1. Physics
  2. Work, Energy and Power
  3. Vertical Circular Motion

Work, Energy and Power · JEE & NEET Physics

Vertical Circular Motion: notes and previous year questions

Speed and tension around an upright circle, the critical speeds √(gr) and √(5gr), the 6mg rule, slack strings, rods, loop tracks and humps.

Vertical Circular Motion in short

  • In a vertical circle the speed changes: energy makes the body fastest at the bottom and slowest at the top.
  • The string's tension is perpendicular to the motion and does no work.
  • Toward the centre, the net force is always mv²/r.
  • Tension is largest at the bottom, mg + mv²/r, and smallest at the top, mv²/r − mg.

1Why the speed changes

Whirl a ball on a string in an upright circle: gravity slows it on the way up and speeds it up on the way down. The string pulls along its length, at 90° to the motion, so it does no work. Only gravity changes the energy, so the ball is fastest at the bottom and slowest at the top.

Two tools work together. Energy gives the speed: at an angle θ\theta from the bottom, the ball has climbed h=r(1−cos⁡θ)h = r(1 - \cos\theta).

v2=vb2−2gr(1−cos⁡θ)v^2 = v_b^2 - 2gr(1 - \cos\theta)

Newton's second law toward the centre gives the force: the net inward force must be mv2/rmv^2/r.

2Tension at the bottom, the top and between

At the bottom, the tension pulls up toward the centre and gravity pulls down, away from it. At the top, both pull down, toward the centre.

Tbottom=mg+mvb2rT_{\text{bottom}} = mg + \frac{mv_b^2}{r}The largest tension: the string most likely breaks here.
Ttop=mvt2r−mgT_{\text{top}} = \frac{mv_t^2}{r} - mgThe smallest tension.

At an angle θ\theta from the bottom, gravity splits into mgcos⁡θmg\cos\theta outward along the string and mgsin⁡θmg\sin\theta along the path (slowing the ball):

T=mgcos⁡θ+mv2rT = mg\cos\theta + \frac{mv^2}{r}Level with the centre (θ = 90°), T = mv²/r.

Example: a 1 kg ball on a 1.25 m string at 5 m/s at the bottom: T=10+25/1.25=30T = 10 + 25/1.25 = 30 N.

3The critical speeds

A string can only pull, so Ttop≥0T_{\text{top}} \ge 0. At the very least Ttop=0T_{\text{top}} = 0, and gravity alone supplies the pull to the centre: mg=mvt2/rmg = mv_t^2/r.

vtop≥grv_{\text{top}} \ge \sqrt{gr}

Energy from the bottom to the top, 2r2r higher: 12vb2=12vt2+2gr\tfrac{1}{2}v_b^2 = \tfrac{1}{2}v_t^2 + 2gr. With vt2=grv_t^2 = gr, 12vb2=52gr\tfrac{1}{2}v_b^2 = \tfrac{5}{2}gr:

vbottom≥5grv_{\text{bottom}} \ge \sqrt{5gr}

4The 6mg rule

At exactly the critical speed, the tension is 6mg6mg at the bottom, 3mg3mg level with the centre and 0 at the top: T=mg(3+3cos⁡θ)T = mg(3 + 3\cos\theta).

For any speed that completes the circle, subtract the two formulas: Tb−Tt=2mg+m(vb2−vt2)/rT_b - T_t = 2mg + m(v_b^2 - v_t^2)/r, and energy gives vb2−vt2=4grv_b^2 - v_t^2 = 4gr.

Tbottom−Ttop=6mgT_{\text{bottom}} - T_{\text{top}} = 6mg

5Not enough speed

Speed at the bottomWhat happens
less than 2gr\sqrt{2gr}stops below the centre and swings back; the string stays tight
exactly 2gr\sqrt{2gr}just reaches the level of the centre
between 2gr\sqrt{2gr} and 5gr\sqrt{5gr}rises above the centre, the string goes slack, and the ball flies as a projectile
5gr\sqrt{5gr} or morecompletes the circle

Below the centre, gravity's part along the string points outward, so the tension is always positive: the string cannot go slack there. Above the centre, gravity pulls toward the centre, and if that is more than mv2/rmv^2/r the string would need to push, so it goes slack.

cos⁡θ=vb2−2gr3gr\cos\theta = \frac{v_b^2 - 2gr}{3gr}Where the string goes slack, with θ measured from the TOP.
h=vb2+gr3gh = \frac{v_b^2 + gr}{3g}The same point, as a height above the bottom.

