Question

A particle slides freely on a flat surface, attached to a massless string that passes through a hole in the surface. At a given moment, the particle is moving in a circle of radius $r_0$ with angular velocity ${\omega }_0$. Assume that the string is pulled through the hole at a constant rate. Find an expression for the tension in the string as a function of the length $r$ of string left on the surface.

Short answer

The tension is $T=\frac{m{r}_0^4{\omega }_0^2}{r^3}$.

Step-by-step solution

Define the System

Initially, note that the particle moves in a circular path because it's attached to a string that is pulled through a hole in the surface. This system has a changing radius over time due to the string being pulled.

Relationship of Angular Velocity and Radius

Due to the conservation of angular momentum, the quantity $L=mvr$ remains constant, where $m$ is the mass of the particle, $v$ is the linear velocity, and $r$ is the radius. Since $v=r\omega$, we have $L=m(r\omega )r=m{r}^2\omega$. Thus, $m{r}_0^2{\omega }_0=m{r}^2\omega$.

Solve for Angular Velocity $\omega$

From the conservation of angular momentum from Step 2, solve for $\omega$: $$\omega =\frac{r_0^2{\omega }_0}{r^2}$$

Relate to Tension in the String

The centripetal force $F_c$ required for circular motion is provided by the tension $T$ in the string, thus $T=m\frac{v^2}{r}=mr{\omega }^2$.

Substitute Angular Velocity into Tension Formula

Substitute $\omega$ from Step 3 into the tension formula from Step 4: $$T=mr{\left(\frac{r_0^2{\omega }_0}{r^2}\right)}^2$$Simplify to get:$$T=\frac{m{r}_0^4{\omega }_0^2}{r^3}$$

Conclusion

The final expression for the tension in the string as a function of the radius $r$ is: $$T=\frac{m{r}_0^4{\omega }_0^2}{r^3}$$

Key concepts

Centripetal force

When a particle moves in a circular path, it requires a force directed towards the center of the circle to maintain its motion. This is the centripetal force. Without centripetal force, the particle would move off in a tangent.

• Centripetal force ensures continuous circular motion.
• For objects in circular motion, centripetal force is often provided by tension, gravity, or friction.

In our case with the string and the particle, the centripetal force is provided by the tension in the string. This means that as the string pulls the particle inwards, it acts as the centripetal force ensuring that the particle maintains its circular path.
The formula for centripetal force is given by $$F_c=\frac{m{v}^2}{r},$$where:

• $m$ is the mass of the particle,
• $v$ is the linear velocity of the particle, and
• $r$ is the radius of the circle.

Thus, in the context of our problem, the tension in the string ($T$) acts to pull the particle towards the center, fulfilling the role of the centripetal force.

Angular velocity

Angular velocity is a measure of how quickly an object rotates around a central point. It is different from linear velocity, which is the speed of an object along a path.

• Angular velocity is denoted by the symbol $\omega$.
• It's measured in radians per second (rad/s).

In our problem, the particle initially rotates with an angular velocity ${\omega }_0$, and as the string is pulled through the hole, this velocity changes.
The relationship between angular velocity and radius can be derived from the conservation of angular momentum. Angular momentum is conserved because there is no external torque acting on the system. This is mathematically represented as:$$m{r}_0^2{\omega }_0=m{r}^2\omega .$$Solving for the new angular velocity $\omega$ gives:$$\omega =\frac{r_0^2{\omega }_0}{r^2}.$$This equation shows that as the radius $r$ decreases (the string is pulled in), the angular velocity $\omega$ increases, ensuring that the total angular momentum remains constant.

Tension in string

The tension in the string is a force that results from pulling or tautness and is crucial in maintaining the particle’s circular path. In this scenario, as the string is pulled through the hole, it continuously changes the circle's radius and affects the tension.

• Tension is equivalent to the centripetal force in this situation.
• It adapts as the string length and particle’s angular velocity change.

The formula for tension can be derived from the centripetal force equation:$$T=m\frac{v^2}{r}.$$Utilizing the relationship between linear velocity $v=r\omega$, and substituting the earlier expression for $\omega$, we get:$$T=mr{\left(\frac{r_0^2{\omega }_0}{r^2}\right)}^2.$$After simplifying, the tension in the string is:$$T=\frac{m{r}_0^4{\omega }_0^2}{r^3}.$$This equation illustrates how the tension in the string increases as the radius of the circle diminishes.
The tension accounts for the need for a stronger centripetal force to keep the particle moving as the circle tightens.