Question

In presence of gravity but no air resistance, to whirl a stone on a horizontal circular path with the help of thread of length $l$, a boy has to tilt the thread at an angle $\theta$ from the vertical and impart a horizontal velocity $v$ perpendicular to the thread. If a proper angel is chosen the thread will sweep a stable cone, axis of which will be a vertical line through the end O of the thread held stationary as shown in the figure. This kind of circular motion is known as conical pendulum. Acceleration due to gravity is g.

(a) Determine the proper angle $\theta$ in terms of $l$, $v$, and $g$.

(b) Determine the radius $r$ of the circular path in terms of $l$, $v$ and $g$.

(c) Describe qualitatively, what will happen if the air resistance is considered?

Answer 1

a) $\tan\theta = \frac{v^2}{l g\sin\theta}$, b) $r = \sqrt{\frac{v^2 (-v^2 + \sqrt{v^4 + 4 g^2 l^2})}{2g^2}}$, c) The stone's velocity will decrease due to air resistance, causing the radius of the circular path and the angle $\theta$ to decrease. The stone will follow a spiral path, moving inwards and downwards, eventually coming to rest below the point of suspension.

Answer 2

The problem describes a conical pendulum and asks for derivations related to its motion under ideal conditions (no air resistance), and then a qualitative description of the effect of air resistance.

Let's denote the mass of the stone as $m$, the length of the thread as $l$, the angle the thread makes with the vertical as $\theta$, the horizontal velocity of the stone as $v$, the radius of the circular path as $r$, and the acceleration due to gravity as $g$.

(a) Determine the proper angle $\theta$ in terms of $l$, $v$, and $g$.

1. Identify Forces: There are two forces acting on the stone:

• Tension (T): Acting along the thread, towards the point of suspension.

• Gravity (mg): Acting vertically downwards.

2. Resolve Forces: Resolve the tension $T$ into its vertical and horizontal components:

• Vertical component: $T_y = T \cos\theta$ (upwards)

• Horizontal component: $T_x = T \sin\theta$ (towards the center of the circular path)

3. Apply Newton's Second Law:

• Vertical Equilibrium: Since the stone moves in a horizontal plane, there is no vertical acceleration. Thus, the net vertical force is zero. $T \cos\theta = mg \quad \text{(Equation 1)}$

• Horizontal Motion (Centripetal Force): The horizontal component of the tension provides the necessary centripetal force ($F_c = \frac{mv^2}{r}$) for the circular motion. $T \sin\theta = \frac{mv^2}{r} \quad \text{(Equation 2)}$

4. Relate Geometric Parameters: From the geometry of the conical pendulum, the radius $r$ of the circular path is related to the length $l$ and the angle $\theta$: $r = l \sin\theta \quad \text{(Equation 3)}$

5. Derive the Angle Relation: Divide Equation 2 by Equation 1: $\frac{T \sin\theta}{T \cos\theta} = \frac{mv^2/r}{mg}$ $\tan\theta = \frac{v^2}{rg}$ Now, substitute the expression for $r$ from Equation 3 into this equation: $\tan\theta = \frac{v^2}{(l \sin\theta)g}$ $\tan\theta = \frac{v^2}{l g\sin\theta}$ This is the required relation for the proper angle $\theta$.

(b) Determine the radius $r$ of the circular path in terms of $l$, $v$ and $g$.

