Question

A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is 2 m. (Assume there is no friction between the block and the hemisphere)

Answer 1

2

Answer 2

The problem asks for the height 'h' at which a small block, sliding down from the top of a smooth hemisphere of radius R, will lose contact with the surface. We are given R = 3 m, and the problem states that 'h' is 2 m. We need to verify this value.

1. Define the system and variables: Let the mass of the block be 'm'. The radius of the hemisphere is R. Let the base of the hemisphere be the reference level for potential energy (PE = 0). The block starts from rest at the top of the hemisphere, which is at a height R from the base. Let 'h' be the height from the base at which the block loses contact. Let 'θ' be the angle made by the radius vector from the center of the hemisphere to the block with the vertical, at the point where it loses contact.

From the geometry, the height of the block from the base, 'h', can be expressed in terms of R and θ: h = R cos θ (Equation 1)

2. Apply Conservation of Mechanical Energy: Since there is no friction, mechanical energy is conserved. Initial state (at the top): Potential Energy PE_initial = mgR Kinetic Energy KE_initial = 0 (starts from rest) Total Initial Energy E_initial = mgR

Final state (at height 'h' where contact is lost): Potential Energy PE_final = mgh Kinetic Energy KE_final = (1/2)mv^2 (where 'v' is the speed of the block) Total Final Energy E_final = mgh + (1/2)mv^2

By conservation of energy: E_initial = E_final mgR = mgh + (1/2)mv^2 Divide by 'm': gR = gh + (1/2)v^2 v^2 = 2g(R - h) (Equation 2)

3. Apply Newton's Second Law for Circular Motion: At the point where the block is at angle θ (and height 'h'), the forces acting on it are:

• Gravitational force mg acting vertically downwards.
• Normal force N acting radially outwards from the center of the sphere.

For the block to move in a circular path, the net force towards the center provides the centripetal force mv^2/R. The component of gravity along the radius towards the center is mg cos θ. So, the radial force equation is: mg cos θ - N = mv^2/R (Equation 3)

4. Condition for losing contact: The block loses contact with the surface when the normal force N becomes zero. Setting N = 0 in Equation 3: mg cos θ = mv^2/R v^2 = gR cos θ (Equation 4)

5. Solve for 'h': Now we have two expressions for v^2 (Equation 2 and Equation 4). Equate them: 2g(R - h) = gR cos θ Divide by 'g': 2(R - h) = R cos θ (Equation 5)

Substitute cos θ = h/R from Equation 1 into Equation 5: 2(R - h) = R (h/R) 2(R - h) = h 2R - 2h = h 2R = 3h h = (2/3)R

6. Substitute the given value of R: Given R = 3 m. h = (2/3) * 3 m h = 2 m

The calculated height 'h' at which the block loses contact is 2 m, which matches the value stated in the question. This confirms the interpretation of 'h' as the height from the base of the hemisphere.

The question asks for the height 'h' at which the block will lose contact, and states it is 2m. Our calculation confirms this.