Question

A stone of mass $m$ is tied at one end of a light inextensible thread. A boy holds the free end of the thread by one hand and then moves the end of the thread at a constant angular velocity $\omega$ on a horizontal circular path of radius $r$. Length of the thread between the stone and the end held by the boy is $l$.

(a) If the experiment is performed in a free space, how will the stone move?

(b) In free space conditions, determine the tensile fore developed in the thread in terms of $r$, $\omega$, $l$ and $m$.

(c) Neglect effect of gravity but consider air, determine radius of the circular path of the stone. Force of air resistance that follows the law $\vec{f} = -k\vec{v}$, where $k$ is a constant and $\vec{v}$ is velocity of the stone. Calculate Radius of Path

(d) Describe qualitatively, what will happen if gravity is present but no air resistance.

(e) Describe qualitatively, what will happen if gravity and air resistance both are.

Answer 1

(a) The stone will move in a horizontal circular path of radius $R_S = r+l$ (or $|r-l|$), centered at the same point as the boy's hand's path, with a constant angular velocity $\omega$.

(b) The tensile force developed in the thread is $T = m (r+l) \omega^2$.

(c) The radius of the circular path of the stone is $R_S = \frac{r m\omega + \sqrt{l^2((m\omega)^2 + k^2) - r^2 k^2}}{\sqrt{(m\omega)^2 + k^2}}$.

(d) If gravity is present but no air resistance, the stone will move in a horizontal circular path at a lower level than the boy's hand. The thread will be inclined downwards, forming a conical pendulum.

(e) If gravity and air resistance are both present, the stone will move in a horizontal circular path at a lower level. The thread will be inclined downwards (due to gravity) and also angled backward tangentially (to overcome air resistance).

Answer 2

The problem describes a stone of mass $m$ tied to a light inextensible thread of length $l$. A boy moves the free end of the thread in a horizontal circular path of radius $r$ with a constant angular velocity $\omega$. We need to analyze the motion of the stone under different conditions.

(a) If the experiment is performed in a free space, how will the stone move?

In free space, there is no gravity and no air resistance. The only force acting on the stone is the tension in the thread. The boy's hand moves in a circle of radius $r$ with angular velocity $\omega$. For the stone to maintain a steady circular motion, it must also move with the same angular velocity $\omega$ around the same center as the boy's hand.

Let the center of the circular path of the boy's hand be the origin. The hand's position vector is $\vec{R}_H$. The stone's position vector is $\vec{R}_S$. The length of the thread is $l$, so $|\vec{R}_S - \vec{R}_H| = l$.

For the stone to move in a stable circular path of radius $R_S$ around the origin with angular velocity $\omega$, the tension in the thread must provide the necessary centripetal force. This means the tension force must be directed towards the center of the stone's circular path (the origin).

The tension force acts along the thread, from the stone towards the hand. For the tension to be directed towards the origin, the hand, the stone, and the origin must be collinear. This implies two possible configurations:

1. The stone is between the origin and the hand (Origin --- Stone --- Hand). In this case, $R_S = r - l$. This is only possible if $r > l$.
2. The hand is between the origin and the stone (Origin --- Hand --- Stone). In this case, $R_S = r + l$.

Assuming the string is taut and the stone is pulled by the hand, the second configuration ($R_S = r+l$) is the most natural.

Thus, the stone will move in a horizontal circular path of radius $r+l$ (or $|r-l|$ in general, but $r+l$ is more common for taut strings) with a constant angular velocity $\omega$, centered at the same point as the circular path of the boy's hand.

(b) In free space conditions, determine the tensile force developed in the thread in terms of $r$, $\omega$, $l$ and $m$.

As established in part (a), the stone performs uniform circular motion with radius $R_S = r+l$ and angular velocity $\omega$. The centripetal force required for this motion is provided by the tension $T$ in the thread.

The centripetal force is given by $F_c = m R_S \omega^2$.

Substituting $R_S = r+l$:

$T = m (r+l) \omega^2$

(c) Neglect effect of gravity but consider air, determine radius of the circular path of the stone. Force of air resistance that follows the law $\vec{f} = -k\vec{v}$, where $k$ is a constant and $\vec{v}$ is velocity of the stone. Calculate Radius of Path

In this case, air resistance $\vec{f} = -k\vec{v}$ is present. For the stone to move in a steady circular path with constant angular velocity $\omega$ (same as the hand), the net force on it must be purely centripetal.

Let the radius of the stone's path be $R_S$. The magnitude of its velocity is $v = R_S \omega$.

The air resistance force has magnitude $f = k v = k R_S \omega$. This force acts tangentially, opposite to the direction of motion.

The tension $\vec{T}$ in the thread must provide both the centripetal force and balance the air resistance. Let $\vec{T}$ make an angle $\alpha$ with the radial direction (pointing towards the center).

The components of tension are:

• Radial component: $T_r = T \cos \alpha$
• Tangential component: $T_\theta = T \sin \alpha$

Applying Newton's second law:

1. Radial direction: $T \cos \alpha = m R_S \omega^2$ (centripetal force)
2. Tangential direction: $T \sin \alpha = k R_S \omega$ (balancing air resistance)

From the tangential equation, $\sin \alpha = \frac{k R_S \omega}{T}$.

