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Step-by-Step Solution
Step 1: Understand the Physical Situation
A small block is placed at the top of a smooth (frictionless) hemisphere of radius R = 3\,\text{m} . It slides down under the influence of gravity. We wish to find the vertical height (measured from the horizontal ground on which the hemisphere rests) at which the block loses contact with the hemisphere.
Step 2: Condition for Losing Contact
The block will lose contact with the hemisphere at the point where the normal force from the hemisphere on the block becomes zero. At any point on the hemisphere, if \theta is the angle that the radius to the block makes with the vertical, then the centripetal force required for circular motion is provided by the component of the block’s weight:
mg \cos \theta = \dfrac{m v^2}{R}
Here, v is the speed of the block at that angle, and g is the acceleration due to gravity.
Step 3: Use Conservation of Energy to Relate Speed and Height
Assume the height of the top of the hemisphere (where the block starts) is H_{\text{top}} above the ground. Since the radius is R=3\,\text{m} , the top is 3\,\text{m} above the center, and the center itself is 3\,\text{m} above the ground (making the top 6\,\text{m} above the ground). However, it is easier first to measure changes in potential energy from the center or from the top. Let us do it from the center for simplicity:
At the top, the block is at a distance R above the center.
At angle \theta , the block is at vertical distance R\cos\theta above the center.
The loss in potential energy from top ( R above center) to the angle \theta is
mg\,(R - R \cos\theta) = mg\,R\,\bigl(1 - \cos\theta\bigr).
This converts into kinetic energy:
\tfrac{1}{2}\,m\,v^2 = mg\,R\,\bigl(1 - \cos\theta\bigr).
Hence,
v^2 = 2\,g\,R\,\bigl(1 - \cos\theta\bigr).
Step 4: Apply the Contact-Loss (Normal = 0) Condition
Set the expression from the centripetal condition equal to the expression from energy conservation:
1) v^2 = g\,R\,\cos\theta,
2) v^2 = 2\,g\,R\,(1 - \cos\theta).
Equating these:
g\,R\,\cos\theta = 2\,g\,R\,[1 - \cos\theta].
Cancel out common factors g\,R (assuming g>0 and R>0 ):
\cos\theta = 2\,[1 - \cos\theta].
\cos\theta = 2 - 2\,\cos\theta.
3\,\cos\theta = 2 \quad\Longrightarrow\quad \cos\theta = \tfrac{2}{3}.
Step 5: Find the Vertical Height above the Ground
The vertical distance of the block from the center of the sphere at this point is
R \cos\theta = R \times \tfrac{2}{3} = \tfrac{2R}{3}.
Since R=3\,\text{m} , this distance above the center is 2\,\text{m} . The center of the hemisphere is itself 3\,\text{m} above the ground (because the hemisphere, of radius 3\,\text{m} , rests on the ground). Therefore, the height of the block above the ground at the instant it loses contact is
3\,\text{m} \;(\text{center above ground}) \;+\; 2\,\text{m} \;=\; 5\,\text{m}.
Step 6: Interpret the Question’s “Height h”
In many standard formulations of this problem, “the height h at which the block loses contact” refers to its vertical coordinate measured from the ground. From our calculation, at the moment of losing contact, the block stands 5\,\text{m} above the ground. However, some presentations of this problem define “h” as the distance below the top of the hemisphere, in which case h would be 1\,\text{m} (because the top is at 6\,\text{m} from the ground, and the block is at 5\,\text{m} ).
Because the question’s figure and standard convention usually take the base of the hemisphere on the ground and ask for the height above that ground, the direct numeric answer is:
\boxed{2\,\text{m} \text{ (if the hemisphere’s center is taken as the base)} \quad\text{or}\quad 5\,\text{m} \text{ (if the ground is the base).}}
Most commonly, for a “solid hemisphere” resting on the ground with radius 3\,\text{m} , the center is 3\,\text{m} from the ground, and the top is 6\,\text{m} from the ground. The point of separation is 2\,\text{m} above the center, i.e., at 5\,\text{m} above the ground, and hence 1\,\text{m} below the top. In many textbook versions, the final simpler quoted number is often “2 m from the top” or “1 m below the top” depending on the precise definition of h.
Final Note on the Most Common Answer Convention
If the hemisphere’s bottom is taken as “height = 0” and the top is at R = 3\,\text{m} , the block loses contact at height 2\,\text{m} above the bottom. That concise result is typically stated as the final numeric answer.
