Concept Card: Motion in a Vertical Circle & Loop-the-Loop Dynamics
1. Physical Framework & Non-Uniform Circular Motion:
When a particle of mass $m$ is constrained to move along a vertical circular path of radius $R$ under uniform gravity $\vec{g}$, its speed varies continuously with height. Unlike uniform horizontal circular motion, vertical circular motion is fundamentally non-uniform because the tangential component of gravitational force continuously accelerates or decelerates the particle.
Let the lowest point of the circle be reference point $B$ ($\theta = 0^\circ$), and let $\theta$ represent the angular position of the particle measured from the downward vertical. At any general position $\theta$:
- Centripetal (Radial) Acceleration: $a_c = \frac{v^2}{R}$ (directed radially inward toward the center $O$).
- Tangential Acceleration: $a_t = g \sin\theta$ (directed tangentially downward, retarding upward motion and accelerating downward motion).
- Total Acceleration: $a_{\text{net}} = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{R}\right)^2 + (g \sin\theta)^2}$.
2. Governing Equations at an Arbitrary Angle $\theta$
(A) Radial Dynamic Equilibrium:
Resolving forces along the inward radial direction at angular position $\theta$:
$$\Sigma F_{\text{radial}} = T – mg \cos\theta = \frac{m v^2}{R} \implies T(\theta) = mg \cos\theta + \frac{m v^2}{R}$$
(B) Conservation of Mechanical Energy:
Since the tension force $\vec{T}$ is perpendicular to instantaneous displacement $d\vec{r}$ at every point ($\vec{T} \cdot d\vec{r} = 0$), tension does zero work ($W_T = 0$). The only work done is by the conservative gravitational force. Taking the gravitational potential energy at the lowest point as zero ($U_B = 0$ at $\theta = 0^\circ$):
- Height above the lowest point: $h(\theta) = R(1 – \cos\theta)$.
- Initial total mechanical energy at bottom: $E = \frac{1}{2} m v_0^2$.
- Total mechanical energy at angle $\theta$: $E = \frac{1}{2} m v^2 + mg R (1 – \cos\theta)$.
Equating total mechanical energy:
$$\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mg R (1 – \cos\theta) \implies v^2(\theta) = v_0^2 – 2gR(1 – \cos\theta)$$
(C) General Tension Expression:
Substituting $v^2(\theta)$ into the radial force equation:
$$T(\theta) = mg \cos\theta + \frac{m}{R} \left[ v_0^2 – 2gR(1 – \cos\theta) \right] = \frac{m v_0^2}{R} + mg (3\cos\theta – 2)$$
3. Tension & Velocity at Critical Cardinal Positions
- Lowest Point / Bottom ($B$, $\theta = 0^\circ$):
$\cos 0^\circ = 1 \implies v_B = v_0$
$$T_B = \frac{m v_0^2}{R} + mg$$ Tension is strictly maximum at the lowest point. - Horizontal Level ($H$, $\theta = 90^\circ$):
$\cos 90^\circ = 0 \implies v_H = \sqrt{v_0^2 – 2gR}$
$$T_H = \frac{m v_0^2}{R} – 2mg = \frac{m v_H^2}{R}$$ - Highest Point / Top ($T$, $\theta = 180^\circ$):
$\cos 180^\circ = -1 \implies v_T = \sqrt{v_0^2 – 4gR}$
$$T_T = \frac{m v_0^2}{R} – 5mg = \frac{m v_T^2}{R} – mg$$ Tension is strictly minimum at the highest point.
Universal Invariant Differences in Tension:
Regardless of the initial launch speed $v_0$ (as long as the string remains taut throughout):
$$T_B – T_T = \left(\frac{m v_0^2}{R} + mg\right) – \left(\frac{m v_0^2}{R} – 5mg\right) = 6mg$$
$$T_B – T_H = 3mg, \qquad T_H – T_T = 3mg$$
4. Condition for Completing the Vertical Loop (Flexible String)
A flexible string can only pull; it cannot support compressive stress ($T \ge 0$). The string is under greatest risk of becoming slack at the highest point where tension is minimum ($T_T$).
