A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

Question

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to (A) (5/4) D  (B) (3/2) D  (C) (7/5) D  (D) D

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ea225d7c9e6f58c888975b43e1c51877-.png" /> As track is frictionless, so total mechanical energy will remain constant T.M.EI = T.M.EF 0 + mgh = (1/2) mvL2 + 0 h = (vL2/2g) For completing the vertical circle, vL ≥ √5 g R h = (5g R/2g) = (5/2) R = (5/4) D

Q. A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $A B = D$ . The height h is equal to

$\frac{5}{4} D$

$\frac{3}{2} D$

$\frac{7}{5} D$

$D$

As track is frictionless, so total mechanical energy will remain constant
$T . M . E_{I} = T . M . E_{F}$
$0 + m g h = \frac{1}{2} m v_{L}^{2} + 0$
$h = \frac{v _{L}^{2}}{2 g}$
For completing the vertical circle, $v_{L} \geq 5 g R$
$h = \frac{5 g R}{2 g} = \frac{5}{2} R = \frac{5}{4} D$

Q. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB=D. The height h is equal to

45​D

23​D

57​D

D

As track is frictionless, so total mechanical energy will remain constant T.M.EI​=T.M.EF​ 0+mgh=21​mvL2​+0 h=2gvL2​​ For completing the vertical circle, vL​≥5gR​ h=2g5gR​=25​R=45​D

Q. A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $A B = D$ . The height h is equal to

$\frac{5}{4} D$

$\frac{3}{2} D$

$\frac{7}{5} D$

$D$

As track is frictionless, so total mechanical energy will remain constant
$T . M . E_{I} = T . M . E_{F}$
$0 + m g h = \frac{1}{2} m v_{L}^{2} + 0$
$h = \frac{v _{L}^{2}}{2 g}$
For completing the vertical circle, $v_{L} \geq 5 g R$
$h = \frac{5 g R}{2 g} = \frac{5}{2} R = \frac{5}{4} D$