A particle of mass m is released from rest at the top of a spherical dome of radius R.
how far below the starting point will the particle leave the surface of the dome?
how i should solve this?
They need to say whether the particle slides or rolls. If it slides, you have to know the friction coefficient. If it rolls, you have to know if it is a solid or hollow sphere.
If it slides without friction, we can do the problem. After descending a vertical distance H, it will have acquired a speed
V = sqrt(2gH)
Let A be the angle than it has cescended, measured from the center of the sphere
It leaves the sphere when the componemt of its weight normal to the sphere, M g cos A, is equal to the centripetal force required to make it follow the circular trajectory. When this happens, the sphere no longer needs to apply a reaction force to the particle to keep it there, and the particle leaves the surface.
So require that
M g cos A = M V^2/R = M *2g H/R
cos A = 2H/R
Geometry also tells you that
H = R (1-cos A)
Therefore
cos A = (1-cos A)
cos A = 1/2
A = 60 degrees
To solve this problem, you can use conservation of energy and consider the different forms of energy involved: gravitational potential energy and kinetic energy.
Step 1: Calculate the potential energy at the starting point.
The potential energy at the starting point is equal to the product of the mass (m), acceleration due to gravity (g), and the height (h) above the bottom point of the dome. Since the particle is released from rest at the top of the dome, the height is equal to the radius of the dome (h = R). Hence, the potential energy at the starting point is given by PE = mgh.
Step 2: Calculate the kinetic energy at the bottom point of the dome.
The kinetic energy at the bottom point of the dome is equal to the potential energy at the starting point (due to conservation of energy). This can be represented as KE = mgh.
Step 3: Equate the potential energy at the starting point and the kinetic energy at the bottom point of the dome.
Setting PE = KE, we have mgh = mgh. Simplifying this equation, the mass (m) and acceleration due to gravity (g) cancel out, leaving h = h. Thus, the particle will leave the surface of the dome at a height equal to the starting point's height, which is R.
Therefore, the particle will leave the surface of the dome a distance equal to the radius of the dome (R) below the starting point.