The analysis begins by considering the forces acting on the ball of mass $m$. The forces are the tension $T$ in the string and gravity $mg$. The ball moves in a horizontal circle of radius $r$ with the string of length $L$ making an angle $\theta$ with the vertical.

The radius of the circular path is $r = L\sin\theta$. Applying Newton's second law, we resolve the forces into vertical and horizontal components. The vertical components are in equilibrium:

$$T\cos\theta = mg \quad (1)$$

The horizontal component of the tension provides the necessary centripetal force $F_c = ma_c$:

$$T\sin\theta = ma_c \quad (2)$$

[Q1] Constant Angular Velocity ($\omega$)

For a constant angular velocity $\omega$, the centripetal acceleration is $a_c = \omega^2 r$. Substituting $r = L\sin\theta$:

$$a_c = \omega^2 L\sin\theta$$

We substitute this into the horizontal force equation (2):

$$T\sin\theta = m(\omega^2 L\sin\theta)$$

Since the pendulum is rotating, $\theta eq 0$, so $\sin\theta eq 0$. We can divide both sides by $\sin\theta$:

$$T = m\omega^2 L$$

This result shows that for a given mass $m$ and constant angular velocity $\omega$, the tension $T$ is directly proportional to the length of the string $L$. Therefore, a longer string will experience a greater tension.

[Q2] Constant Linear Velocity ($v$)

For a constant linear velocity $v$, the centripetal acceleration is $a_c = v^2/r$. Substituting $r = L\sin\theta$:

$$a_c = \frac{v^2}{L\sin\theta}$$

We substitute this into the horizontal force equation (2):

$$T\sin\theta = m\left(\frac{v^2}{L\sin\theta}\right)$$ $$T = \frac{mv^2}{L\sin^2\theta}$$

This expression for tension depends on both $L$ and $\theta$. To understand how $T$ depends on $L$, we must first find the relationship between $L$ and $\theta$. We can use the vertical force equation (1), $T = \frac{mg}{\cos\theta}$, and equate the two expressions for $T$:

$$\frac{mg}{\cos\theta} = \frac{mv^2}{L\sin^2\theta}$$

Solving for $L$:

$$L = \frac{v^2}{g} \frac{\cos\theta}{\sin^2\theta}$$

Let's analyze this relationship between $L$ and $\theta$.

• As the angle $\theta$ approaches $0$ (nearly vertical), $\cos\theta \to 1$ and $\sin^2\theta \to 0$, causing $L \to \infty$.
• As the angle $\theta$ approaches $90^\circ$ (nearly horizontal), $\cos\theta \to 0$, causing $L \to 0$.

This shows that a shorter string length $L$ corresponds to a larger angle $\theta$.

Now, let's examine the tension using equation (1):

$$T = \frac{mg}{\cos\theta}$$

Since a shorter string $L$ implies a larger angle $\theta$, the value of $\cos\theta$ will be smaller (as $\cos\theta$ is a decreasing function for $\theta \in [0, 90^\circ]$). A smaller value in the denominator results in a larger tension $T$. Therefore, a shorter string will experience a greater tension.