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

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(B) The loss in potential energy $(PE)$ between point $A$ and point $D$ is equal to the gain in kinetic energy $(KE)$ between $A$ and $D$.$mg(h - 2r)

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The loss in potential energy $(PE)$ between point $A$ and point $D$ is equal to the gain in kinetic energy $(KE)$ between $A$ and $D$.$mg(h - 2r) = \frac{1}{2}mv^2$$v^2 = 2g(h - 2r) \quad .....(i)$For the block to complete the circular loop,the condition at the highest point $D$ is that the normal force must be at least zero,which implies the centripetal force must be at least equal to the gravitational force:$\frac{mv^2}{r} \geq mg$$v^2 \geq rg \quad .....(ii)$Substituting the value of $v^2$ from equation $(i)$ into equation $(ii)$:$2g(h - 2r) \geq rg$$2(h - 2r) \geq r$$2h - 4r \geq r$$2h \geq 5r$$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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