Question

A particle of mass m is suspended with an inextensible light thread of length l from a fixed-point O. The particle is given a horizontal velocity of $v_0$ when it is in rest vertically below the point of suspension as shown in the figure.

$v = \sqrt{v_0^2-2gl(1-sin\theta)}$

(a) Write expression for speed of the particle and tensile force in the string when the thread makes an angle $\theta$ with the vertical. Assume that the particle remains on the circular path.

(b) What should be the minimum speed at the highest point of the circular path?

(c) What should be the minimum speed $v_0$ to complete circular path?

(d) Find the range of speed $v_0$ so that the particle will oscillate.

(e) Find the range of speed $v_0$ so that the particle will remain always on the circular path.

(f) What will happen, if $\sqrt{2gl} < v_0 < \sqrt{5gl}$?

(g) Where will the particle leave the circular path, if $\sqrt{2gl} < v_0 < \sqrt{5gl}$?

Answer 1

The problem deals with the vertical circular motion of a particle suspended by an inextensible light thread. We will use the principles of conservation of mechanical energy and Newton's second law for circular motion.

Let $m$ be the mass of the particle, $l$ be the length of the thread, and $v_0$ be the initial horizontal velocity at the lowest point. The angle $\theta$ is measured from the vertically downward direction.

(a) Expression for speed of the particle and tensile force in the string when the thread makes an angle $\theta$ with the vertical.

Speed of the particle ($v$): We apply the principle of conservation of mechanical energy. Let the lowest point be the reference for potential energy ($PE=0$).

• Initial state (at the lowest point, $\theta = 0^\circ$): Potential Energy ($PE_i$) = $0$ Kinetic Energy ($KE_i$) = $\frac{1}{2}mv_0^2$ Total Initial Energy ($E_i$) = $\frac{1}{2}mv_0^2$
• Final state (at angle $\theta$): The vertical height gained by the particle is $h = l - l\cos\theta = l(1-\cos\theta)$. Potential Energy ($PE_f$) = $mgh = mgl(1-\cos\theta)$ Kinetic Energy ($KE_f$) = $\frac{1}{2}mv^2$ Total Final Energy ($E_f$) = $\frac{1}{2}mv^2 + mgl(1-\cos\theta)$

By conservation of mechanical energy, $E_i = E_f$: $\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgl(1-\cos\theta)$ Multiply by $\frac{2}{m}$: $v_0^2 = v^2 + 2gl(1-\cos\theta)$ Solving for $v$: $v^2 = v_0^2 - 2gl(1-\cos\theta)$ $v = \sqrt{v_0^2 - 2gl(1-\cos\theta)}$ (Note: The expression provided in the question, $v = \sqrt{v_0^2-2gl(1-sin\theta)}$, contains a typo. It should be $\cos\theta$ as derived above).

Tensile force in the string ($T$): We apply Newton's second law in the radial direction (towards the center of the circular path, O). The forces acting along the string are the tension $T$ (inwards) and the component of gravity $mg\cos\theta$ (outwards). The net radial force provides the centripetal force: $T - mg\cos\theta = \frac{mv^2}{l}$ $T = mg\cos\theta + \frac{mv^2}{l}$ Substitute the expression for $v^2$: $T = mg\cos\theta + \frac{m}{l} [v_0^2 - 2gl(1-\cos\theta)]$ $T = mg\cos\theta + \frac{mv_0^2}{l} - 2mg(1-\cos\theta)$ $T = mg\cos\theta + \frac{mv_0^2}{l} - 2mg + 2mg\cos\theta$ $T = \frac{mv_0^2}{l} + 3mg\cos\theta - 2mg$ $T = \frac{m}{l}(v_0^2 + 3gl\cos\theta - 2gl)$

(b) What should be the minimum speed at the highest point of the circular path?