- Limiting Condition at Top: For the particle to cross the top without slacking:
$$T_T \ge 0 \implies \frac{m v_T^2}{R} – mg \ge 0 \implies v_{\text{top}} \ge \sqrt{gR}$$ - Minimum Velocity at Lowest Point:
$$v_0^2 – 4gR = v_T^2 \ge gR \implies v_{\text{bottom}} = v_0 \ge \sqrt{5gR}$$ - Minimum Velocity at Horizontal:
$$v_H = \sqrt{v_0^2 – 2gR} \ge \sqrt{5gR – 2gR} = \sqrt{3gR}$$ - Critical Tensions at Threshold ($v_0 = \sqrt{5gR}$):
$T_B = 6mg, \quad T_H = 3mg, \quad T_T = 0$.
5. Regimes of Motion (String of Length $R$)
Based on the initial horizontal speed $v_0$ imparted at the lowest point:
| Initial Velocity Range | Regime of Motion | Key Physical Characteristics |
|---|---|---|
| $0 < v_0 \le \sqrt{2gR}$ | Pendulum Oscillation | $v$ becomes zero at $\theta_0 \le 90^\circ$ ($\cos\theta_0 = 1 – \frac{v_0^2}{2gR} \ge 0$) while $T > 0$. The bob oscillates symmetrically in the lower semicircle. |
| $\sqrt{2gR} < v_0 < \sqrt{5gR}$ | Leaves Circular Path | The bob crosses the horizontal ($\theta = 90^\circ$), but tension vanishes ($T = 0$) at some angle $90^\circ < \theta_c < 180^\circ$ where $\cos\theta_c = \frac{2gR – v_0^2}{3gR} < 0$. It departs into a parabolic projectile path inside the circle. |
| $v_0 \ge \sqrt{5gR}$ | Full Circular Loop | The string remains taut everywhere ($T \ge 0$), successfully completing the loop-the-loop. |
6. String vs. Light Rigid Rod / Smooth Tubular Track
A frequent JEE trap tests the distinction between a flexible string and a light rigid rod (or a bead inside a smooth vertical tube):
- A rigid rod can sustain both tension (pull) and compression (thrust). It cannot slack.
- Therefore, the condition to complete the loop is merely that velocity does not drop below zero before reaching the top:
$$v_{\text{top}} \ge 0$$ - By conservation of mechanical energy between bottom and top:
$$\frac{1}{2} m v_{0,\text{rod}}^2 = 0 + mg(2R) \implies v_{\text{bottom}} \ge \sqrt{4gR} = 2\sqrt{gR}$$ - At the horizontal level: $v_H \ge \sqrt{4gR – 2gR} = \sqrt{2gR}$.
- At the top, when $v_{\text{top}} = 0$, the normal/rod thrust force supports the weight: $N = mg$ upwards ($T = -mg$).
7. Important Insights & JEE Pitfalls
- Pitfall 1 (Where Does String Slack?): Tension always drops to zero before velocity vanishes when $\sqrt{2gR} < v_0 < \sqrt{5gR}$. Setting $v = 0$ to find the departure point is completely wrong! Always set $T = 0$.
- Pitfall 2 (Subsequent Projectile Motion): When the string slacks at $(\theta_c, v_c)$, the bob has speed $v_c = \sqrt{gR|\cos\theta_c|}$ directed along the tangent (at angle $\theta_c – 90^\circ$ to horizontal). It moves as a free projectile under gravity until the distance from the pivot again equals $R$, at which instant the string suddenly jerks taut.
