12

5 Answers

Dhruvit Raithatha
32 Points
6 years ago
\\\text{Velocity at top of circle is } \sqrt{5gr} \\ \text{Energy lost to reach height D from height H } = mg(H-D) \\ \text{Energy gained in the form of KE } = \frac{1}{2}mv^2 = \frac{1}{2}m(5gr) = \frac{5mgD}{4} \\\\ \text {Equating K.E. and P.E.}, \\\\ \frac{5mgD}{4} = mg(H-D) \\H = \frac{9D}{4}
Lavina singh
22 Points
6 years ago
The answer is 5/4D for the above sum and not 9/4D..............................................................................................
Dhruvit Raithatha
32 Points
6 years ago
Yep. Sorry. My mistake. Velocity is √(5gr) at bottom. Replace mg(H-D) with mgH and you should get the answer. Cheers.
Lavina singh
22 Points
6 years ago
Got it !Thanks....................................................................................................................................................................
Shaikh tuba
15 Points
4 years ago
If a body slides down in circular loop then after solving we get height as 5r/2
Therefore considering diameter, h=5r/2×1/2
 
Hence,h=5D/4

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