परिपत्र गति

√5√2gl ??? guessing :P

Why theres no topic [3]

http://targetiit.com/iit_jee_forum/posts/no_topic_1407.html

i think because of that hindi titile [4]

above picture is rotated 45° clockwise.. :P

Now v is to be found out..

1/2mv^2 - 1/2m (5√2gl) = mg√2l(1-cos45°)

1/2mv²+m√2gl(1-1/√2)=1/2m(5√2gl)

[7]

1/2mv² - 1/2m (5√2gl) = mgl(1-cos45°)

question achha hai :)

ok final ans :P

v = √2gl(1+2√2)

Where's the pic? [7]

no.. not the correct answer :)

How did you get √5gl?

isn't the ans will simply be √5gl,because work done by horizontal force on bob will be zero from lowest to highest point ????/

correct me if i am wrong

so whats the min velocity imparted at position shown so that it just completes the circle..

hint is in ur reply [1] or solve subjectively [1]

Opse [3] just editing

Wait a min .. [1]
just drawing..