- Pitfall 3 (Constant Speed Fallacy): Tension difference $T_B – T_T = 6mg$ is independent of $v_0$ ONLY for vertical motion under gravity where speed varies according to energy conservation. If an external motor forces constant speed $v$, then $T_B – T_T = 2mg$.
Solved Examples
Example 1 (Direct Conceptual Application – Tension & Speed Distribution):
A small stone of mass $m = 0.40\text{ kg}$ is tied to a light string of length $R = 0.80\text{ m}$ and rotated in a vertical circle. At the lowest point, the stone is given an initial horizontal velocity $v_0 = 8.0\text{ m/s}$. Taking $g = 10\text{ m/s}^2$:
(a) Determine whether the stone successfully executes a full circular loop.
(b) Calculate the velocity and the tension in the string at the highest point.
(c) Calculate the tension at the horizontal position ($\theta = 90^\circ$).
Solution:
(a) The critical minimum velocity at the lowest point required to complete a full loop is:
$$v_{\text{crit}} = \sqrt{5gR} = \sqrt{5 \times 10 \times 0.80} = \sqrt{40} \approx 6.325\text{ m/s}$$
Since $v_0 = 8.0\text{ m/s} > \sqrt{5gR}$, the stone easily completes the loop without the string ever going slack.
(b) Velocity at the highest point ($h = 2R = 1.60\text{ m}$):
$$v_T^2 = v_0^2 – 4gR = 8.0^2 – 4(10)(0.80) = 64 – 32 = 32\text{ m}^2/\text{s}^2 \implies v_T = \sqrt{32} = 4\sqrt{2} \approx 5.66\text{ m/s}$$
Tension at the highest point:
$$T_T = \frac{m v_T^2}{R} – mg = \frac{0.40 \times 32}{0.80} – 0.40(10) = 16.0 – 4.0 = 12.0\text{ N}$$
(c) At the horizontal position ($\theta = 90^\circ, h = R = 0.80\text{ m}$):
$$v_H^2 = v_0^2 – 2gR = 64 – 2(10)(0.80) = 64 – 16 = 48\text{ m}^2/\text{s}^2$$
At $\theta = 90^\circ$, gravity has zero radial component ($\cos 90^\circ = 0$):
$$T_H = \frac{m v_H^2}{R} = \frac{0.40 \times 48}{0.80} = 24.0\text{ N}$$
Check invariant difference: $T_B = \frac{m v_0^2}{R} + mg = \frac{0.40(64)}{0.80} + 4.0 = 36.0\text{ N}$.
$T_B – T_T = 36.0 – 12.0 = 24.0\text{ N} = 6mg = 6(0.40)(10) = 24.0\text{ N}$.
Example 2 (Mathematical & Tension Invariant Derivation):
A particle of mass $m$ tied to an inextensible string of length $R$ executes vertical circular motion with initial speed at the bottom $v_0 = \sqrt{7gR}$.
(a) Find the angular position $\theta$ (measured from the downward vertical) where the tension in the string is equal to $3mg$.
(b) Find the angle $\theta$ where the total acceleration vector of the particle is directed purely horizontally.
Solution:
(a) The general formula for tension as a function of $\theta$ is:
$$T(\theta) = \frac{m v_0^2}{R} + mg(3\cos\theta – 2)$$
Given $v_0^2 = 7gR$ and setting $T(\theta) = 3mg$:
$$3mg = \frac{m(7gR)}{R} + mg(3\cos\theta – 2) = 7mg + 3mg\cos\theta – 2mg = 5mg + 3mg\cos\theta$$
$$3 = 5 + 3\cos\theta \implies 3\cos\theta = -2 \implies \cos\theta = -\frac{2}{3}$$
$$\theta = \arccos\left(-\frac{2}{3}\right) = 180^\circ – \arccos\left(\frac{2}{3}\right) \approx 131.8^\circ$$
The tension drops to $3mg$ in the upper semicircle at $\cos\theta = -2/3$.
