Figure shows a smooth track in a vertical plane, consisting of two circular arcs A B and B C of the same radius R=2 m. The common tangent to the arcs, B D, is horizontal. A small block placed on the track at a height h above B D leaves the track at exactly the same depth below. Determine h in cm.

Question

Figure shows a smooth track in a vertical plane, consisting of two circular arcs A B and B C of the same radius R=2 m. The common tangent to the arcs, B D, is horizontal. A small block placed on the track at a height h above B D leaves the track at exactly the same depth below. Determine h in cm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/57ca2a5db3a6e074db933279f7faa795-.png" /> Condition for leaving contact m g((R-h/R))=(m v2/R) .... (i) From work energy theorem 2 m g h=(1/2) m v2 m v2=4 m g h .... (ii) From (i) and (ii), we get h=(R/5)=40 cm

Q. Figure shows a smooth track in a vertical plane, consisting of two circular arcs AB and BC of the same radius R=2m. The common tangent to the arcs, BD, is horizontal. A small block placed on the track at a height h above BD leaves the track at exactly the same depth below. Determine h in cm.

Condition for leaving contact mg(RR−h​)=Rmv2​ .... (i) From work energy theorem 2mgh=21​mv2 mv2=4mgh .... (ii) From (i) and (ii), we get h=5R​=40cm

Condition for leaving contact
$m g (\frac{R - h}{R}) = \frac{m v ^{2}}{R}$ .... (i)
From work energy theorem
$2 m g h = \frac{1}{2} m v^{2}$
$m v^{2} = 4 m g h$ .... (ii)
From (i) and (ii), we get
$h = \frac{R}{5} = 40 c m$