6String, rod and loop track

A rigid rod can push as well as pull, so the ball can pass over the top at almost zero speed. It only needs the energy to get there: 12vb2=2gr\tfrac{1}{2}v_b^2 = 2gr.

A car inside a loop track is pushed inward by the track, just as a string pulls, so it follows the string's rules: N+mg=mv2/rN + mg = mv^2/r at the top.

String or loop trackRigid rod
Can it push?noyes
Least speed at the topgr\sqrt{gr}0
Least speed at the bottom5gr\sqrt{5gr}4gr=2gr\sqrt{4gr} = 2\sqrt{gr}
Can it go slack?yesno

A 1 m rod needs 40≈6.32\sqrt{40} \approx 6.32 m/s at the bottom; a 1 m string needs 50≈7.07\sqrt{50} \approx 7.07 m/s.

7Over a hump

At the top of a hump of radius rr, the road can only push up. The net force down toward the centre is mg−N=mv2/rmg - N = mv^2/r:

N=mg−mv2rN = mg - \frac{mv^2}{r}

The faster the car, the lighter it presses. Contact needs N≥0N \ge 0, so here gr\sqrt{gr} is a maximum speed: faster than that, the car leaves the road. For r=20r = 20 m, v≤200≈14.1v \le \sqrt{200} \approx 14.1 m/s.

  • In a dip, the road must push harder: N=mg+mv2/rN = mg + mv^2/r, and you feel heavier.
  • Over a 10 m hump at 5 m/s: N=m(10−2.5)=0.75 mgN = m(10 - 2.5) = 0.75\,mg.

Summary

Key ideas

  • In a vertical circle the speed changes: energy makes the body fastest at the bottom and slowest at the top.
  • The string's tension is perpendicular to the motion and does no work.
  • Toward the centre, the net force is always mv²/r.
  • Tension is largest at the bottom, mg + mv²/r, and smallest at the top, mv²/r − mg.
  • A string needs at least √(gr) at the top and √(5gr) at the bottom.
  • For a full circle, the bottom tension is always 6mg more than the top tension.
  • Below √(2gr) the body swings back; between √(2gr) and √(5gr) the string goes slack above the centre.
  • A rigid rod can push, so it needs only 2√(gr) at the bottom; a loop track behaves like a string.
  • Over a hump the normal force is mg − mv²/r, so √(gr) is the maximum speed for keeping contact.

Every equation

Speed at angle θ from the bottom
v2=vb2−2gr(1−cos⁡θ)v^2 = v_b^2 - 2gr(1 - \cos\theta)
Tension at angle θ
T=mgcos⁡θ+mv2rT = mg\cos\theta + \frac{mv^2}{r}
Tension at the bottom
Tb=mg+mvb2rT_b = mg + \frac{mv_b^2}{r}
Tension at the top
Tt=mvt2r−mgT_t = \frac{mv_t^2}{r} - mg
Least speed at the top (string)
vt=grv_t = \sqrt{gr}
Least speed at the bottom (string)
vb=5grv_b = \sqrt{5gr}
Tension difference
Tb−Tt=6mgT_b - T_t = 6mg
Just reaching the centre level
vb=2grv_b = \sqrt{2gr}
Slack point (θ from the top)
cos⁡θ=vb2−2gr3gr\cos\theta = \frac{v_b^2 - 2gr}{3gr}
Slack height above the bottom
h=vb2+gr3gh = \frac{v_b^2 + gr}{3g}
Least speed at the bottom (rod)
vb=2grv_b = 2\sqrt{gr}
Over a hump
N=mg−mv2rN = mg - \frac{mv^2}{r}
In a dip
N=mg+mv2rN = mg + \frac{mv^2}{r}

Previous year questions with solutions

Real JEE and NEET questions on vertical circular motion. Try each one before you open the solution.

