# Tension in a pendulum string?

## Homework Statement

Hi all!
I was wondering what the tension is in the string of a pendulum, because I think sparknotes is wrong on this.
Sparknotes says that:

"Choose a coordinate system: We want to calculate the forces acting on the pendulum at any given point in its trajectory. It will be most convenient to choose a y-axis that runs parallel to the rope. The x-axis then runs parallel to the instantaneous velocity of the bob so that, at any given moment, the bob is moving along the x-axis.
Draw free-body diagrams: Two forces act on the bob: the force of gravity, F = mg, pulling the bob straight downward and the tension of the rope, , pulling the bob upward along the y-axis. The gravitational force can be broken down into an x-component, mg sin, and a y-component, mg cos. The y component balances out the force of tension—the pendulum bob doesn’t accelerate along the y-axis—so the tension in the rope must also be mg cos. Therefore, the tension force is maximum for the equilibrium position and decreases with . The restoring force is mg sin , so, as we might expect, the restoring force is greatest at the endpoints of the oscillation, and is zero when the pendulum passes through its equilibrium position."

So it seems to be saying that the only force acting on the pendulum at any point is the tangential component of its weight... But I thought thatthere had to be a net force inwards because the pendulum almost has circular motion, and the net force would be greatest at the botton since it has the greatest velocity and then zero at its maximum height when it has zero velocity. I kinda doubt myself because I can't figure out where the tangential component of the weight, mgsinθ, fits into this. There is no tangential force in cetripetal motion... Is this the force that causes the change in velocity of the circular motion (if that makes any sense at all)?

## Homework Equations

I'm also confused because I figured out the tension at any point to be 3mgcosθ-2mgcos(θmax), which seems to work because then the 'centripetal force' at the bottom of the swing is 2mg(cosθ-cos(θmax)) and at the max height is zero, which corresponds to the changes in velocity...

## The Attempt at a Solution

help!

Thank you in advance :)]

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
verty
Homework Helper
The tension must vary with speed, by the following argument. Suppose we put a motor in the center of the pendulum and make it spin really quick, there is a speed at which the string will break, the tension will be too high. But if the tension is just the weight, it could spin arbitrarily fast without breaking.

Now at very high speeds the force of gravity will be vanishingly small compared to the tension. So as ##\omega \to \infty##, ##{T \over r \omega^2} \to 1##. At low speeds, ##\omega ≈ 0##, the tension will be close to ##mg \cos\theta##.

I haven't found the formula for the tension but it must have this behaviour.

AlephZero
Homework Helper
The tension must vary with speed
That is correct, but when deriving the equation for simple harmonic motion, you assume the amplitude of the motion is very small, so the angular velocity is small and the forces arising from the angular velocity can be ignored compared with the weight.

If you expand the sine and cosine of the angle of the string as a power series,
##\sin \theta = \theta - \dfrac{\theta^3}{6} + \cdots##, ##\cos \theta = 1 - \dfrac{\theta^2}{2} + \cdots##, you assume ##\theta## is small and you make the approximations ##\sin \theta = \theta##, ##\cos \theta = 1##.

An analysis of the motion that is correct for "large" angles and "large" angular velocities is too complicated for a first course in dynamics, because the students are unlikely to know enough calculus to solve the equations.

It gives the period as ##T = T_0\left(1 + \dfrac {\theta^2}{16} + \dfrac{11\theta^4}{3072} + \dfrac{173 \theta^6}{368640} + \cdots\right)## where ##T_0## is the period for small amplitude and ##\theta## is the (large) amplitude of the swing, in radians.

A pretty good approximation to the "accurate" formula, with about 1% error when ##\theta < 90^{\circ}##, iis ##T = \dfrac{T_0}{ \sqrt{\cos \frac{\theta}{2}}}##.

http://hyperphysics.phy-astr.gsu.edu/hbase/pendl.html
http://cds.cern.ch/record/899791/files/0510206.pdf?version=1
http://physics.j3science.com/images/d/d9/PTE000081-Large_Angle_Pendulum.pdf [Broken]

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