A particle is at rest at the apex A of a smooth fixed hemisphere whose base is horizontal. The hemisphere has centre O and radius a. The particle is then displaced very slightly from rest and moves on the surface of the hemisphere. At the point P on the surface where angle AOP = ¦Á the particle has speed v. Find an expression for v in terms of a, g and ¦Á.
According to the analysis here, the particle does leave the sphere 1/3 of the way down:
http://www.feynmanlectures.info/solutions/particle_on_sphere_sol_1.pdf
To find an expression for the speed of the particle (v) in terms of a, g, and ¦Á, we can start by analyzing the forces acting on the particle at point P.
At point P, the particle experiences two forces:
1. The gravitational force (mg) acting vertically downwards towards the center of the hemisphere.
2. The normal force (N) exerted by the hemisphere perpendicular to the surface at point P.
Since the particle is moving on the surface of the hemisphere, there is no frictional force acting on it.
By considering the forces acting on the particle, we can write the following equation using Newton's second law:
N - mgcos(¦Á) = mv² / a
Here, cos(¦Á) represents the component of the gravitational force along the surface of the hemisphere. The difference between the normal force and this gravitational force component is what provides the necessary centripetal force for the particle to move in a circular path.
Now, let's solve the equation for v:
N = mgcos(¦Á) + mv² / a
Since the particle is at rest at the apex A initially, N = mg.
mg = mgcos(¦Á) + mv² / a
mg - mgcos(¦Á) = mv² / a
mg(1 - cos(¦Á)) = mv² / a
v² = ag(1 - cos(¦Á))
Taking the square root of both sides, we get:
v = √(ag(1 - cos(¦Á)))
So, the expression for the speed of the particle (v) in terms of a, g, and ¦Á is √(ag(1 - cos(¦Á))).
To find an expression for the speed of the particle in terms of a, g, and ¦Á, we can analyze the forces acting on the particle at point P.
1. First, let's consider the weight of the particle. Since the hemisphere is smooth, there is no friction, and the only vertical force acting on the particle is its weight, which can be represented by mg, where m is the mass of the particle and g is the acceleration due to gravity.
2. At point P, there is also a centripetal force acting on the particle due to its circular motion along the surface of the hemisphere. This force is directed towards the center of the hemisphere and can be represented by mv²/r, where v is the speed of the particle and r is the radius of the circular path it follows.
3. Additionally, there is a normal force acting on the particle at point P, which is perpendicular to the surface of the hemisphere. This force acts as a reaction to the weight of the particle and provides the necessary centripetal force to keep the particle moving in a circular path.
4. By using trigonometry, we can see that the angle between the vertical line and the line from the center of the hemisphere to point P is ¦Á.
Considering the forces acting on the particle, we can set up the following equation:
mg = mv²/r
Now, we need to find the relationship between r and a using the triangle formed by the vertical line, the line from the center of the hemisphere to point P, and the line connecting the apex of the hemisphere to point P. From this triangle, we can see that the length of the line from the center of the hemisphere to the point P is a * sin(¦Á).
So, we can substitute r with a * sin(¦Á) in our equation:
mg = mv²/(a * sin(¦Á))
Simplifying the equation, we get:
g = v²/(a * sin(¦Á))
Rearranging this equation, we can solve for v:
v = √(g * a * sin(¦Á))
Therefore, the expression for the speed of the particle in terms of a, g, and ¦Á is v = √(g * a * sin(¦Á)).
Use conservation of energy. At angle A (measured from vertical), the particles elevation has decreased by a (1 - cosA).
m g (1-cosA) = (1/2) m V^2
V = sqrt[2*(1-cosA)/g]
This assumes that the particle remains in contact with the hemisphere. At some point it may (or may not) be fast enough to leave the surface. That is a separate problem.