What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

a) √(gR)
b) √(2gR)
c) √(3gR)
d) √(5gR)

To determine the minimum velocity with which a body of mass mm must enter a vertical loop of radius RR in order to complete the loop, we need to consider the forces acting on the body as it moves along the loop, particularly at the top of the loop.
Key Concepts:
1. Centripetal Force: The force required to keep an object moving in a circular path is provided by the centripetal force, which is given by:
Fc=mv2RF_c = \frac{mv^2}{R}
where mm is the mass of the object, vv is the speed, and RR is the radius of the loop.
2. At the Top of the Loop: For the object to complete the loop, at the top of the loop, the gravitational force mgmg and the centripetal force must balance. The minimum condition for the object to remain in motion along the loop is that the centripetal force required to keep it in the circular path should be provided entirely by the gravitational force at the top.
3. Forces at the Top of the Loop: At the top of the loop, the centripetal force required is:
Fc=mv2RF_c = \frac{mv^2}{R}
And the gravitational force mgmg is acting downward. At the minimum speed, the gravitational force must provide the entire centripetal force. Therefore, at the top of the loop, the following condition must hold:
mv2R=mg\frac{mv^2}{R} = mg
Simplifying this equation:
v2=gRv^2 = gR v=gRv = \sqrt{gR}
Step 1: Minimum Speed at the Top of the Loop
The minimum speed at the top of the loop for the object to just complete the loop (i.e., maintain contact with the track) is v=gRv = \sqrt{gR}.
Step 2: Speed at the Bottom of the Loop
However, the object is not moving with this speed throughout the loop. It is accelerating as it moves upward from the bottom of the loop to the top. At the bottom of the loop, the object has the highest speed due to conservation of mechanical energy (assuming no friction or energy losses).
The total mechanical energy at the bottom of the loop consists of kinetic energy and potential energy:
Etotal=12mv2+0E_{\text{total}} = \frac{1}{2}mv^2 + 0
At the top of the loop, the total energy is the sum of kinetic and potential energies. The potential energy at the top is higher due to the height 2R2R (the diameter of the loop):
Etop=12m(vtop)2+2mgRE_{\text{top}} = \frac{1}{2}m(v_{\text{top}})^2 + 2mgR
By conservation of mechanical energy, the total energy at the bottom of the loop equals the total energy at the top of the loop:
12mv2=12m(vtop)2+2mgR\frac{1}{2}mv^2 = \frac{1}{2}m(v_{\text{top}})^2 + 2mgR
Substitute vtop=gRv_{\text{top}} = \sqrt{gR}:
12mv2=12m(gR)+2mgR\frac{1}{2}mv^2 = \frac{1}{2}m(gR) + 2mgR
Simplifying:
12mv2=12mgR+2mgR\frac{1}{2}mv^2 = \frac{1}{2}mgR + 2mgR 12mv2=52mgR\frac{1}{2}mv^2 = \frac{5}{2}mgR v2=5gRv^2 = 5gR v=5gRv = \sqrt{5gR}
Conclusion:
The minimum velocity with which the body must enter the loop in order to complete it is v=5gRv = \sqrt{5gR}.
Thus, the correct answer is:
d) 5gR\sqrt{5gR}.