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).
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Given,
- Diameter (2R) = 360 cm = 3.6 m
- Radius (R) = 1.8 m
- Mass of the sphere (m) = 500 g = 0.5 kg
- AE = $\frac {\text R}{3} = \frac {1.8}{3}$ = 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 UA at A relative to B is given by
UA = m x g x R
Substituting the values we get,
UA = 0.5 x 10 x 1.8
⇒ UA = 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 'vB'.
Then,
Kinetic energy at B = UA
$$\begin{aligned}\Rightarrow \frac{1}{2} \text m {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow \frac{1}{2} \times 0.5 \times {\text v_ \text B}^2 = 9 \\[1em] \Rightarrow {\text v_ \text B}^2 = \frac{2\times 9}{0.5} = 36 \\[1em] \Rightarrow \text v_ \text B = \sqrt{36} = 6 \text { m s}^{-1} \\[1em]\end{aligned}$$
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.