The highest point of the circular path corresponds to $\theta = 180^\circ$. At this point, the string must remain taut, meaning the tension $T \ge 0$. Applying Newton's second law at the highest point: The forces are tension $T$ (downwards, towards O) and gravity $mg$ (downwards, towards O). $T + mg = \frac{mv^2}{l}$ For the minimum speed ($v_{min}$) at the highest point, the tension $T$ just becomes zero ($T=0$). $0 + mg = \frac{mv_{min}^2}{l}$ $v_{min}^2 = gl$ $v_{min} = \sqrt{gl}$

(c) What should be the minimum speed $v_0$ to complete circular path?

To complete the circular path, the particle must reach the highest point ($\theta = 180^\circ$) with at least the minimum speed $v_{min} = \sqrt{gl}$ found in part (b). Using the speed expression from part (a) at $\theta = 180^\circ$: $v^2 = v_0^2 - 2gl(1-\cos180^\circ)$ $v^2 = v_0^2 - 2gl(1-(-1))$ $v^2 = v_0^2 - 2gl(2)$ $v^2 = v_0^2 - 4gl$ For the particle to complete the circle, $v^2$ at the highest point must be at least $gl$: $v_0^2 - 4gl \ge gl$ $v_0^2 \ge 5gl$ $v_0 \ge \sqrt{5gl}$ The minimum speed $v_0$ to complete the circular path is $\sqrt{5gl}$.

(d) Find the range of speed $v_0$ so that the particle will oscillate.

For the particle to oscillate, it must reach a certain maximum height and then swing back, without the string ever becoming slack. This means the speed $v$ becomes zero ($v=0$) at some angle $\theta_{max}$ before the string becomes slack ($T=0$). If $v=0$, from part (a): $0 = v_0^2 - 2gl(1-\cos\theta_{max})$ $v_0^2 = 2gl(1-\cos\theta_{max})$ For oscillation, the particle should not pass the horizontal position ($\theta_{max} < 90^\circ$). This implies $\cos\theta_{max} > 0$. $1-\cos\theta_{max} < 1$ $v_0^2 < 2gl$ $v_0 < \sqrt{2gl}$ Also, for motion to occur, $v_0 > 0$. Let's check the tension at $v=0$: $T = mg\cos\theta_{max} + \frac{m}{l}(0) = mg\cos\theta_{max}$ Since $\theta_{max} < 90^\circ$, $\cos\theta_{max} > 0$, so $T > 0$. This confirms the string remains taut. Thus, the range of speed $v_0$ for oscillation is: $0 < v_0 < \sqrt{2gl}$

(e) Find the range of speed $v_0$ so that the particle will remain always on the circular path.

"Always on the circular path" implies that the particle completes a full circle. From part (c), the minimum speed $v_0$ for this is $\sqrt{5gl}$. There is no upper limit for this condition, as any speed greater than or equal to $\sqrt{5gl}$ will ensure completion of the circle. So, the range of speed $v_0$ is: $v_0 \ge \sqrt{5gl}$

(f) What will happen, if $\sqrt{2gl} < v_0 < \sqrt{5gl}$?

This is the intermediate range of $v_0$.

• If $v_0 > \sqrt{2gl}$, then $v_0^2 > 2gl$. From $v_0^2 = 2gl(1-\cos\theta_{max})$, this would imply $1-\cos\theta_{max} > 1$, so $\cos\theta_{max} < 0$. This means if the speed were to become zero, it would happen at $\theta_{max} > 90^\circ$.
• If $v_0 < \sqrt{5gl}$, then $v_0^2 < 5gl$. This means the particle does not have enough energy to reach the top point ($\theta = 180^\circ$) with sufficient speed to keep the string taut.

In this range, the particle will pass the horizontal position ($\theta = 90^\circ$). However, the tension in the string will become zero ($T=0$) before the particle reaches the highest point or before its speed becomes zero. When the tension becomes zero, the string slackens, and the particle leaves the circular path to follow a parabolic trajectory under gravity. So, if $\sqrt{2gl} < v_0 < \sqrt{5gl}$, the particle will leave the circular path.

