A body slides down a curved track which is one quadrant of a circle of radius R. If it starts from rest at the top of the track and there is no friction, find the speed at the bottom of the track is R equals 5 m.
apply energy conservation to get : (10)^1/2
To find the speed of the body at the bottom of the track, we can use the conservation of mechanical energy principle. Since there is no friction, the total mechanical energy of the body will remain constant throughout its motion.
At the top of the track, the body has only potential energy, which can be calculated as mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the top of the track.
As the body slides down the curved track, it loses some potential energy and gains an equal amount of kinetic energy, given by (1/2)mv^2, where v is the velocity of the body at any point on the track.
At the bottom of the track, all the potential energy is converted into kinetic energy, so we can equate the potential energy at the top of the track to the kinetic energy at the bottom of the track:
mgh = (1/2)mv^2
Since the mass of the body cancels out, we can simplify the equation to:
gh = (1/2)v^2
We are given that the radius of the track, R, is 5 m. The height, h, of the top of the track is also equal to R.
Substituting these values into the equation, we have:
g(5) = (1/2)v^2
Rearranging the equation, we can solve for v^2:
v^2 = 10g
Finally, taking the square root of both sides, we get:
v = √(10g)
The value of the acceleration due to gravity, g, is approximately 9.8 m/s². Plugging this value into the equation, we find:
v ≈ √(10 * 9.8)
v ≈ 9.9 m/s
So the speed of the body at the bottom of the track, with a radius of 5 m, is approximately 9.9 m/s.