We have the relation from part (a): $\tan\theta = \frac{v^2}{l g\sin\theta}$ We also know that $\tan\theta = \frac{\sin\theta}{\cos\theta}$. Substitute this into the equation: $\frac{\sin\theta}{\cos\theta} = \frac{v^2}{l g\sin\theta}$ Rearrange to solve for $\cos\theta$: $l g \sin^2\theta = v^2 \cos\theta$ We know that $\sin^2\theta = 1 - \cos^2\theta$. Substitute this into the equation: $l g (1 - \cos^2\theta) = v^2 \cos\theta$ $l g - l g \cos^2\theta = v^2 \cos\theta$ Rearrange into a quadratic equation in terms of $\cos\theta$: $l g \cos^2\theta + v^2 \cos\theta - l g = 0$ Let $x = \cos\theta$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-v^2 \pm \sqrt{(v^2)^2 - 4(lg)(-lg)}}{2(lg)}$ $x = \frac{-v^2 \pm \sqrt{v^4 + 4 l^2 g^2}}{2lg}$ Since $\theta$ is an angle from the vertical in a conical pendulum, $0 < \theta < \pi/2$, which means $\cos\theta$ must be positive. Therefore, we take the positive root: $\cos\theta = \frac{-v^2 + \sqrt{v^4 + 4 l^2 g^2}}{2lg}$ Now, we need to find $r$ in terms of $l, v, g$. We know $r = l \sin\theta$. We can find $\sin\theta$ using $\sin\theta = \sqrt{1 - \cos^2\theta}$. Alternatively, we can use the relation derived earlier: $g r^2 = v^2 \sqrt{l^2 - r^2}$ (from substituting $r = l\sin\theta$ and $\cos\theta = \frac{\sqrt{l^2-r^2}}{l}$ into $l g \sin^2\theta = v^2 \cos\theta$). Square both sides: $(g r^2)^2 = (v^2 \sqrt{l^2 - r^2})^2$ $g^2 r^4 = v^4 (l^2 - r^2)$ $g^2 r^4 = v^4 l^2 - v^4 r^2$ Rearrange into a quadratic equation in terms of $r^2$: $g^2 r^4 + v^4 r^2 - v^4 l^2 = 0$ Let $X = r^2$. Using the quadratic formula: $X = \frac{-v^4 \pm \sqrt{(v^4)^2 - 4(g^2)(-v^4 l^2)}}{2g^2}$ $X = \frac{-v^4 \pm \sqrt{v^8 + 4 g^2 v^4 l^2}}{2g^2}$ $X = \frac{-v^4 \pm v^2 \sqrt{v^4 + 4 g^2 l^2}}{2g^2}$ Since $X = r^2$ must be positive, we take the positive root: $r^2 = \frac{-v^4 + v^2 \sqrt{v^4 + 4 g^2 l^2}}{2g^2}$ Therefore, the radius $r$ is: $r = \sqrt{\frac{v^2 (-v^2 + \sqrt{v^4 + 4 g^2 l^2})}{2g^2}}$

(c) Describe qualitatively, what will happen if the air resistance is considered?

If air resistance is considered, a drag force ($F_d$) will act on the stone. This force always opposes the direction of the stone's motion.

1. Tangential Deceleration: The velocity of the stone is horizontal and tangential to the circular path. Therefore, the air resistance force will act tangentially, opposing the velocity. This will cause the magnitude of the horizontal velocity ($v$) of the stone to continuously decrease over time.

2. Change in Angle and Radius: As the horizontal velocity $v$ decreases, the required centripetal force ($F_c = \frac{mv^2}{r}$) also decreases. To maintain the circular motion, the horizontal component of tension ($T \sin\theta$) must also decrease. Since the vertical component of tension ($T \cos\theta = mg$) must still balance gravity, as $T \sin\theta$ decreases, the angle $\theta$ must decrease. A decrease in $\theta$ means the stone drops to a lower horizontal plane, and the radius $r = l \sin\theta$ will also decrease.

3. Spiral Path: As the velocity decreases and the angle $\theta$ decreases, the stone will no longer maintain a stable circular path at a constant radius. Instead, it will follow a spiral path, gradually moving inwards and downwards, until it eventually comes to rest directly below the point of suspension (i.e., $\theta \to 0$ and $v \to 0$).

Summary of effects:

• The speed of the stone will decrease.

• The radius of the circular path will decrease.

• The angle $\theta$ will decrease (the stone will drop).

• The motion will no longer be a stable circle but a decaying spiral.