(g) Where will the particle leave the circular path, if $\sqrt{2gl} < v_0 < \sqrt{5gl}$?

The particle leaves the circular path when the tension $T$ in the string becomes zero. From part (a), the tension is $T = \frac{m}{l}(v_0^2 + 3gl\cos\theta - 2gl)$. Setting $T=0$: $v_0^2 + 3gl\cos\theta - 2gl = 0$ $3gl\cos\theta = 2gl - v_0^2$ $\cos\theta = \frac{2gl - v_0^2}{3gl}$ In the range $\sqrt{2gl} < v_0 < \sqrt{5gl}$:

• Since $v_0 > \sqrt{2gl} \implies v_0^2 > 2gl$, the numerator $(2gl - v_0^2)$ is negative.
• Since $v_0 < \sqrt{5gl} \implies v_0^2 < 5gl \implies 2gl - v_0^2 > 2gl - 5gl = -3gl$. Therefore, $-3gl < (2gl - v_0^2) < 0$. Dividing by $3gl$: $-1 < \frac{2gl - v_0^2}{3gl} < 0$ So, $-1 < \cos\theta < 0$. This means the angle $\theta$ at which the particle leaves the path is between $90^\circ$ and $180^\circ$. The particle will leave the circular path at an angle $\theta = \arccos\left(\frac{2gl - v_0^2}{3gl}\right)$.

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Summary of Results:

• (a) Speed ($v$) and Tension ($T$) at angle $\theta$: $v = \sqrt{v_0^2 - 2gl(1-\cos\theta)}$ $T = \frac{m}{l}(v_0^2 + 3gl\cos\theta - 2gl)$

• (b) Minimum speed at the highest point: $v_{min} = \sqrt{gl}$

• (c) Minimum speed $v_0$ to complete circular path: $v_{0,min} = \sqrt{5gl}$

• (d) Range of speed $v_0$ for oscillation: $0 < v_0 < \sqrt{2gl}$

• (e) Range of speed $v_0$ to remain always on the circular path: $v_0 \ge \sqrt{5gl}$

• (f) What happens if $\sqrt{2gl} < v_0 < \sqrt{5gl}$? The particle will leave the circular path (string slackens).

• (g) Where will the particle leave the circular path if $\sqrt{2gl} < v_0 < \sqrt{5gl}$? At an angle $\theta = \arccos\left(\frac{2gl - v_0^2}{3gl}\right)$, where $90^\circ < \theta < 180^\circ$.

Answer 2

The problem is solved using the principles of conservation of mechanical energy and Newton's second law for circular motion.

1. Speed ($v$): Derived using conservation of energy: initial kinetic energy at the bottom equals kinetic energy plus potential energy at angle $\theta$.
2. Tension ($T$): Derived using Newton's second law in the radial direction: net radial force ($T - mg\cos\theta$) provides the centripetal force ($mv^2/l$). The derived $v^2$ is substituted.
3. Minimum speed at highest point: At the highest point, tension is minimum. For circular motion, tension must be $\ge 0$. Minimum speed occurs when $T=0$.
4. Minimum $v_0$ for complete circle: The particle must reach the highest point with at least the minimum speed found in (b). This condition is applied to the energy conservation equation.
5. Range for oscillation: Particle oscillates if it stops ($v=0$) before reaching the horizontal level ($\theta=90^\circ$) and the string remains taut ($T>0$). This condition on $v=0$ defines the range for $v_0$.
6. Range for always on circular path: This means completing the full circle, so $v_0$ must be greater than or equal to the minimum $v_0$ found in (c).
7. What happens if $\sqrt{2gl} < v_0 < \sqrt{5gl}$: This is an intermediate range where the particle has enough energy to pass the horizontal level but not enough to complete the full circle without the string slackening. The string slackens when tension becomes zero.
8. Where particle leaves path: The particle leaves the circular path when the tension in the string becomes zero. The angle $\theta$ at which $T=0$ is determined using the tension equation.