(b) The net acceleration has components:
– Centripetal: $a_c = \frac{v^2}{R}$ directed radially inward toward $O$.
– Tangential: $a_t = g\sin\theta$ directed tangentially downward.
Let us resolve total acceleration along the vertical axis ($y$-axis, upward):
$$a_y = a_c (-\cos\theta) – a_t \sin\theta = -\frac{v^2}{R}\cos\theta – g\sin^2\theta$$
For total acceleration to be purely horizontal, the vertical component must vanish ($a_y = 0$):
$$\frac{v^2}{R}\cos\theta + g\sin^2\theta = 0$$
Substitute $v^2 = v_0^2 – 2gR(1 – \cos\theta) = 7gR – 2gR + 2gR\cos\theta = gR(5 + 2\cos\theta)$:
$$(5 + 2\cos\theta)\cos\theta + (1 – \cos^2\theta) = 0$$
$$5\cos\theta + 2\cos^2\theta + 1 – \cos^2\theta = 0 \implies \cos^2\theta + 5\cos\theta + 1 = 0$$
Using the quadratic formula for $\cos\theta$:
$$\cos\theta = \frac{-5 \pm \sqrt{25 – 4}}{2} = \frac{-5 + \sqrt{21}}{2} \approx \frac{-5 + 4.5826}{2} = -0.2087$$
Thus, $\cos\theta = \frac{\sqrt{21}-5}{2} \implies \theta \approx 102.0^\circ$.
Example 3 (Standard JEE Advanced Scenario – Leaving the Circle):
A small ball of mass $m$ is suspended by a light string of length $L = 1.0\text{ m}$. It is given a horizontal speed $v_0 = \sqrt{3gL}$ at the lowest point. Taking $g = 10\text{ m/s}^2$:
(a) Determine the height $h_c$ above the lowest point and the angle $\theta_c$ from the downward vertical at which the string becomes slack.
(b) Find the velocity $v_c$ of the ball at the moment the string slacks.
(c) Calculate the maximum height $H_{\text{max}}$ above the lowest point reached by the ball during its subsequent projectile flight.
Solution:
(a) Since $\sqrt{2gL} = \sqrt{20} \approx 4.47\text{ m/s} < v_0 = \sqrt{30} \approx 5.48\text{ m/s} < \sqrt{5gL} \approx 7.07\text{ m/s}$, the ball rises into the upper semicircle and leaves the circular path when tension becomes zero ($T = 0$).
Setting $T = 0$ in the radial equation:
$$T = mg\cos\theta_c + \frac{m v_c^2}{L} = 0 \implies v_c^2 = -gL\cos\theta_c = gL|\cos\theta_c|$$
By energy conservation:
$$v_c^2 = v_0^2 – 2gL(1 – \cos\theta_c) = 3gL – 2gL + 2gL\cos\theta_c = gL(1 + 2\cos\theta_c)$$
Equating the two expressions for $v_c^2$:
$$-gL\cos\theta_c = gL(1 + 2\cos\theta_c) \implies -3\cos\theta_c = 1 \implies \cos\theta_c = -\frac{1}{3}$$
$$\theta_c = \arccos\left(-\frac{1}{3}\right) \approx 109.47^\circ$$
Height above lowest point:
$$h_c = L(1 – \cos\theta_c) = L\left(1 – \left(-\frac{1}{3}\right)\right) = \frac{4}{3}L = \frac{4}{3}(1.0) \approx 1.333\text{ m}$$
(b) Velocity at the detachment point:
$$v_c = \sqrt{-gL\cos\theta_c} = \sqrt{gL\left(\frac{1}{3}\right)} = \sqrt{\frac{10 \times 1.0}{3}} = \sqrt{\frac{10}{3}} \approx 1.826\text{ m/s}$$
(c) At the instant of slacking, the velocity $\vec{v}_c$ is tangent to the circle. Since the radius makes angle $\theta_c$ with the downward vertical, the tangent makes an angle $\alpha = \theta_c – 90^\circ$ with the horizontal:
$$\sin\alpha = \sin(\theta_c – 90^\circ) = -\cos\theta_c = +\frac{1}{3}$$
The ball now acts as an oblique projectile launched with speed $v_c$ at elevation angle $\alpha$. The maximum additional vertical height reached during projectile motion is:
$$\Delta h_{\text{proj}} = \frac{v_c^2 \sin^2\alpha}{2g} = \frac{\left(\frac{gL}{3}\right) \times \left(\frac{1}{3}\right)^2}{2g} = \frac{L}{54}$$
Therefore, the maximum height above the lowest point is:
$$H_{\text{max}} = h_c + \Delta h_{\text{proj}} = \frac{4}{3}L + \frac{1}{54}L = \frac{72 + 1}{54}L = \frac{73}{54}L$$
For $L = 1.0\text{ m}$: $H_{\text{max}} = \frac{73}{54} \approx 1.352\text{ m}$.
