ABCDE is a channel in the vertical plane, par | vedclass

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Question

Loss in $PE$ between $A$ and $D=$ gain in $KE$ between $A$ and $D$

${\mathop{m mg}
olimits} ({m{h}} - 2{m{r}}) = \frac{1}{2}{m{m}}\left( {

Vedclass · Accepted Answer

$h \geq \frac{5}{2}r$

Vedclass · Answer

Loss in $PE$ between $A$ and $D=$ gain in $KE$ between $A$ and $D$

${\mathop{m mg}
olimits} ({m{h}} - 2{m{r}}) = \frac{1}{2}{m{m}}\left( {{{m{v}}^2} - 0} ight)\quad \left( {{{m{K}}_{m{A}}} = 0} ight)$

$\mathrm{v}^{2} =2 \mathrm{g}(\mathrm{h}-2 \mathrm{r})$       .....$(i)$

If the block is to complete the loop path then at $D$

$\frac{\mathrm{mv}^{2}}{\mathrm{r}} \geq \mathrm{mg}$

$\mathrm{v}^{2} \geq \mathrm{rg}$      .....$(ii)$

From Eqs. $(i)$ and $(ii)$

$2g(h - 2r) \ge rg$

$h  \geq \frac{5}{2} r$

$ABCDE$ is a channel in the vertical plane, part $BCDE$ being circular with radius $r$ . A block is released from $A$ and slides without friction and without rolling. The block will complete the loop if $h$ is

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