Question

A, B, C and D are four points on a hemispherical cup placed inverted on the ground. Diameter BC = 360 cm and AE = `R/3` (R is the radius of the cup). A small spherical mass of 500 g at rest at the point A slides down along the smooth surface of the cup. Assuming there is no loss of energy, calculate its:

1. Potential Energy at A relative to B.
2. Speed at point B (lowest point).
3. Kinetic Energy at D (g = 10 ms⁻²).

Answer 1

Given: Diameter = 360 cm = 3.6 m

Radius (R) = 1.8 m

Mass of the sphere (m) = 500 g = 0.5 kg

AE = `R/3 = 1.8/3`​ = 0.6 m

Acceleration due to gravity (g) = 10 ms⁻²

Condition: No energy loss so energy is conserved.

a. As, point B is on the ground, then

Height of point A relative to point B = radius of hemisphere (R) = 1.8 m

Potential energy U_A at A relative to B is given by,

U_A = m × g × R

Substituting the values we get,

U_A = 0.5 × 10 × 1.8

⇒ U_A = 9 J

Hence, the potential energy at point A relative to point B is 9 J.

b. As there are no energy losses, total mechanical energy is the same at all points of the path due to conservation of mechanical energy and all the potential energy at point A will be converted to kinetic energy at point B.

Let the speed at point B be ‘v_B’.

Then,

Kinetic energy at B = U_A

⇒ `1/2 (mv_B)^2`

⇒ `1/2 xx 0.5 xx (v_B)^2`

⇒ `(v_B)^2 = (2 xx 9)/0.5`

⇒ `(v_B)^2` = 36

⇒ v_B = `sqrt36`

⇒ v_B = 6 ms⁻¹

So, the speed at point B is 6 ms⁻¹.

c. As points D and E are on the same height, at these points their potential and kinetic energies are equal, i.e.,

Potential energy at point D = Potential energy at point E

Kinetic energy at point D = Kinetic energy at point E

Now,

Distance of point D from the ground = R − AE = 1.8 − 0.6 = 1.2 m

Thus, potential energy at point D = mg × 1.2 = 0.5 × 10 × 1.2 = 6 J

Mechanical energy at point D = Potential energy at point D + Kinetic energy at point D

As, mechanical energy is conserved,

Mechanical energy at point D = Potential energy at point A

⇒ Potential energy at point D + Kinetic energy at point D = 9

⇒ 6 + Kinetic energy at point D = 9

⇒ Kinetic energy at point D = 9 − 6 = 3 J

So, kinetic energy at point D is 3 J.