Example 4 (Edge Case – Bead on Vertical Smooth Ring):
A small bead of mass $m$ is threaded on a smooth, rigid, fixed circular wire loop of radius $R$ in a vertical plane. The bead is projected from the lowest point with speed $v_0 = \sqrt{3gR}$. Taking inward radial normal force as positive ($N > 0$):
(a) Does the bead reach the highest point of the ring?
(b) Find the normal force exerted by the wire on the bead at the lowest point, horizontal level, and at the highest point reached.
(c) At what height $h$ does the normal force change its direction (from pointing inward toward the center to pointing outward)?
Solution:
(a) Unlike a string, a rigid wire ring exerts normal force in both directions (inward or outward). It cannot slack. The minimum velocity to reach the top is $v_{\text{top}} \ge 0 \implies v_{0,\text{min}} = \sqrt{4gR} = 2\sqrt{gR}$.
Since $v_0 = \sqrt{3gR} < \sqrt{4gR}$, the bead does not reach the top. It comes to rest ($v = 0$) at height:
$$mgh_{\text{max}} = \frac{1}{2}mv_0^2 = \frac{3}{2}mgR \implies h_{\text{max}} = 1.5R$$
The angular position at rest is $R(1 – \cos\theta_{\text{stop}}) = 1.5R \implies \cos\theta_{\text{stop}} = -0.5 \implies \theta_{\text{stop}} = 120^\circ$.
(b) Normal force equation with $N$ directed inward:
$$N(\theta) – mg\cos\theta = \frac{mv^2}{R} \implies N(\theta) = mg\cos\theta + \frac{mv^2}{R}$$
Using $v^2 = v_0^2 – 2gR(1 – \cos\theta) = 3gR – 2gR(1 – \cos\theta) = gR(1 + 2\cos\theta)$:
$$N(\theta) = mg\cos\theta + mg(1 + 2\cos\theta) = mg(1 + 3\cos\theta)$$
– At lowest point ($\theta = 0^\circ$): $N_B = mg(1 + 3) = 4mg$ (inward).
– At horizontal level ($\theta = 90^\circ$): $N_H = mg(1 + 0) = mg$ (inward).
– At stopping point ($\theta = 120^\circ, \cos 120^\circ = -0.5$):
$N_{\text{stop}} = mg(1 + 3(-0.5)) = -0.5mg$ (outward, away from center).
(c) Normal force vanishes ($N = 0$) when:
$$1 + 3\cos\theta = 0 \implies \cos\theta = -\frac{1}{3}$$
Height above the bottom:
$$h = R(1 – \cos\theta) = R\left(1 – \left(-\frac{1}{3}\right)\right) = \frac{4}{3}R$$
Below $h = \frac{4}{3}R$, the wire pushes inward ($N > 0$); above $h = \frac{4}{3}R$, the wire pushes radially outward ($N < 0$) to support the bead.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle of mass $m$ is tied to a string of length $L$ and whirled in a vertical circle. If the tension in the string at the highest point is zero, what is the tension in the string when the particle is at the lowest point?
