An object is sliding from the top of the smooth inclined plane of height \( h \) from rest and it just completes a vertical circle of diameter 20 cm. Then the minimum height \( h \) of smooth inclined plane is:
Show Hint
For any object to complete a vertical circle, the required release height on a smooth incline is always \( 2.5 \) times the radius of the circle.
Step 1: Identify the critical condition.
To "just complete" a vertical circle, the slowest point is the top. There, the minimum speed satisfies gravity alone providing the centripetal force: $mg = \dfrac{mv_{top}^2}{r}$, giving $v_{top}^2 = gr$. Step 2: Set up energy conservation.
The incline is smooth, so mechanical energy is conserved from the release height $h$ to the top of the loop, which sits at height $2r$. Step 3: Write the energy balance.
\[ mgh = mg(2r) + \tfrac{1}{2}m v_{top}^2 \] Step 4: Substitute the top-speed condition.
Put $v_{top}^2 = gr$: $mgh = 2mgr + \tfrac{1}{2}m(gr) = 2.5\,mgr$. Cancelling $mg$ gives $h = 2.5\,r$. Step 5: Get the radius.
The diameter is $20$ cm, so the radius is $r = 10$ cm $= 0.1$ m. Step 6: Compute the height.
$h = 2.5\times 0.1 = 0.25$ m.
\[ \boxed{0.25\ \text{m}} \]
