IIT JEE , AIEEE Forum - Physics , Mechanics

boy

Figure shows a light rod of length L rigidly attached to a small heavy block of mass m at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth 'h' below the initial position of the hook and the hook gets into the ringas it reaches there. What should be the minimum value of h such taht the block moves in a complete circle about the ring ?

The answer is --- h = L

But why the answer does not come if we write the equation at top position as---- mv²/L = mg...
Is this equation not correct..???
Is it not the condition for motion in a vertical circle.???

But on writing the equation based on energy conservation principle we get---
1/2 x mv² = mgL....{initial K.E. = final P.E.}
This gives v² = 2Lg
On using this relation in the equation --- mgh = 1/2 x mv²........The value of h comes out to be L.....
But why does the equation mv²/L = mg does not yield any answer....??????

pratham_bittu

see in this question we should not get confused and directly apply conservation of mechanical energy
PE lost while coming to a vertical position wrt ring
mgh+mgl
now the mass again rises to a height 2L(vertical to the ring)
so lets equate energies
mgh+mgl=mg(2l)
mgh=mgl
h=l

pratham_bittu

well ur equqtion doesnot yielsd the answer bcoz
kinetic energy =1/2 mv^2
and u have equated
mV^2=mgl
so u had to equate
1/2 mv^2=mg(2l)
now u get
v=2 *root of gl

elessar_iitkgp

Initial state of the mass: At rest at a height h above the ring.
Final state: It should reach the topmost point of a circle of radius L.
Assuming the horizontal level of the ring to be the zero PE level,
mgh = mgL + (1/2)mv²
v² = 2g(h - L)
where v is the speed at the topmost point of the circle of radius L.
For reaching the topmost point
v

0
2g(h - L)

0
h

L
Hence h_min = L

gopal_duggirala

valuable piece of advice is that when ur coserving mechanical energy always conserve total energy.

mgh+0.5(m)(0)=-mg(l)+mg(2l)(since for the mass to go around the circle it has to gain the p.e. and rise to a height of 2l)

mgh=mg(2l-l)

h=l

pls correct me if iam wrong

edison

you are right elessar_iitkgp

elessar_iitkgp

Thank you edison

Prakriteesh

mv²/L = mg does not yield any answer because it does not contain the h term in it. Also you must include the contact force by the ring in this equation. i.e

mv²/L = mg - N (in the vertically extreme position.)

Now from energy calculations we have found that for minimum height, v needs to be minimum. For this purpose, N will maximise and become equal to mg.

So, mv²/L = mg - mg = 0.

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