- #1

brotherbobby

- 699

- 163

**Statement of the problem :**

*A ball shown in the figure is allowed to swing in a vertical plane like a simple pendulum. Answer the following :*

(a) Is the angular momentum of the ball conserved?

(a) Is the angular momentum of the ball conserved?

**No**, the angular momentum ##L = mvl##, where

*m*is the mass of the ball and

*v*is its speed at an instant. Note, this is an example of (non-uniform) circular motion whereby vector

**v**##\perp## radius vector

**r**at all instants. Clearly,

*v*changes, reaching its maximum value when ##\theta = 0## and hence the angular momentum L also changes, becoming maximum at the mean position.

*(b) Calculate the direction of*

**L**at some time. Does it change?**No**. The motion takes place in a plane. From the definition of

**L = r ##\times## p**, using either the right hand cork screw rule or taking the convenient mean position of the pendulum to evaluate (

**r**##\rightarrow - \hat y##,

**p**##\rightarrow \hat x##, ##-\hat y \times \hat x = \hat z##), we find that the direction of

**L**is along the

**+z**

**axis, out of the page**.

*(c) What force acting on the pendulum gives a zero torque about an axis perpendicular to the motion plane and through the point of support?*

**Tension**

*. By the definition of torque, ##\tau = r \times F \Rightarrow \tau = l \hat l \times -T \hat l = 0##, where ##\hat l## is the unit vector along the rope out from the point of support.*

**T***(d) Calculate the torque due to the weight of the ball about this axis at an angle ##\theta##.*

From definition, ##\tau = r \times F##. At the angle ##\theta##, the only contributing force to the torque is the "vertical" component of weight, ##mg \sin \theta## (The tension and the "horizontal" component of weight ##mg \cos \theta## both lie along the rope). Hence the magnitude of torque ##\mathbb{\tau = mg \sin \theta l}##.

*(e) Calculate the magnitude of the rate of change of angular momentum of the pendulum bob at ##\theta##.*

If ##L = mvl## [see (a)], then ##\frac {dL} {dt} = m \frac {dv} {dt} l = m g \sin \theta l = mgl \sin \theta##, as the rate of change of

*v,*##\frac {dv}{dt} = g \sin \theta##, in a direction perpendicular to the rope and along the motion of the bob.

Thank you for you interest.