A small solid sphere, with radius 0.25 cm and mass 0.61 g rolls without slipping on the inside of a large fixed hemisphere with radius 17 cm and a vertical axis of symmetry. The sphere starts at the top from rest.
(a) What is its kinetic energy at the bottom?
See:
http://www.jiskha.com/display.cgi?id=1256606932
To find the kinetic energy of the small solid sphere at the bottom of the hemisphere, we can use the principle of conservation of energy.
The total mechanical energy of the sphere at any point in its motion is given by the sum of its kinetic energy (KE) and potential energy (PE) at that point. At the top of the hemisphere, the sphere has only potential energy, and at the bottom, it has only kinetic energy.
Given:
Radius of the small solid sphere (r) = 0.25 cm = 0.0025 m
Mass of the small solid sphere (m) = 0.61 g = 0.00061 kg
Radius of the large fixed hemisphere (R) = 17 cm = 0.17 m
Now, let's find the potential energy at the top of the hemisphere. The potential energy is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
At the top of the hemisphere, the height (h) is equal to two times the radius of the large hemisphere (R) since the sphere starts at the top.
So, h = 2R = 2 * 0.17 m = 0.34 m
Now, substituting the values into the equation, we have:
PE = mgh = 0.00061 kg * 9.8 m/s^2 * 0.34 m = 0.002 mJ (millijoules)
Since the sphere starts from rest at the top, it has no kinetic energy initially.
So, at the bottom of the hemisphere, all the potential energy is converted into kinetic energy.
Therefore, the kinetic energy at the bottom can be calculated by subtracting the potential energy (0.002 mJ) from the total mechanical energy at the bottom (which is only kinetic energy).
Thus, the kinetic energy at the bottom of the hemisphere is 0.002 mJ.