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Grade 12Mechanics

A stone of mass m is whirled in a vertical circle with a string of length l. At an instant, the stone is at its lowest position and has velocity u. Find the change in magnitude of its velocity when the string is horizontal.

Profile image of Rupali Sehgal
11 Years agoGrade 12
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3 Answers

Profile image of Naveen Kumar
11 Years ago
Initial kinetic energy of block (at bottom)=(1/2)*m*u2
let
At the time when the string is horizontal,the velocity is V(say) and so the kinetic energy=(1/2)*m*V2
during the process,gain iun potential energy=mg*l
Applying energy conservation,
(1/2)*m*u2 = (1/2)*m*V2........+.........mg*l
V=sq.root(u2-2g*l)
and hence, u-V=u-sq.root(u2-2g*l)
Profile image of Rupali Sehgal
11 Years ago
sir from the options given, answer is  sq root 2(u2-gl). How does that come?
Profile image of Mallikarjun Maram
11 Years ago
u and v are perpendicular to each other.  Hence magnitude of their difference will be equal to sqrt(u2 + u2 – 2gl) = sqrt(2(u2 – gl))