Suppose we have a simple pendulum swinging between two extreme positions. At the extreme position its speed becomes zero. As per this reason can I say that at extreme positions radial acceleration (v^2/r) is zero so tension in the string has to be equal and opposite to the mgcos☆ , where angle ☆ is angular position of the pendulum bob from the vertical. Then at this instant can I add that mgsin☆ is the resultant of tension T and gravitational pull (mg). Am I correct in saying that there is no radial acceleration at extreme positions or I have misunderstood something?

  • 2
    $\begingroup$ Try to use math notation in your post. Enclose math expressions in dollar signs such that $1 + \sin \theta$ renders like $1 + \sin \theta$. $\endgroup$ Mar 20 at 17:07

2 Answers 2


Yes you are right - although you may have made a sin/cos error or defined angles differently than me. Since the velocity/angular velocity is zero, the tension in the string doesn't need to pull against the centrifugal force to keep the pendulum along the correct path. It only needs to counteract the extent to which gravity is pulling it on it's path. If $\theta$ is the angle the string is from vertical, then $mg\cos(\theta)$ is the component of gravity in the direction parallel to the string - so that's also the tension in the string. The string provides tension such that the ball at the end of the string doesn't accelerate away from the string.


No, the nature of the string or rope of the pendulum means constant length, and this translates to zero radial acceleration.

Suppose that wasn't the case and you had the coordinates of the mass described by the radial distance $r$ and angle $\theta$

$$\begin{aligned} x & = r \sin \theta \\ y & = -r \cos \theta \end{aligned}$$

by direct differentiation you get the following expression of the acceleration of the mass

$$\begin{aligned} \ddot{x} & = (r \ddot{\theta}+2 \dot{r} \dot{\theta}) \cos \theta + (\ddot{r}-r \dot{\theta}^2) \sin \theta \\ \ddot{y} & = (r \ddot{\theta}+2 \dot{r} \dot{\theta}) \sin \theta + (\ddot{r}-r \dot{\theta}^2) \cos \theta \end{aligned}$$

Consider the balance of forces with tension $T$ and weight $W= m g$ you have

$$\begin{aligned} -T \sin \theta & = m \ddot{x} \\ T \cos \theta - m g & = m \ddot{y} \end{aligned}$$

With solution

$$ \begin{aligned} \ddot{r} &= g \cos \theta - \frac{T - m r \dot{\theta}^2}{m} \\ \ddot{\theta} & = -\frac{g \sin \theta}{r} - \frac{2 \dot{r} \dot{\theta}}{r} \end{aligned}$$

with your question now is $\ddot{r}$ zero or not.

The answer has to do with the nature of $T$. The above equations are what you have if the string has elasticity and tension is a function of radial position $T = k ( r - \ell)$. This would result in an oscillation of $r$ around some equilibrium deflection.

But a simplification exists here as this oscillation would occur much faster than the swinging of the pendulum (stiffness $k \gg m \dot{\theta}^2$) and it settles to a value of $r = \ell$. For lower values of stiffness, this value is a bit higher, but we can assume infinite stiffness for now and that the string is inextensible.

The value of tension $T$ is exactly what is needed to keep the string inextensible and thus, $\dot{r}=0$

Now the result of the equations of motion can be used to find the tension $T$ when we can assume $r=\ell$ and thus $\dot{r}=0$ and $\ddot{r}=0$

$$\begin{aligned} T & = m g \cos \theta + m \ell \dot{\theta}^2 \\ \ddot{\theta} &= -\frac{g}{\ell} \sin \theta \end{aligned}$$

If at the lowest point the mass as speed $v_0 = r \dot{\theta}_0$ then the maximum angle is found from integration of the second equation as

$$ \cos \theta_{\rm max} = 1 - \frac{v_0^2}{2 g \ell} $$

At that instant, with $\dot{\theta}=0$ the solution is

$$\begin{aligned} T &= m \left( g - \frac{v_0^2}{2 \ell} \right) \\ \ddot{\theta} &= -\frac{v_0}{\ell} \sqrt{ \frac{g}{\ell} - \left( \frac{v_0}{2 \ell} \right)^2} \end{aligned}$$

Neither of which is zero. Also, note that the tension at this moment is always less than the tension on the bottom of the stroke which is $T_{\rm max} = m g + \frac{m v_0^2}{\ell}$

In summary, if we ignore the elastic properties of the string then we must assume constant length and thus zero radial acceleration at all times.

But if we include the effects of stiffness then you have a more complex 2-DOF oscillator with the pendulum swinging to and from and at the same time the radial position stretching and contracting with a different frequency.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.