Q.One end of a string of length $l$ is connected to a particle of mass $m$ and the other 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:

Showing the 12 most recent of 47 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A bike race is conducted on a circular track of radius 60 m. If the slowest bike, weighing 120 kg, moved around the track at a constant speed of 108 kmph, then the time taken by it to complete one lap and its acceleration are respectively (A) 3.49 s, $1.11\ ms^{-2}$ (B) 12.56 s, $15\ ms^{-2}$ (C) 3.49 s, $0\ ms^{-2}$ (D) 2 s, $9.8\ ms^{-2}$

   ›Reveal solutionSolution

   Constant-speed circular motion still has centripetal acceleration; period and $a=v^2/r$ give 12.56 s and 15 m/s².

   Concept and Intuition

   Uniform circular motion has zero tangential acceleration (speed is constant) but a nonzero centripetal acceleration directed toward the center, since the velocity's direction is always changing. A common trap is to think "constant speed" means "zero acceleration" — it doesn't, for circular paths.

   Step-by-Step Solution

   1. Convert speed: $108\ kmph=108\times\dfrac{5}{18}=30\ ms^{-1}$.
   2. Circumference $=2\pi r=2\pi(60)=120\pi$ m.
   3. Time for one lap $=\dfrac{120\pi}{30}=4\pi\approx12.566$ s. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A stone of mass 400 g tied to one end of a string is rotated in a horizontal circle of radius 125 m. If the string can withstand a maximum tension of $50\pi^2\ N$, then the minimum time period with which the stone can be rotated is (A) 3 s (B) 2 s (C) 4 s (D) 6 s

   ›Reveal solutionSolution

   Setting the string's maximum tension equal to the required centripetal force gives the maximum angular speed, hence the minimum period, of 2 s.

   Concept and Intuition

   A shorter period means faster rotation, which needs more centripetal force (tension). So the minimum period corresponds exactly to the maximum tension the string can bear — beyond that speed, the string snaps.

   Step-by-Step Solution

   1. Centripetal force needed: $T=m\omega^2r$.
   2. Maximum allowed: $T_{max}=50\pi^2$ N $=m\omega_{max}^2r$.
   3. $\omega_{max}^2=\dfrac{50\pi^2}{mr}=\dfrac{50\pi^2}{(0.4)(125)}=\dfrac{50\pi^2}{50}=\pi^2$.
   4. $\omega_{max}=\pi\ rad/s$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A small bead is placed on a thin circular loop of radius 25 cm which is rotating about its vertical diameter with a constant velocity of $10\ rad\,s^{-1}$. The angle made by the radius vector joining the center of the loop to the bead makes with the vertically downward direction is (Neglect friction and take acceleration due to gravity $= 10\ ms^{-2}$) (A) $\cos^{-1}(0.4)$ (B) $\cos^{-1}(0.6)$ (C) $\sin^{-1}(0.4)$ (D) $\sin^{-1}(0.6)$

   ›Reveal solutionSolution

   The bead's equilibrium position on the rotating loop satisfies $\cos\theta=g/(R\omega^2)$; plugging in the given numbers gives $\cos\theta=0.4$, so $\theta=\cos^{-1}(0.4)$.

   Concept and Intuition

   A bead free to slide (frictionless) on a circular wire loop that spins about its vertical diameter settles where the net of gravity and the loop's normal reaction supplies exactly the centripetal force needed for its circular path (whose radius is $R\sin\theta$, since the bead moves in a horizontal circle of that radius as the loop spins). Resolving forces along and perpendicular to the wire at the bead's position yields the classic $\cos\theta=g/(R\omega^2)$ relation.

