Physics Lab
Class XI/Chapter 5: Laws of Motion/Vertical Circle and Pseudo Forces

Vertical Circle and Pseudo Forces

A stone whirled in a vertical circle does not move uniformly — gravity speeds it up on the way down and slows it on the way up. The string tension also varies dramatically. To analyze accelerating frames cleanly (like inside a turning car) we introduce pseudo forces.

Concept

Vertical circle. A body of mass mm tied to a string of length LL moves in a vertical circle. At a general angle θ\theta from the lowest point, write Newton's second law along the radius (toward the centre):

Tmgcosθ=mv2LT - mg\cos\theta = \frac{mv^2}{L}

(at angle θ\theta measured from the bottom, with cosθ\cos\theta being the component of weight along the string toward the centre when at the bottom and away at the top).

At the top of the circle: Ttop+mg=mvtop2/LT_{\text{top}} + mg = mv_{\text{top}}^2/L. The string can only pull (not push), so Ttop0T_{\text{top}} \geq 0. The minimum speed at the top occurs when Ttop=0T_{\text{top}} = 0:

vtop, min=gLv_{\text{top, min}} = \sqrt{gL}

Minimum speed at the bottom (so the stone barely completes the loop): by energy conservation,

12mvbot2=12mvtop2+mg(2L)\tfrac{1}{2}m v_{\text{bot}}^2 = \tfrac{1}{2}m v_{\text{top}}^2 + mg(2L)

With vtop2=gLv_{\text{top}}^2 = gL,

vbot, min=5gLv_{\text{bot, min}} = \sqrt{5gL}

Tension difference (any complete vertical circle): TbotTtop=6mgT_{\text{bot}} - T_{\text{top}} = 6mg.

Pseudo forces. Newton's laws hold only in inertial frames. In a frame accelerating with acceleration a0\vec{a}_0, to apply F=ma\vec{F}=m\vec{a} we must add a pseudo force Fpseudo=ma0\vec{F}_{\text{pseudo}} = -m\vec{a}_0 on each body. Centrifugal force (in a rotating frame) and the Coriolis force are pseudo forces.

Derivation

Minimum speed at the top. At the top, both gravity and tension point downward (toward the centre):

T+mg=mvtop2LT + mg = \frac{mv_{\text{top}}^2}{L}

For the string to remain taut, T0T \geq 0, hence

mvtop2Lmgvtop2gL\frac{mv_{\text{top}}^2}{L} \geq mg \quad \Longrightarrow \quad v_{\text{top}}^2 \geq gL

Minimum speed at the bottom. Conservation of mechanical energy between bottom and top of the circle (height difference 2L2L):

12mvbot2=12mvtop2+2mgL\tfrac{1}{2}mv_{\text{bot}}^2 = \tfrac{1}{2}mv_{\text{top}}^2 + 2mgL

Set vtop2=gLv_{\text{top}}^2 = gL:

vbot2=gL+4gL=5gLv_{\text{bot}}^2 = gL + 4gL = 5gL

vbot, min=5gL\boxed{v_{\text{bot, min}} = \sqrt{5gL}}

Tension difference. Bottom: Tbotmg=mvbot2/LT_{\text{bot}} - mg = mv_{\text{bot}}^2/L. Top: Ttop+mg=mvtop2/LT_{\text{top}} + mg = mv_{\text{top}}^2/L. Subtract:

TbotTtop=2mg+m(vbot2vtop2)L=2mg+m4gLL=6mgT_{\text{bot}} - T_{\text{top}} = 2mg + \frac{m(v_{\text{bot}}^2 - v_{\text{top}}^2)}{L} = 2mg + \frac{m \cdot 4gL}{L} = 6mg

Worked Example

A 0.5 kg stone is tied to a 1 m string and rotated in a vertical circle. (a) What is the minimum speed at the top so the string stays taut? (b) What is the tension at the bottom in that limiting case? Take g=10m/s2g = 10 \, \text{m/s}^2.

Solution:

(a) vtop, min=gL=10×1=103.16m/sv_{\text{top, min}} = \sqrt{gL} = \sqrt{10 \times 1} = \sqrt{10} \approx 3.16 \, \text{m/s}.

(b) At the bottom, vbot2=5gL=50m2/s2v_{\text{bot}}^2 = 5gL = 50 \, \text{m}^2/\text{s}^2. Tension:

Tbot=mg+mvbot2L=(0.5)(10)+(0.5)(50)1=5+25=30NT_{\text{bot}} = mg + \frac{mv_{\text{bot}}^2}{L} = (0.5)(10) + \frac{(0.5)(50)}{1} = 5 + 25 = 30 \, \text{N}

The tension at the bottom is six times the weight (5 N) plus the top tension (zero) — confirming TbotTtop=6mgT_{\text{bot}} - T_{\text{top}} = 6mg.

Common Confusions

  • Centrifugal force is real in the rotating frame. In an inertial frame it doesn't exist; in a co-rotating frame you must include it to apply F=ma\vec{F}=m\vec{a}.
  • String tension can never be negative. If math says T<0T < 0, the string goes slack and the body follows projectile motion.
  • Energy conservation is the cleanest tool for relating speeds at different points in a vertical circle (since gravity is conservative).
  • Pseudo forces are not Newtonian forces — they have no reaction partner. They are mathematical patches for non-inertial frames.

Key Takeaways

  • Minimum speed at top of vertical circle: vtop=gLv_{\text{top}} = \sqrt{gL}.
  • Minimum speed at bottom: vbot=5gLv_{\text{bot}} = \sqrt{5gL}.
  • Tension difference between bottom and top: 6mg6mg, independent of speed.
  • Pseudo force in a frame with acceleration a0\vec{a}_0 is ma0-m\vec{a}_0.
  • Centrifugal and Coriolis forces are pseudo forces in rotating frames.

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