A body of mass m moves in a vertical circular loop of radius R. To just complete the loop, what should be the minimum value of v^2 at the lowest point in terms of g and R?

Introduction / Context:
The motion of a body in a vertical circular loop is a classic problem in mechanics. It combines concepts of energy conservation and circular motion. To just complete the loop without losing contact at the top, the body must have a certain minimum speed. This question asks for the minimum value of v^2 at the lowest point expressed in terms of g and R, where R is the radius of the loop.

Given Data / Assumptions:

• Mass of the body is m.
• The body moves in a vertical circle of radius R.
• g is the acceleration due to gravity.
• We are interested in the minimum speed at the lowest point so that the body just maintains contact at the top of the loop.
• Air resistance is neglected and the track is rigid and frictionless.

Concept / Approach:
Two conditions are used. First, at the top of the loop, the normal reaction must not become negative, so the centripetal force requirement is m v_top^2 / R ≥ m g, giving a minimum v_top^2 = g R. Second, mechanical energy is conserved between the bottom and the top of the loop. The loss in kinetic energy equals the gain in potential energy when the body rises by 2 R from bottom to top. Using these two ideas together, we relate v_bottom^2 to v_top^2 and then find the required expression for v_bottom^2, which is the v^2 asked in the question.

Step-by-Step Solution:
Step 1: Apply the condition at the top of the loop. At minimum speed, centripetal force is provided entirely by weight, so m v_top^2 / R = m g. Step 2: Simplify this to obtain v_top^2 = g R. Step 3: Use conservation of mechanical energy between bottom and top. At bottom: kinetic energy is (1/2) m v_bottom^2 and potential energy can be taken as zero reference. At top: kinetic energy is (1/2) m v_top^2 and potential energy is m g (2 R). Step 4: Write energy conservation: (1/2) m v_bottom^2 = (1/2) m v_top^2 + m g (2 R). Step 5: Cancel m from all terms and multiply the entire equation by 2: v_bottom^2 = v_top^2 + 4 g R. Step 6: Substitute v_top^2 = g R from Step 2. Then v_bottom^2 = g R + 4 g R = 5 g R. Step 7: Therefore, to just complete the loop, the minimum value of v^2 at the lowest point must be 5 g R.

Verification / Alternative check:
Many standard textbooks present this result as a key formula: minimum speed at lowest point v_bottom = sqrt(5 g R). Squaring both sides gives v_bottom^2 = 5 g R, which matches what we obtained by detailed energy and force analysis. This consistency with known results is a strong verification of the calculation.

Why Other Options Are Wrong:
0: A zero speed at the bottom obviously would not move the body at all, so the loop would not be completed. gR: This corresponds to v_top^2, the minimum speed squared at the top, not at the bottom. 2gR: This value is too low; using it would not satisfy both energy conservation and the contact condition at the top simultaneously.

Common Pitfalls:
A typical mistake is to forget that the body rises by a height of 2 R from bottom to top, using R instead of 2 R in potential energy. Another error is either ignoring the contact condition at the top or using it incorrectly. Always use both the centripetal force condition at the top and energy conservation between bottom and top to derive the correct expression for v_bottom^2.

Final Answer:
The required minimum value of v^2 at the lowest point is 5gR.