   Step-by-Step Solution

   1. Let the bead sit at angle $\theta$ from the downward vertical. Its horizontal circular path (due to the loop's rotation) has radius $r = R\sin\theta$.
   2. The forces on the bead are gravity $mg$ (downward) and the normal reaction $N$ from the wire, directed along the radius of the loop (toward or away from the center $O$).
   3. Resolve $N$ into vertical and horizontal components. Balancing forces:
      • Vertical: $N\cos\theta = mg$
      • Horizontal (centripetal, toward the vertical axis): $N\sin\theta = m\omega^2 R\sin\theta$
   4. From the horizontal equation (cancel $\sin\theta$, assuming $\theta\neq0$): $N = m\omega^2 R$.
   5. Substitute into the vertical equation: $m\omega^2R\cos\theta = mg \Rightarrow \cos\theta = \dfrac{g}{\omega^2R}$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A particle moves in horizontal circle. If the speed is doubled, centripetal force becomes (A) Same (B) Double (C) Four times (D) Half

   ›Reveal solutionSolution

   This tests the $v^2$ dependence in the centripetal force formula. The answer is (C), four times.

   Concept and Intuition

   Centripetal force is what continuously redirects a particle's velocity to keep it on a circular path. Its formula, $F=\dfrac{mv^2}{r}$, shows it depends on the square of the speed — so any change in speed gets amplified quadratically in the force.

   Step-by-Step Solution

   1. Centripetal force: $F=\dfrac{mv^2}{r}$, with $m$ and $r$ (radius) unchanged.
   2. If speed doubles, $v\to2v$: $F'=\dfrac{m(2v)^2}{r}=\dfrac{4mv^2}{r}=4F$.
   3. Hence the new centripetal force is four times the original.

   Common Mistakes …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A body of mass 1 kg is attached to one end of a string of 1 m length. It is rotated in a vertical circle with a constant speed of $4\ ms^{-1}$. When the object is at the highest point of the vertical circle, tension in the string is ($g=10\ ms^{-2}$) (A) 6 N (B) 8 N (C) 10 N (D) 16 N

   ›Reveal solutionSolution

   At the top of a vertical circle, both the string tension and gravity point toward the centre; their sum supplies the required centripetal force $mv^2/r$. Solving for $T$ gives 6 N.

   Concept and Intuition

   Unlike the bottom of the loop (where tension must overcome gravity to still provide the net centripetal force), at the top of the loop both gravity and tension act in the same direction — toward the centre of the circle (straight down at that point). So they add together to produce the centripetal force, and tension is whatever is left after gravity's contribution: $T = \frac{mv^2}{r} - mg$. Since speed here is given as constant (this is a driven, constant-speed circular motion, not a free swing purely under gravity), we can directly plug in $v=4\,ms^{-1}$ without worrying about energy conservation around the loop.

   Step-by-Step Solution

   1. At the highest point, both $T$ (string pulls the bob toward the centre, i.e., downward here) and $mg$ (downward) point toward the centre.
   2. Newton's second law along the radial direction: $T + mg = \dfrac{mv^2}{r}$. …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.An artificial satellite of mass $m$ revolves around a planet in a circular orbit of radius $R$ under the influence of an attractive central force given by $F \propto R^{-5/2}$. How does the orbital velocity ($V_e$) and time period ($T$) depend on the orbital radius ($R$)? (A) $V_e \propto R^{-3/4}, \ T \propto R^{7/4}$ (B) $V_e \propto R^{-5/4}, \ T \propto R^{3/2}$ (C) $V_e \propto R^{-1/2}, \ T \propto R^{5/4}$ (D) $V_e \propto R^{-3/4}, \ T \propto R^{-7/4}$

   ›Reveal solutionSolution

   This tests deriving orbital velocity and period scaling from a non-inverse-square central force law using $F=mV^2/R$. Answer: $V\propto R^{-3/4}$, $T\propto R^{7/4}$.

   Concept and Intuition

   For any central attractive force providing circular motion, the force supplies the centripetal requirement $mV^2/R$. Given how $F$ scales with $R$, we can directly deduce how $V$ (and hence $T$) scales with $R$ — this generalizes Kepler's laws beyond the inverse-square case.

