Physics · Laws Of Motion · NEET
Centripetal force always points toward the centre. At the bottom, the centre is UP, so tension (up) must overcome gravity (down) AND supply the centripetal pull: T_bottom = mg + mv²/r. At the top, the centre is DOWN, so gravity (down) already helps pull inward, and the string only supplies the rest: T_top = mv²/r − mg. Bottom adds mg, top subtracts it, so bottom tension is larger.
Write Newton's second law along the string toward the centre. At the bottom (centre above): T − mg = mv²/r, so T = mg + mv²/r. At the top (centre below): T + mg = mv²/r, so T = mv²/r − mg. The sign of mg flips because gravity points away from the centre at the bottom and toward the centre at the top.
No. In a real vertical circle the speed is NOT constant — this is non-uniform circular motion. The body slows down going up (gravity does negative work) and speeds up coming down. Using energy conservation between top and bottom (height difference 2r): v_bottom² = v_top² + 4gr. So v_bottom is always larger than v_top.
Subtract the two formulas: T_bottom − T_top = (mg + mv_b²/r) − (mv_t²/r − mg) = 2mg + m(v_b² − v_t²)/r. Since v_b² − v_t² = 4gr, this becomes 2mg + m(4gr)/r = 2mg + 4mg = 6mg. So the tension at the bottom exceeds the tension at the top by exactly 6mg (for a string that just completes the loop, and 6mg in general when energy is conserved).
At the lowest point. Tension is greatest there because T_bottom = mg + mv²/r and the speed is also maximum at the bottom. So the wire or string snaps at the lowest point of the vertical circle — this is a direct NEET 2019 question.
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T_bottom = mg + mv²/r, where m is mass, v is the speed at the bottom, r is the radius and g is gravity. Tension and centripetal force both point up (toward the centre), so weight is added.
T_top = mv²/r − mg, where v is the speed at the top. Since gravity already points toward the centre (downward) at the top, the string supplies only the remaining inward force.
Zero. When the string is about to go slack, T_top = 0, which gives the minimum speed at the top v_top = √(gr). Below this speed the string cannot stay taut and the body cannot complete the circle. This links to the next topic: minimum speed to complete a vertical loop.
When energy is conserved, T_bottom − T_top = 6mg. This comes from 2mg (the two weight terms) plus m(v_b² − v_t²)/r = 4mg (from the height difference 2r). It is a common one-line NEET result.
No. The speed changes with height because gravity does work, so it is non-uniform circular motion. Only horizontal circular motion (like a conical pendulum on a level circle) keeps constant speed.