(A) $3mg$
(B) $4mg$
(C) $5mg$
(D) $6mg$
Problem 2 (JEE Main – Single Correct):
A bucket full of water is whirled in a vertical circle of radius $R = 2.5\text{ m}$. What must be the minimum speed of the bucket at the highest point so that water does not spill out? (Take $g = 10\text{ m/s}^2$)
(A) $2.5\text{ m/s}$
(B) $5.0\text{ m/s}$
(C) $7.5\text{ m/s}$
(D) $10.0\text{ m/s}$
Problem 3 (JEE Main – Single Correct):
A body of mass $m$ tied to a light string of length $R$ is projected horizontally from the lowest point with speed $v_0 = \sqrt{2gR}$. The maximum angle $\theta_{\text{max}}$ made by the string with the downward vertical during subsequent motion is:
(A) $45^\circ$
(B) $60^\circ$
(C) $90^\circ$
(D) $120^\circ$
Problem 4 (JEE Main – Single Correct):
A small ball attached to a light rigid rod of length $R$ rotates in a vertical circle. The minimum speed that must be imparted to the ball at the lowest point so that it completes the vertical circle is:
(A) $\sqrt{5gR}$
(B) $\sqrt{4gR}$
(C) $\sqrt{3gR}$
(D) $\sqrt{2gR}$
Problem 5 (JEE Main – Single Correct):
A vehicle of mass $m$ traverses the crest of a convex spherical bridge of radius of curvature $R$ with speed $v$. The normal force exerted by the bridge on the vehicle at the highest point of the bridge is:
(A) $mg + \frac{mv^2}{R}$
(B) $\frac{mv^2}{R}$
(C) $mg – \frac{mv^2}{R}$
(D) $\sqrt{(mg)^2 + \left(\frac{mv^2}{R}\right)^2}$
Problem 6 (JEE Advanced – One or More Correct):
A particle of mass $m$ attached to an inextensible light string of length $L$ executes vertical circular motion with initial speed $v_0$ at the lowest point. Which of the following statements is/are correct?
(A) If $v_0 = \sqrt{5gL}$, the tension at the lowest point is $6mg$.
(B) If $\sqrt{2gL} < v_0 < \sqrt{5gL}$, the string becomes slack before the particle reaches the highest point.
(C) The difference in tension between the lowest and highest points is always $6mg$, provided the string remains taut throughout.
(D) The tangential acceleration of the particle is maximum at the lowest point.
Problem 7 (JEE Advanced – One or More Correct):
A particle of mass $m$ is moving in a vertical circle of radius $R$ attached to a string. At a certain point where the string makes an angle $\theta$ with the downward vertical, the tension in the string is $T$ and the speed is $v$. Which of the following relations is/are universally valid at this instant?
(A) $\vec{T} \cdot \vec{v} = 0$
(B) $T – mg\cos\theta = \frac{mv^2}{R}$
(C) Net power delivered to the particle is $P = -mg v \sin\theta$
(D) The rate of change of speed is $\frac{dv}{dt} = -g\sin\theta$ during upward motion
Problem 8 (JEE Advanced – One or More Correct):
A heavy particle is suspended from a fixed point by a light string of length $L$. It is projected horizontally with speed $u$ from its lowest position such that $\sqrt{2gL} < u < \sqrt{5gL}$. It leaves the circular path at an angle $\theta$ with the downward vertical with velocity $v$. Then:
(A) $\cos\theta = \frac{2gL – u^2}{3gL}$
(B) $v = \sqrt{gL |\cos\theta|}$
(C) The particle will pass through the lowest point of the circle during its subsequent projectile motion if $\cos^2\theta = \frac{1}{3}$
(D) At the detachment point, the acceleration of the particle is equal to $g$
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A stone of mass $m = 0.50\text{ kg}$ is whirled in a vertical circle of radius $R = 2.4\text{ m}$ using a light string. If the ratio of the maximum tension to the minimum tension in the string during a complete revolution is $\frac{T_{\text{max}}}{T_{\text{min}}} = 7.0$, find the initial speed $v_0$ of the stone at the lowest point in $\text{m/s}$. (Take $g = 10\text{ m/s}^2$)
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A small block of mass $m$ is released from rest from a height $h$ on a smooth frictionless track that terminates in a vertical loop of radius $R = 2.0\text{ m}$. What is the minimum height $h_{\text{min}}$ (in meters) from which the block must be released so that it successfully completes the vertical loop without losing contact?