   Step-by-Step Solution

   1. Let $F = kR^{-5/2}$ for some constant $k$.
   2. Centripetal force condition: $\dfrac{mV^2}{R} = kR^{-5/2}$.
   3. Solve for $V^2$: $V^2 = \dfrac{k}{m}R^{1}\cdot R^{-5/2} = \dfrac{k}{m}R^{-3/2}$.
   4. So $V \propto R^{-3/4}$.
   5. Time period: $T = \dfrac{2\pi R}{V} \propto \dfrac{R}{R^{-3/4}} = R^{1+3/4} = R^{7/4}$. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A wire of length 2.5 m is fixed at one end and a box of mass 4 kg is tied at the other end. If the wire rotates in a horizontal circle about the fixed end with $\dfrac{2}{\pi}$ rotations per second, then the tension in the wire is (A) 16 N (B) 32 N (C) 64 N (D) 160 N

   ›Reveal solutionSolution

   This tests converting rotational frequency to angular velocity and then applying the centripetal-force formula for a horizontally-rotating conical-type wire. Answer: (D).

   Concept and Intuition

   When a mass is whirled in a horizontal circle at the end of a wire/string, the wire's tension is entirely responsible for providing the centripetal force (there's no other horizontal force). So $T=m\omega^2 r$ directly, once we know $\omega$.

   Step-by-Step Solution

   1. Convert frequency to angular velocity: $f = \dfrac{2}{\pi}\ \text{rev/s}$, so $\omega = 2\pi f = 2\pi\times\dfrac{2}{\pi} = 4\ \text{rad/s}$.
   2. The radius of the circular path equals the wire length, $r=2.5\ \text{m}$ (horizontal circle about the fixed end).
   3. Centripetal force needed $=$ tension in the wire: $T = m\omega^2 r$.
   4. Substitute: $T = 4\ \text{kg}\times(4\ \text{rad/s})^2\times2.5\ \text{m} = 4\times16\times2.5 = 160\ \text{N}$.

   Common Mistakes …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A body of mass 10 kg is attached to one end of a wire of length 0.3 m and area of cross-section $10^{-6}$ m$^2$. If the maximum stress the wire can withstand is $2.7 \times 10^7$ N m$^{-2}$, then the maximum angular velocity with which the wire-body system can be rotated in a horizontal circle about the other end of the wire is (A) 4 rad s$^{-1}$ (B) 5 rad s$^{-1}$ (C) 9 rad s$^{-1}$ (D) 3 rad s$^{-1}$

   ›Reveal solutionSolution

   The wire's breaking tension sets the maximum centripetal force it can supply; equating that to $m\omega^2L$ gives the maximum angular velocity, 3 rad/s.

   Concept and Intuition

   A wire whirled in a horizontal circle supplies the centripetal force needed to keep the attached mass moving in a circle. That force cannot exceed what the wire can withstand before it snaps, which is governed by the maximum stress (force per unit area) the material tolerates.

   Step-by-Step Solution

   1. Maximum tension the wire can bear: $T_{max} = \sigma_{max}\times A = 2.7\times10^7 \times 10^{-6} = 27$ N.
   2. This tension provides the centripetal force for circular motion of radius equal to the wire length $L=0.3$ m: $T_{max}=m\omega_{max}^2L$. …

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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If a stone of mass 0.5 kg tied to one end of a wire is whirled in a circular path of radius 2 m with a speed 40 rev/min in a horizontal plane, then the tension in the wire is nearly (A) 14.8 N (B) [AMBIGUOUS] (C) 17.5 N (D) 20.8 N

   ›Reveal solutionSolution

   The wire's tension is the centripetal force $m\omega^2 r$; converting rev/min to rad/s and substituting gives about 17.5 N.

   Concept and Intuition

   For a stone whirled in a horizontal circle, the string tension is the only horizontal force, and it must supply exactly the centripetal force needed to keep the stone moving in its circular path: $T=m\omega^2 r = \dfrac{mv^2}{r}$.

   Step-by-Step Solution

   1. Convert angular speed: $40\ \text{rev/min} = \dfrac{40\times2\pi\,\text{rad}}{60\,\text{s}} = \dfrac{4\pi}{3}\,\text{rad/s} \approx 4.189\,\text{rad/s}$.
   2. Centripetal force (tension): $T = m\omega^2 r$. …

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10. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.A car is moving on a horizontal curved road of radius 50 m. If the friction coefficient between tyres and road is 0.34, the approximate maximum speed of the car will be nearly (A) $3.4\ \mathrm{ms}^{-1}$ (B) $22.4\ \mathrm{ms}^{-1}$ (C) $13\ \mathrm{ms}^{-1}$ (D) $17\ \mathrm{ms}^{-1}$

    ›Reveal solutionSolution

    Friction alone provides the centripetal force on a flat curve; solving $\mu mg = mv^2/R$ gives $v\approx13$ m/s.

