A small mass m slides without friction along the looped apparatus shown in Fig. 6-39. If the object is to remain on the track, even at the top of the circle (whose radius is r), from what minimum height h must it be released? (Answer in terms of r.)
I can only imagine what the figure is, and my crystal ball is not working.
I assume h is measured from the bottom of the loop, so the PE going into the mass is h-2r measured at the top. So that is equal to KE at the top.
mg(h-2r)=1/2 m v^2
so solve that for v^2
Now, at the top, the force on the track is mg-mv^2/r, and that to stay on the track must be non negative.0<mg-mv^2/r
so put in v^2 from before, and solve for h.
Thanks that helps alot
To determine the minimum height h from which the object must be released to remain on the track at the top of the circle, we need to consider the forces acting on the object at that point.
At the top of the circle, the object is moving in a circular path, so there must be a net inward force pointing towards the center of the circle. This inward force is provided by the gravitational force and the normal force.
1. Gravitational force: The gravitational force always acts downward and can be calculated using the formula Fg = mg, where m is the mass of the object and g is the acceleration due to gravity.
2. Normal force: At the top of the circle, the normal force acts inward and balances the outward centrifugal force. The magnitude of the normal force can be calculated as N = mv^2/r, where v is the velocity of the object at the top of the circle.
For the object to remain on the track, the normal force (N) must not become zero. Therefore, the minimum height h can be determined by finding the height at which the normal force equals zero.
Since the normal force is given by N = mv^2/r, we can set N = 0 and solve for the velocity v:
mv^2/r = 0
Since the mass m cannot be zero, the only way for the equation to hold is if the velocity v is zero. This means that the object must have zero velocity at the top of the circle.
To find the velocity of the object at the top of the circle, we can consider conservation of energy. The potential energy at a height h above the ground is given by mgh, where h is the height and g is the acceleration due to gravity. At the top of the circle, all the potential energy is converted to kinetic energy.
Therefore, at the top of the circle, the potential energy equals the kinetic energy:
mgh = (1/2)mv^2
Simplifying the equation, we have:
gh = (1/2)v^2
Since v = 0 at the top of the circle, we can solve for h:
gh = 0
This equation indicates that the object would remain on the track at the top of the circle regardless of the height from which it is released.
Therefore, there is no specific minimum height h for the object to remain on the track at the top of the circle. It can be released from any height and still remain on the track.
To determine the minimum height h from which the object must be released to stay on the track, we need to consider the gravitational potential energy at the top of the loop.
The key here is that for the object to stay on the track, the gravitational force must provide sufficient centripetal force to keep the object in circular motion.
At the top of the loop, the gravitational force and the normal force combine to provide the centripetal force needed. The normal force acts perpendicular to the surface and doesn't contribute to the centripetal force.
To calculate the gravitational potential energy at the top of the loop, we can equate it to the difference in gravitational potential energy between the minimum height h and the top of the loop.
The gravitational potential energy at a height h is given by U = mgh, where m is the mass, g is the gravitational acceleration, and h is the height.
At the top of the loop, the gravitational potential energy is zero since it is the reference level.
Equating the two gravitational potential energies, we have:
mgh = 0
Solving for h, we find:
h = 0 / (mg)
Since the mass m and the gravitational acceleration g are positive values, we can divide both sides of the equation by m and g to get:
h = 0
Therefore, the minimum height from which the object must be released is h = 0.
In other words, the object must be released from a height just enough for it to reach the top of the loop and maintain contact with the track without any additional height requirement.