Q1JEE Main 2025One correct option

A particle is released from height SS above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

  1. AS4,3gS2\frac{S}{4},\frac{3\mathrm{gS}}{2}
  2. BS2,3gS2\frac{S}{2},\frac{3\mathrm{gS}}{2}
  3. CS4,3gS2\frac{S}{4},\sqrt{\frac{3\mathrm{gS}}{2}}
  4. DS2,3gS2\frac{S}{2},\sqrt{\frac{3\mathrm{gS}}{2}}
Show answer and solution

Answer: Option C

Split the total mgSmgS into four equal parts: K=3UK = 3U means three of them are kinetic and one is potential. So U=mgh=14mgSU = mgh = \tfrac{1}{4}mgS and h=S4h = \dfrac{S}{4}. The particle has fallen 3S4\dfrac{3S}{4}, so 12mv2=mg×3S4\tfrac{1}{2}mv^{2} = mg \times \dfrac{3S}{4} and v=3gS2v = \sqrt{\dfrac{3gS}{2}}. Options A and B give 3gS2\dfrac{3gS}{2} without the square root — that is v2v^{2}, not a speed, and its units give it away. B and D put the particle at S/2S/2, which is where K=UK = U, not K=3UK = 3U.

Q2NEET 2025One correct option

A bob of heavy mass mm is suspended by a light string of length ll. The bob is given a horizontal velocity v0v_{0} at its lowest point. If the string gets slack at some point PP making an angle θ\theta from the horizontal, above the point of suspension, the ratio of the speed vv of the bob at point PP to its initial speed v0v_{0} is:

  1. A(cos⁡θ2+3sin⁡θ)12{(\frac{\cos \theta }{2+3\sin \theta })}^{\frac{1}{2}}
  2. B(sin⁡θ2+3sin⁡θ)12{(\frac{\sin \theta }{2+3\sin \theta })}^{\frac{1}{2}}
  3. C(sin⁡θ)12(\sin \theta {)}^{\frac{1}{2}}
  4. D(12+3sin⁡θ)12{(\frac{1}{2+3\sin \theta })}^{\frac{1}{2}}
Show answer and solution

Answer: Option B

Slack means T=0T = 0 at P, where the part of the weight pointing at the pivot is mgsin⁡θmg\sin\theta: mgsin⁡θ=mv2lmg\sin\theta = \dfrac{mv^{2}}{l}, so v2=glsin⁡θv^{2} = gl\sin\theta. P is l(1+sin⁡θ)l(1 + \sin\theta) above the starting point, so energy gives v02=v2+2gl(1+sin⁡θ)=glsin⁡θ+2gl+2glsin⁡θ=gl(2+3sin⁡θ)v_{0}^{2} = v^{2} + 2gl(1 + \sin\theta) = gl\sin\theta + 2gl + 2gl\sin\theta = gl(2 + 3\sin\theta). Dividing, v2v02=sin⁡θ2+3sin⁡θ\dfrac{v^{2}}{v_{0}^{2}} = \dfrac{\sin\theta}{2 + 3\sin\theta}, and the ratio of the speeds is its square root. Option D has lost the sin⁡θ\sin\theta on top — it is what writing v2=glv^{2} = gl at P gives, as though P were the top of the circle.

Q3JEE Main 2025One correct option

A bead of mass ' mm ' slides without friction on the wall of a vertical circular hoop of radius ' RR '. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ' RR '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' RR ', would be (spring constant is ' kk ', gg is accleration due to gravity)

  1. A2Rg+kR2m\sqrt{2Rg+\frac{{\mathrm{kR}}^{2}}{ m}}
  2. B3Rg+kR2m\sqrt{3\mathrm{Rg}+\frac{{\mathrm{kR}}^{2}}{ m}}
  3. C2Rg+4kR2m\sqrt{2\mathrm{Rg}+\frac{4{\mathrm{kR}}^{2}}{ m}}
  4. D2gR+kR2m2\sqrt{\mathrm{gR}+\frac{{\mathrm{kR}}^{2}}{ m}}
Show answer and solution

