Circular Motion of Mass m in a Uniform Plane: Analyzing Forces

[UrbanXrisis]

A ball of mass m attached to a string of length L moves in a circle in a uniform vertical plane. When the ball is at the top of its path, the tension in the string is equal to its weight. Assume that air resistance is negligible.

1. Total force of the ball at the top in terms of m, L, and g:

Okay, total force in a circular motion diagram is:
Centripetal Force= mv^2/r

Centripetal Force=mg+tension
Since tension is the same as weight on the top of the circle, then:
Centripetal Force=mg+mg=2mg

is this the correct way of thinking?

2. speed at the ball at the top?

2mg=mv^2/r
v=2gr

correct?

3. speed at the bottom?

it would be the same as on the top right?

4. Tension at the bottom:

T=mg correct?

5. Speed of the ball when the string is horizonal:

It would be the same correct?

 

[fannemel]

3 and 5 are not correct. Consider using conservation of energy.

 

[UrbanXrisis]

what is conservation of energy? Could you expand on this a little more?

is it Total force=KE+PE?

 

[Tom McCurdy]

KEi+Ui=KEf+Uf

Energy must be conserved

 

[Tom McCurdy]

U=potential energy

 

[Tom McCurdy]

Set either Ui or Uf to zero... zero being the bottom of the loop sing U=mgy and y would be zero

 

[fannemel]

Conservation of energy means in this case (actually most of the time friction and such forces aren't applied) that:

$$\frac{1}{2}mv^2 + mgh = k$$

where k is a constant

if you use that you will find that

$$\frac{1}{2}mv^2_t + mgh = \frac{1}{2}mv^2_b$$

where h = 2r (the difference in height between the two points)

you can use conservation of energy in the same way to calculate the speed when the string is horizontal.

oh, one more thing. 4. is also not correct you must insert the new speed calculated at the bottom to get it correct.

 

[Pyrrhus]

Actually to be more precise with what Fannelmel (his definition only considers potential gravitational energy) is saying

Conservation of Mechanical Energy says:

This works for conservative systems (For Example: those without friction)

$$\Delta \Omega + \Delta K = 0$$

I see you are not familiar with this, that's because you don't need it to solve the problem.

Now for the problem analysis:

At the top the ball will be affected by the tension and its weight pointing the same way, so

$$T + mg = m \frac{v^2}{r}$$

Its speed at the top will be

$$\sqrt{\frac{Tr}{m} + gr} = v$$

There is still more to do because the problem states at top $T = mg$

At the bottom the will be affected by the tension and its weight pointing the opposite ways, so

$$T - mg = m \frac{v^2}{r}$$

so its speed will be

$$\sqrt{\frac{Tr}{m} - gr} = v$$

When the ball is horizontal (forms an angle of 90 degrees with the vertical)

$$T = m \frac{v^2}{r}$$