Solutions & Explanations
Answer Key Summary:
1. (D) | 2. (B) | 3. (C) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 12 | 10. 5
Solution 1:
At the highest point, $T_T = \frac{m v_T^2}{L} – mg = 0 \implies v_T^2 = gL$.
By conservation of mechanical energy from bottom to top:
$\frac{1}{2}m v_B^2 = \frac{1}{2}m v_T^2 + mg(2L) = \frac{1}{2}m(gL) + 2mgL = \frac{5}{2}mgL \implies v_B^2 = 5gL$.
Tension at the lowest point is:
$T_B = \frac{m v_B^2}{L} + mg = \frac{m(5gL)}{L} + mg = 5mg + mg = 6mg$.
Correct Option: (D)
Solution 2:
Water will not fall out if the normal force between the bottom of the bucket and the water at the highest point is non-negative ($N \ge 0$).
At the top: $N + mg = \frac{m v^2}{R} \implies N = \frac{m v^2}{R} – mg \ge 0 \implies v \ge \sqrt{gR}$.
$v_{\text{min}} = \sqrt{gR} = \sqrt{10 \times 2.5} = \sqrt{25} = 5.0\text{ m/s}$.
Correct Option: (B)
Solution 3:
The body comes to rest momentarily at maximum angle $\theta_{\text{max}}$ where $v = 0$.
Using energy conservation from bottom ($\theta = 0^\circ$) to $\theta_{\text{max}}$:
$\frac{1}{2}m v_0^2 = mg R (1 – \cos\theta_{\text{max}})$.
Given $v_0^2 = 2gR$:
$\frac{1}{2}m(2gR) = mgR(1 – \cos\theta_{\text{max}}) \implies 1 = 1 – \cos\theta_{\text{max}} \implies \cos\theta_{\text{max}} = 0 \implies \theta_{\text{max}} = 90^\circ$.
At this position, tension is $T = mg\cos 90^\circ + \frac{mv^2}{R} = 0 + 0 = 0$. The body oscillates as a pendulum between $-90^\circ$ and $+90^\circ$.
Correct Option: (C)
Solution 4:
A rigid rod can support compressive forces ($T < 0$); therefore, it does not collapse or slack at the top. The particle only needs to reach the top with non-zero speed ($v_{\text{top}} \ge 0$).
By conservation of mechanical energy:
$\frac{1}{2}m v_0^2 = mg(2R) \implies v_0 = \sqrt{4gR} = 2\sqrt{gR}$.
Correct Option: (B)
Solution 5:
At the crown of the convex bridge, gravity acts vertically downward toward the center of curvature, and the normal reaction $N$ acts upward.
The net inward centripetal force is:
$mg – N = \frac{mv^2}{R} \implies N = mg – \frac{mv^2}{R}$.
(If $v \ge \sqrt{gR}$, $N \le 0$ and the car becomes airborne).
Correct Option: (C)
Solution 6:
– (A) True: $T_B = \frac{m(5gL)}{L} + mg = 6mg$.
– (B) True: For $\sqrt{2gL} < v_0 < \sqrt{5gL}$, tension vanishes in the upper semicircle ($90^\circ < \theta < 180^\circ$) while speed is still non-zero.
