# Homework Help: Centripetal acceleration turntable problem

1. Jul 29, 2015

### Zynoakib

1. The problem statement, all variables and given/known data
A small container of water is placed on a turntable
inside a microwave oven, at a radius of 12.0 cm from
the center. The turntable rotates steadily, turning one
revolution in each 7.25 s. What angle does the water
surface make with the horizontal?

2. Relevant equations
F = ma
a = v^2/ r

3. The attempt at a solution
Find the velocity of rotation:
0.12 x 2 x pi/ 7.25= 0.1 m/s

Find the centripetal acceleration:
(0.1)^2/ 0.12 = 0.09 ms^-2

Centripetal acceleration = 0.09 ms^-2 = horizontal acceleration
Gravitational acceleration = 9.8 ms^-2 = vertical acceleration

Angle make wit the horizontal surface
arctan (9.8/0.09) = 89 degree

I know it should have been arctan (0.09/9.8) = 0.526 degree (the actual answer), but in this way the centripetal acceleration would become the vertical acceleration whereas the gravitational acceleration would become horizontal acceleration.
What's wrong?

Thanks!

Last edited: Jul 29, 2015
2. Jul 29, 2015

### TSny

Did you mean to type arctan (.09/9.8)?
Anyway, is the water surface parallel to the effective total acceleration, or is the water surface perpendicular to the acceleration?

3. Jul 29, 2015

### Zynoakib

Yes, my mistake. Corrected.
The water surface should be parallel to the resultant acceleration but how is it related to my problem?

4. Jul 29, 2015

### TSny

Suppose the turntable is not rotating so that there is no centripetal acceleration. The effective acceleration is now just gravity. What angle does the acceleration of gravity make with respect to the horizontal? What angle does the surface of the water make with respect to the horizontal in this case? Is the surface of the water parallel to the acceleration of gravity or perpendicular to the acceleration of gravity?

When the turntable is rotating, then in the frame of reference of the rotating table you can introduce a "net effective acceleration vector" by combining the downward acceleration of gravity with the outward, horizontal "centrifugal" acceleration $v^2/r$. You should think about how the water surface is oriented relative to this net effective acceleration.

Last edited: Jul 30, 2015
5. Jul 30, 2015

### Zynoakib

if the table is not rotating, there should only be gravitational acceleration acting perpendicular to the horizontal water surface.

If the table is rotating, there water should move outward, forming a slope pointing toward the center like this, but that would be arctan (9.8/0.09) = 89 degree instead of the other way

6. Jul 30, 2015

### Qwertywerty

Do you know the formula for pressure change - ΔP = hρg ?

7. Jul 30, 2015

### TSny

The 89 degrees is the angle that the "net effective acceleration vector" makes to the horizontal. But that is not the angle that the water surface makes to the horizontal.

How is the surface of a lake oriented relative to the direction of the acceleration of gravity? What angle does the acceleration of gravity make relative to the horizontal and what angle does the surface of the lake make relative to the horizontal?

8. Jul 31, 2015

### Zynoakib

No, I don't

You mean something like that? where F should be replaced by the centripetal acceleration and P should be replaced by the gravitational acceleration?

9. Jul 31, 2015

### Qwertywerty

Tan θ has a direct relation to height y upon distance x ( Tan θ = y/x ) , not to the accelerations .

10. Jul 31, 2015

### TSny

No.
I think I’ve caused confusion by trying to approach the problem using the idea of an “effective acceleration of gravity”. I thought that was maybe the way you were approaching the problem in your first post.

Instead, we can go back to basics and apply Newton’s laws in the inertial frame of the earth. Consider a small chunk of water at the surface as shown in the diagram below. The chunk moves in uniform circular motion where the center of the circle, C, lies on the axis of rotation of the turntable.

There are two forces acting on the chunk: gravity (mg) and a normal force (N) that is due to the water surrounding the chunk. The net force acting on the chunk is the vector sum of these two forces, which is shown as the green vector in the figure at the far right. The net force must point towards C and it must have a magnitude $mv^2/r$.

Try to use this force diagram to determine the angle $\theta$ that the surface of the water makes to the horizontal.

#### Attached Files:

• ###### Turntable 1.png
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11. Jul 31, 2015

### haruspex

No, it can relate to magnitudes of forces (and thus accelerations) too.
But as TSny points out, centripetal force is a resultant, not an applied force. The applied forces are gravity and the normal force (acting on a parcel of surface water from the surrounding water).

12. Jul 31, 2015

### Qwertywerty

You mean like mg = N cos(θ) and N sin(θ) = mac ?

What I saw the OP doing was taking tan(θ) as g/ac , and so I tried to tell him that that's not how it works .

13. Jul 31, 2015

### haruspex

But it was then corrected to ac/g, which is right.

14. Jul 31, 2015

### Qwertywerty

By the OP ? Which post ?

15. Jul 31, 2015

### haruspex

By TSny. The OP's error is not in using tan theta as the ratio of the two accelerations but in getting the ratio the wrong way round. (Maybe by being confused about normal and parallel.) But the basic idea of equating the tangent to that ratio is valid.

16. Jul 31, 2015

### Qwertywerty

I think he got tan(θ) as g/ac , the same way you might use tan(θ) as y/x - ( vertical and horizontal distances ) .

Thus my post .

17. Jul 31, 2015

### haruspex

Ok, but your wording gave the impression you objected to applying arctan to the ratio of the accelerations on principle.

18. Jul 31, 2015

### Qwertywerty

I got it . However , I can't edit my post now .

19. Aug 3, 2015

### Zynoakib

Thank you guys so much!! Especially for the illustartion.

20. Aug 3, 2015

### Zynoakib

Hope you can clarify one thing for me, does that mean I cannot use centripetal acceleration and trigonometric ratio to calculate the angle and other applied force acting on an object? Just because centripetal force is a resultant force?

21. Aug 3, 2015

### haruspex

It depends how you arrived at the tan theta = ratio of accelerations equation. If it was a learnt response you need to be careful that the circumstances are appropriate. If you learnt it in the context of two applied forces but then use it where you have one applied and one resultant then it could turn out wrong. It would be more persuasive if you were to derive it in context from a standard sum of forces equation.

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