the small body of mass m slides on a frictionless loop the height above the bottom path of the circular loop is three times the radius of the circular loop.DETERMINE THE VELOCITY OF THE MASS BEFORE IT REACHES TH BOTTOM PATH OF THE LOOP
This picture doesn't register with me. he is sliding on a loop whose height is greater than 3 radii?
To determine the velocity of the mass before it reaches the bottom path of the loop, we can use the principle of conservation of energy.
First, let's find the height of the circular loop in terms of the radius. It is given that the height above the bottom path of the circular loop is three times the radius. So, the height (h) can be expressed as:
h = 3r
The total mechanical energy of the system is conserved, which means that the initial mechanical energy before the mass reaches the bottom path of the loop is equal to the final mechanical energy at that point.
The initial mechanical energy (Ei) is the sum of the kinetic energy (KE) and potential energy (PE) at the initial position:
Ei = KEi + PEi
Since the body starts from rest at the top of the loop, the initial kinetic energy (KEi) is zero. The only energy at the initial position is the potential energy (PEi), which is equal to the gravitational potential energy:
PEi = mgh
where m is the mass of the body, g is the acceleration due to gravity, and h is the height.
The final mechanical energy (Ef) at the bottom path of the loop is the sum of the kinetic energy (KEf) and potential energy (PEf):
Ef = KEf + PEf
At the bottom path of the loop, the potential energy is zero since the height is zero. Therefore:
Ef = KEf
The kinetic energy (KE) can be expressed as:
KE = (1/2)mv^2
where v is the velocity of the body.
Using the conservation of energy principle:
Ei = Ef
mgh = (1/2)mv^2
Canceling the mass 'm' on both sides:
gh = (1/2)v^2
Substituting the height 'h' with 3r:
3gr = (1/2)v^2
Simplifying the equation:
v^2 = 6gr
Taking the square root:
v = √(6gr)
Therefore, the velocity of the mass before it reaches the bottom path of the loop is √(6gr).