    Concept and Intuition

    On an unbanked circular road, friction between tyres and road supplies the centripetal force needed to keep the car on the curve. The maximum speed without skidding occurs when friction is at its maximum (limiting) value, $f_{max}=\mu N = \mu mg$.

    Step-by-Step Solution

    1. Centripetal force needed: $F_c = \dfrac{mv^2}{R}$.
    2. Maximum available friction: $f_{max}=\mu mg$.
    3. Setting $F_c=f_{max}$: $\dfrac{mv^2}{R}=\mu mg \Rightarrow v=\sqrt{\mu g R}$. …

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11. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.A motor car is moving at $40\ \mathrm{ms}^{-1}$ on a circular road of radius 400 m. If its speed is increasing at the rate of $3\ \mathrm{ms}^{-2}$, then its acceleration is (A) $3\ \mathrm{ms}^{-2}$ (B) $2.7\ \mathrm{ms}^{-2}$ (C) $5\ \mathrm{ms}^{-2}$ (D) $3.3\ \mathrm{ms}^{-2}$

    ›Reveal solutionSolution

    Combine the centripetal (radial) and tangential accelerations as perpendicular vectors; the resultant magnitude is 5 m/s².

    Concept and Intuition

    In non-uniform circular motion, total acceleration has two perpendicular components: centripetal (radial, due to changing direction, $v^2/r$) and tangential (due to changing speed, given directly). The net acceleration is their vector sum via Pythagoras since they are perpendicular.

    Step-by-Step Solution

    1. Centripetal acceleration: $a_c=\dfrac{v^2}{r}=\dfrac{40^2}{400}=\dfrac{1600}{400}=4\ \mathrm{ms^{-2}}$.
    2. Tangential acceleration: $a_t=3\ \mathrm{ms^{-2}}$ (given). …

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12. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.A particle of mass 'm' having charge $(-q)$ moves in a circular orbit of radius 'r' around a fixed charge $(+Q)$. The relation between time period T and radius of the orbit 'r' is (A) $r^2 = \dfrac{Qq}{4\pi\varepsilon_0 m}T^3$ (B) $r^3 = \dfrac{Qq}{16\pi^3\varepsilon_0 m}T^2$ (C) $r^3 = \dfrac{Qq}{16\pi^3\varepsilon_0 m}T$ (D) $r = \dfrac{Qq}{16\pi^2 m\varepsilon_0}T^2$

    ›Reveal solutionSolution

    This tests deriving a Kepler-like $r^3\propto T^2$ relation for a charge orbiting a fixed opposite charge, by equating Coulomb's law to the centripetal force requirement — directly analogous to gravitational orbits.

    Concept and Intuition

    This is the electrostatic analogue of planetary orbital motion: instead of gravity providing the centripetal force, the Coulomb attraction between the fixed $+Q$ and orbiting $-q$ does. Setting Coulomb's force equal to $mv^2/r$ (with $v$ expressed via the orbital period $T$) and eliminating $v$ in favor of $T$ produces a relation between orbital radius and period, structurally identical to Kepler's third law.

    Step-by-Step Solution

    1. Coulomb force provides centripetal force: $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r^2}=\dfrac{mv^2}{r}$.
    2. Express speed via period: $v=\dfrac{2\pi r}{T}$, so $v^2=\dfrac{4\pi^2r^2}{T^2}$.
    3. Substitute: $\dfrac{Qq}{4\pi\varepsilon_0 r^2}=\dfrac{m}{r}\cdot\dfrac{4\pi^2r^2}{T^2}=\dfrac{4\pi^2mr}{T^2}$.
    4. Rearranging: $Qq\,T^2 = 4\pi\varepsilon_0\cdot4\pi^2mr^3 = 16\pi^3\varepsilon_0 m r^3$. …

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