Physics · Work, Energy And Power · NEET
At the topmost point, both gravity (mg, downward) and the string/track tension (T, also downward toward the centre) point to the centre. The net of these is the centripetal force: T + mg = m v_top^2 / R. The slowest the body can go is when the string just goes slack, so T = 0. Then mg = m v_top^2 / R, which gives v_top^2 = gR, so v_top = sqrt(gR). If speed were less than this, gravity would be more than the needed centripetal force and the body would fall off the circular path before reaching the top.
Use conservation of mechanical energy between the lowest point and the top. The top is a height 2R above the bottom (a full diameter). So (1/2) m v_bottom^2 = (1/2) m v_top^2 + mg(2R). Put v_top^2 = gR: (1/2) v_bottom^2 = (1/2)(gR) + 2gR = (5/2)gR. Multiply by 2: v_bottom^2 = 5gR, so v_bottom = sqrt(5gR). No friction is assumed, so all kinetic energy lost going up becomes potential energy.
For a string or the inside of a smooth track it is the same: the support can only push/pull toward the centre, so the critical condition at the top is T = 0, giving v_top = sqrt(gR) and v_bottom = sqrt(5gR). A rigid rod is different: a rod can push outward, so the ball can be momentarily at rest at the top (v_top = 0). For a rod the minimum bottom speed is v_bottom = sqrt(4gR) = 2 sqrt(gR). NEET questions usually mean a string or loop, so use sqrt(5gR) unless a rod is stated.
Tension is maximum at the lowest point because both the speed is largest there and gravity now points away from the centre, so the string must supply the full centripetal force plus support the weight: T_bottom = mg + m v_bottom^2 / R. At the top T_top = m v_top^2 / R - mg. For the minimum-speed case, T_bottom = mg + m(5gR)/R = 6mg and T_top = 0. So the difference in tension between bottom and top is always 6mg.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R at the lowest point so that it can complete the loop?
A body initially at rest, sliding along a frictionless track from a height h, just completes a vertical circle of diameter AB = D. The height h is equal to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
At the top it is v_top = sqrt(gR), and at the lowest point it is v_bottom = sqrt(5gR), where R is the radius and g is gravitational acceleration. These come from setting tension = 0 at the top and then using energy conservation.
sqrt(gR) is the minimum speed needed only at the top. To reach the top with that speed, the body must be faster at the bottom because it has to climb a height of 2R. Energy conservation adds 4gR to the speed-squared, giving v_bottom^2 = gR + 4gR = 5gR.
A rigid rod can push outward, so the body can be momentarily at rest at the top (v_top = 0). Then the minimum bottom speed is v_bottom = sqrt(4gR) = 2 sqrt(gR), not sqrt(5gR).
Tension is maximum at the lowest point, T_bottom = mg + m v_bottom^2 / R, and minimum at the top, T_top = m v_top^2 / R - mg. For the just-completing case the values are 6mg at the bottom and 0 at the top; the difference is always 6mg.
No. The sqrt(5gR) result assumes a smooth (frictionless) path so mechanical energy is conserved. With friction you must subtract the work done against friction from the energy equation, and the required starting speed or height increases.
