# Circular Motion (2)

1. Feb 21, 2015

### Jimmy87

1. The problem statement, all variables and given/known data

A pendulum with a bob on the end is attached to a stand. The stand has a rod sticking out such that when the string of the pendulum strikes it, it starts to undergo circular motion. Consider the bob being released from a height such that when it strikes the rod, it only just manages to complete one rotation (i.e. any smaller height and it would fail to complete a full rotation). If the tension in the string it zero when it is at its highest point discuss, without calculation, the tension in the string at its lowest point in terms of the weight of the bob.

2. Relevant equations
ac = v^2/r

3. The attempt at a solution

I would say that the tension in the string for question two must be higher than twice the weight. This is because the centripetal force at the top is mg if the bob is weightless. Assuming it remained at mg would mean that the centripetal force would be twice mg at the bottom. However, using conservation of energy, the bob must have more kinetic energy at the bottom and therefore more velocity and therefore a greater centripetal force than 2mg. So in conclusion this is not circular motion hence Fc at the bottom is more than 2mg, is that right?

2. Feb 21, 2015

### Staff: Mentor

As the bob rises some of its K.E. is converted into potential energy. You may be able to quantify this.

3. Feb 21, 2015

### Staff: Mentor

In terms of mg and r, what is the kinetic energy at the top of the arc? What is the kinetic energy at the bottom of the arc? What is v2/r at the bottom of the arc?

Chet

4. Feb 21, 2015

### Jimmy87

If the kinetic energy is changing does this mean that this will be non-uniform circular motion?

5. Feb 21, 2015

### Staff: Mentor

The bob is continually slowing as it rises, but predictably.

6. Feb 21, 2015

### Jimmy87

If it is continually slowing does this mean the circular motion is non-uniform?

7. Feb 21, 2015

### Staff: Mentor

Yes, it does mean that.

8. Feb 21, 2015

### Jimmy87

I have no idea how to do that in terms of m, g and r. Could you give some guidance? Thanks

9. Feb 21, 2015

### Staff: Mentor

Well, if, as you said, $m\frac{v^2}{r}=mg$, then what is $m\frac{v^2}{2}$ equal to? What is the change in potential energy between the top and bottom of the arc? What is the change in kinetic energy between the top and bottom of the arc?

Chet