Physics · Laws Of Motion · NEET
It is NOT in equilibrium. The bob moves in a horizontal circle at constant speed, so it has a centripetal acceleration a = v²/r directed toward the centre. That is why the horizontal forces do NOT cancel — the net inward force Tsinθ is exactly what keeps the bob turning. Only the vertical direction is balanced (Tcosθ = mg), because the bob does not rise or fall. Treat vertical as balanced and horizontal as Newton's second law (Tsinθ = mv²/r).
Because the string is slanted. The vertical part of the tension, Tcosθ, must equal mg to hold the bob up. Since cosθ is less than 1, T = mg/cosθ must be larger than mg. The steeper the cone (bigger θ), the smaller cosθ, so the bigger the tension. This is why a fast-whirled bob can snap a string that easily holds its static weight.
No. The time period T_p = 2π√(L cosθ/g) has no mass m in it. Mass cancels out because both the centripetal force and gravity scale with m. A heavy bob and a light bob whirled at the same angle θ take the same time per revolution. Mass only affects the tension (T = mg/cosθ), not the period.
θ is measured from the vertical — the angle between the string and the vertical line through the fixed point. So Tcosθ is the vertical (upward) part and Tsinθ is the horizontal (inward) part. A common slip is measuring θ from the horizontal and swapping sin and cos. Always draw the string coming down from the top point and mark θ against the vertical.
As the bob is whirled faster it flies outward, so θ grows and the string becomes more horizontal. Then cosθ shrinks, so T_p = 2π√(L cosθ/g) shrinks — each loop is completed faster. In the limit θ → 90° the string is horizontal, cosθ → 0, and the period → 0, which needs infinite speed and tension (physically impossible).
A bob is whirled in a horizontal circle by a string at an initial angular speed ω, and the tension in the string is T. If the angular speed becomes 2ω at the same radius, the tension becomes:
One end of a string of length l is connected to a particle of mass m and the other end to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, the net force on the particle (directed towards the centre) is (T = tension in the string):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T_p = 2π√(L cosθ/g), where L is the string length, θ is the angle from the vertical, and g is gravity. Equivalently T_p = 2π√(h/g), where h = L cosθ is the vertical height of the fixed point above the plane of the circle.
T = mg/cosθ. It is always greater than the weight mg because the string is slanted. You can also write T = m√(g² + (v²/r)²) using the two force components, or T = mrω²/sinθ.
The horizontal component of the tension, Tsinθ, points toward the centre and supplies the centripetal force: Tsinθ = mv²/r = mrω². The vertical component Tcosθ balances the weight.
No. Mass cancels out in the derivation, so T_p = 2π√(L cosθ/g) is independent of mass. Only the tension depends on mass.
A simple pendulum swings back and forth in a vertical plane; a conical pendulum sweeps a horizontal circle so the string traces a cone. Both give T_p = 2π√(effective length/g), but the conical pendulum's effective length is the height L cosθ, not the full length L.
