Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
Answer:
Keeping the speed fixed and decreasing the radius by a factor of 4
Explanation:
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :
We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"
It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,
R' = R/4
New centripetal acceleration will be,
So, the centripetal acceleration of the ball can be increased by a factor of 4.
75 percent (calculated percentage %) of what number equals 27? Answer: 36.
ONEEEEEEEEEEEEEEEEEEEEEEEEE
It becomes a different element