From the radial equation, $\cos \alpha = \frac{m R_S \omega^2}{T}$.

Dividing the two equations:

$\tan \alpha = \frac{k R_S \omega}{m R_S \omega^2} = \frac{k}{m \omega}$

Now consider the geometry of the system. The hand is at radius $r$ from the center, and the stone is at radius $R_S$. The thread of length $l$ connects them. The thread makes an angle $\alpha$ with the radial line connecting the stone to the center.

Let the center be the origin. The stone is at $(R_S, 0)$ at some instant. The hand is at $(r \cos \alpha, r \sin \alpha)$ relative to the radial line of the stone.

No, the angle $\alpha$ is the angle between the thread and the line connecting the stone to the center. Since the tension has a tangential component, the hand must be "ahead" of the stone in terms of angular position.

Using the law of cosines in the triangle formed by the origin, the hand, and the stone:

$l^2 = r^2 + R_S^2 - 2 r R_S \cos(\alpha)$

From the force diagram, the angle $\alpha$ is given by $\tan \alpha = \frac{k}{m \omega}$.

So, $\cos \alpha = \frac{1}{\sqrt{1 + \tan^2 \alpha}} = \frac{1}{\sqrt{1 + (k/(m\omega))^2}} = \frac{m\omega}{\sqrt{(m\omega)^2 + k^2}}$.

Substitute $\cos \alpha$ into the law of cosines equation:

$l^2 = r^2 + R_S^2 - 2 r R_S \frac{m\omega}{\sqrt{(m\omega)^2 + k^2}}$

This equation relates $R_S$ to $r, l, m, \omega, k$.

To find $R_S$, we can rearrange this into a quadratic equation for $R_S$:

$R_S^2 - \left( \frac{2 r m\omega}{\sqrt{(m\omega)^2 + k^2}} \right) R_S + (r^2 - l^2) = 0$

This is a quadratic equation of the form $AR_S^2 + BR_S + C = 0$.

$R_S = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$R_S = \frac{\frac{2 r m\omega}{\sqrt{(m\omega)^2 + k^2}} \pm \sqrt{\left(\frac{2 r m\omega}{\sqrt{(m\omega)^2 + k^2}}\right)^2 - 4(r^2 - l^2)}}{2}$

$R_S = r \frac{m\omega}{\sqrt{(m\omega)^2 + k^2}} \pm \sqrt{r^2 \frac{(m\omega)^2}{(m\omega)^2 + k^2} - (r^2 - l^2)}$

$R_S = r \frac{m\omega}{\sqrt{(m\omega)^2 + k^2}} \pm \sqrt{\frac{r^2 (m\omega)^2 - r^2((m\omega)^2 + k^2) + l^2((m\omega)^2 + k^2)}{(m\omega)^2 + k^2}}$

$R_S = r \frac{m\omega}{\sqrt{(m\omega)^2 + k^2}} \pm \sqrt{\frac{-r^2 k^2 + l^2((m\omega)^2 + k^2)}{(m\omega)^2 + k^2}}$

$R_S = \frac{r m\omega \pm \sqrt{l^2((m\omega)^2 + k^2) - r^2 k^2}}{\sqrt{(m\omega)^2 + k^2}}$

For a real solution, $l^2((m\omega)^2 + k^2) - r^2 k^2 \ge 0$.

Since the string is pulling the stone, the stone will be "outside" the hand's circle, so we choose the '+' sign.

Radius of Path: $R_S = \frac{r m\omega + \sqrt{l^2((m\omega)^2 + k^2) - r^2 k^2}}{\sqrt{(m\omega)^2 + k^2}}$

(d) Describe qualitatively, what will happen if gravity is present but no air resistance.

If gravity is present and there is no air resistance, the stone will no longer move in a horizontal circle. The tension in the thread must now support the weight of the stone ($mg$) in the vertical direction, in addition to providing the centripetal force in the horizontal direction.

The thread will make an angle with the horizontal plane. This setup is similar to a conical pendulum.

The stone will move in a horizontal circular path, but it will be at a certain depth below the boy's hand. The thread will form a cone, with its apex at the boy's hand's position (which itself is moving in a circle).

If the boy's hand moves in a horizontal circle of radius $r$, the stone will also move in a horizontal circle. The center of the stone's circle will be vertically below the center of the boy's hand's circle.

The tension in the string will have a vertical component balancing gravity and a horizontal component providing the centripetal force.

(e) Describe qualitatively, what will happen if gravity and air resistance both are.

If both gravity and air resistance are present, the situation becomes more complex.

Similar to part (d), gravity will cause the stone to move in a horizontal circle at a lower level, with the thread making an angle with the horizontal.

Air resistance, acting opposite to the velocity, will have a tangential component that tends to slow down the stone. To maintain the constant angular velocity $\omega$ (which is dictated by the boy's hand), the thread will have to pull the stone in the tangential direction as well, just like in part (c).

So, the thread will not only be inclined downwards due to gravity but also be slightly angled "backward" (tangentially) relative to the direction of motion to overcome the air resistance. The stone will still move in a horizontal circular path, but the thread will be inclined downwards and backward from the radial direction relative to the stone's motion.