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 500 g at rest at the point A, slides down along the smooth surface of the cup. Assuming that there is no loss of energy, calculate its:

(a) Potential Energy at A relative to B.

(b) Speed at the point B (lowest point).

(c) Kinetic Energy at D (g = 10 m s^-2).

Given,

• Diameter (2R) = 360 cm = 3.6 m
• Radius (R) = 1.8 m
• Mass of the sphere (m) = 500 g = 0.5 kg
• AE = $\dfrac {\text R}{3} = \dfrac {1.8}{3}$ = 0.6 m
• Acceleration due to gravity (g) = 10 m s^-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 x g x R

Substituting the values we get,

U_A = 0.5 x 10 x 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 so total mechanical energy is 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, speed at point B be 'v_B'.

Then,

Kinetic energy at B = U_A

$\Rightarrow \dfrac{1}{2} \text m {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow \dfrac{1}{2} \times 0.5 \times {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow {\text v_ \text B}^2 = \dfrac{2\times 9}{0.5} = 36 \\[1em] \Rightarrow \text v_ \text B = \sqrt{36} = 6 \text { m s}^{-1} \\[1em]$

So, speed at point B is 6 m s^-1.

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

Potential energy at point D = Potential energy at point E

And

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 x 1.2 = 0.5 x 10 x 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.