Answer: Option B

At the top the spring spans the whole diameter, 2R2R, so it is stretched by RR and stores 12kR2\tfrac{1}{2}kR^{2}. It is back to its natural length when the bead is a distance RR from the bottom point. That chord of length RR makes an equilateral triangle with the centre, so it subtends 60∘60^{\circ} there, and the bead is R(1−cos⁡60∘)=R2R(1 - \cos 60^{\circ}) = \dfrac{R}{2} above the bottom. So the bead has fallen from 2R2R to R2\dfrac{R}{2}, a drop of 3R2\dfrac{3R}{2}, and the spring has given up all its energy: 12mv2=12kR2+mg×3R2\tfrac{1}{2}mv^{2} = \tfrac{1}{2}kR^{2} + mg \times \dfrac{3R}{2}, so v=3gR+kR2mv = \sqrt{3gR + \dfrac{kR^{2}}{m}}. The hoop pushes along a radius, at right angles to the bead's motion, so it does no work. Option A takes the drop as RR, reading the spring's length as a height.

Practice questions, easy to hard

Three questions from the vertical circular motion practice ladder: one easy, one medium, one hard.

Q4Numerical answer

At the lowest point θ=0\theta = 0 and cos⁡θ=1\cos\theta = 1, so the general equation reads T−mg=mv2rT - mg = \dfrac{mv^{2}}{r}, giving T=mg+mvbot2rT = mg + \dfrac{mv_{\mathrm{bot}}^{2}}{r}. The string has to hold the weight up and bend the path as well, so the tension there is always more than the weight.

A bob of mass 0.5 kg0.5\ \mathrm{kg} on a string of length 1 m1\ \mathrm{m} passes the lowest point at 4 m/s4\ \mathrm{m/s}. What is the tension there, in N\mathrm{N}? (Take g=10 m/s2g = 10\ \mathrm{m/s^{2}})

Show answer and solution

Answer: 13 N

T=mg+mv2r=0.5×10+0.5×161=5+8=13 NT = mg + \dfrac{mv^{2}}{r} = 0.5 \times 10 + \dfrac{0.5 \times 16}{1} = 5 + 8 = 13\ \mathrm{N}. The weight alone would be 5 N5\ \mathrm{N}, so the string is carrying nearly three times that. A string rated to hold a hanging bob can still snap the moment that bob is swung, and the lowest point is where it goes.

Q5One correct option

A bob that is not fast enough for the full circle does not simply stop — it stops being on the circle. From T=mv2r+mgcos⁡θT = \dfrac{mv^{2}}{r} + mg\cos\theta, the tension can only reach zero where cos⁡θ\cos\theta is negative, which means above the horizontal through the centre; below that line the string always stays taut. Write the position as an angle ϕ\phi above that horizontal, so cos⁡θ=−sin⁡ϕ\cos\theta = -\sin\phi.

At the point where the string just goes slack, what is v2v^{2}?

  1. Agrcos⁡ϕgr\cos\phi
  2. B2grsin⁡ϕ2gr\sin\phi
  3. Cgrsin⁡ϕgr\sin\phi
  4. Dgrgr, whatever ϕ\phi is
Show answer and solution

Answer: Option C

Setting T=0T = 0 leaves mv2r=−mgcos⁡θ=mgsin⁡ϕ\dfrac{mv^{2}}{r} = -mg\cos\theta = mg\sin\phi, so v2=grsin⁡ϕv^{2} = gr\sin\phi: the radial part of the weight is doing the whole of the turning on its own. Check the ends — at ϕ=90∘\phi = 90^{\circ} this is the top and v2=grv^{2} = gr, the condition already met, and at ϕ=0\phi = 0 it needs v=0v = 0, which is the bob arriving level with the centre with nothing left. Past that point the string is no longer pulling on anything and the bob leaves the circle as a projectile, so the circular equations stop applying the instant TT hits zero.