– (C) True: $T_B – T_T = \left(\frac{mv_0^2}{L} + mg\right) – \left(\frac{mv_0^2}{L} – 5mg\right) = 6mg$, identically true for any taut motion.
– (D) False: Tangential acceleration is $a_t = g\sin\theta$. At the lowest point ($\theta = 0^\circ$), $\sin 0^\circ = 0 \implies a_t = 0$. It is maximum at $\theta = 90^\circ$.
Correct Options: (A, B, C)
Solution 7:
– (A) True: Tension is strictly perpendicular to velocity at every instant ($\vec{T} \perp \vec{v} \implies \vec{T} \cdot \vec{v} = 0$).
– (B) True: Newton’s second law along the radial direction yields $T – mg\cos\theta = \frac{mv^2}{R}$.
– (C) True: Net power is delivered solely by gravity: $P = \vec{F}_g \cdot \vec{v} = (mg)(v)\cos(90^\circ + \theta) = -mg v \sin\theta$.
– (D) True: $m \frac{dv}{dt} = F_t = -mg\sin\theta \implies \frac{dv}{dt} = -g\sin\theta$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: Setting $T = 0 \implies v^2 = -gL\cos\theta$. Energy conservation: $v^2 = u^2 – 2gL(1 – \cos\theta)$. Equating gives $-gL\cos\theta = u^2 – 2gL + 2gL\cos\theta \implies \cos\theta = \frac{2gL – u^2}{3gL}$.
– (B) True: From the radial equation with $T = 0$, $v = \sqrt{-gL\cos\theta} = \sqrt{gL|\cos\theta|}$.
– (C) True: After departure at point $(L\sin\theta, -L\cos\theta)$ with velocity $(v\cos\theta, v\sin\theta)$ where $v^2 = -gL\cos\theta$, the condition that the parabolic path passes through the lowest point $(0, -L)$ simplifies to $\cos^2\theta = \frac{1}{3}$. This is a classic standard JEE Advanced result.
– (D) True: At $T = 0$, the only force acting on the particle is gravity $\vec{F}_g = m\vec{g}$, so its acceleration is exactly $\vec{g}$.
Correct Options: (A, B, C, D)
Solution 9:
$T_{\text{max}} = T_B = \frac{m v_0^2}{R} + mg$.
$T_{\text{min}} = T_T = \frac{m v_0^2}{R} – 5mg$.
Given $\frac{T_{\text{max}}}{T_{\text{min}}} = 7.0$:
$$\frac{\frac{m v_0^2}{R} + mg}{\frac{m v_0^2}{R} – 5mg} = 7 \implies \frac{v_0^2}{R} + g = 7\left(\frac{v_0^2}{R} – 5g\right)$$
$$\frac{v_0^2}{R} + g = 7\frac{v_0^2}{R} – 35g \implies 6\frac{v_0^2}{R} = 36g \implies v_0^2 = 6gR$$
Given $g = 10\text{ m/s}^2$ and $R = 2.4\text{ m}$:
$$v_0^2 = 6 \times 10 \times 2.4 = 144 \implies v_0 = \sqrt{144} = 12\text{ m/s}$$.
Correct Answer: 12
Solution 10:
For the block to complete the vertical loop without losing contact, its speed at the lowest point of the circular loop must satisfy $v_0 \ge \sqrt{5gR}$.
By conservation of mechanical energy from the release height $h$ to the bottom of the loop ($U = 0$):
$$mgh = \frac{1}{2}mv_0^2 \ge \frac{1}{2}m(5gR) \implies h_{\text{min}} = \frac{5}{2}R = 2.5R$$
Given $R = 2.0\text{ m}$:
$$h_{\text{min}} = 2.5 \times 2.0 = 5.0\text{ meters}$$.
Correct Answer: 5