# Physics lab question

1. Dec 18, 2005

### yosup231

I have to do the lab on

http://www.la.mvla.net/Curric/science/APPhysics/labs/mlab7.htm

ive calculated the linear speed, centripetal acceleration, period of rotation, mass oif the toy plane, the length of string, and the radius of rotation but ive calculated everything experimentally. The lab also asks to compare the theoretical values to the experimental values. How do i calculate the theoretical period of rotation.

Last edited by a moderator: Apr 21, 2017
2. Dec 18, 2005

### Staff: Mentor

Apply Newton's 2nd law to the toy plane, realizing that the plane is executing circular motion and thus is centripetally accelerated. (Write the equations for vertical and horizontal forces. Solve them together and you'll be able to calculate the speed in terms of L and R.)

3. Dec 18, 2005

### yosup231

Confused??

maybe im dumb or something because i still have no idea how to THEORETICALLY calculate the period of rotation

4. Dec 18, 2005

### Staff: Mentor

The period depends on the length of the string and the angle it makes (or, in other words, on L and R). Did you do what I suggested above? Once you find the speed (using Newton's 2nd law) you can use it to calculate the period.

There's no single answer, if that's what you are thinking. It depends on how fast the plane moves (which will change the angle).

5. Dec 18, 2005

### yosup231

ugh??

obviously physics isnt my best subject.... assuming the length of the string is .49m and the radius of rotation is .3 m what do i do next to calculate the period (im sure you already told me what to do i just don't understand stand it so if you could put it in laymens terms it would be very appreciated)

6. Dec 18, 2005

### Staff: Mentor

You're going to calculate the speed that the plane must have under those conditions.

(1) Identify all the forces on the plane (there are only two)
(2) Apply Newton's 2nd law to the vertical and horizontal components of the forces.

Don't forget that the acceleration is centripetal. (What's the formula for the centripetal acceleration in terms of radius and speed?)

You'll solve those two equations to find the speed of the plane. Then use that to solve for the period, which is just the time it takes the plane to make one pass around the circumference of the circle.

7. Dec 18, 2005

### yosup231

ok so...

1.the two forces on the plane are weight and tension
2 f=ma
3 centripetal acceleration is m(v^2)/r

im not sure how to solve these equations since i dont know what the velocity is

8. Dec 18, 2005

### Staff: Mentor

So far, so good. Now find the vertical and horizontal components of the forces (draw yourself a diagram) and apply F = ma to each direction.

When you solve the equations, you'll find the speed.

9. Dec 18, 2005

### yosup231

umm

well i think the force of weight is only vertical so when the formula f=ma is used you get (.0515kg)(-9.8m/s^2)=-.5047N

and for the tension imnot sure i know the length of the string is .49 m and the radius is .3m but im not exactly sure what to do with them since it forms an angle of 30 would it .3sin30=.15N and .49cos30=.424N

10. Dec 18, 2005

### Staff: Mentor

Don't be too quick to plug in numbers; just use symbols until the last step.

The weight of the plane = mg.
Call the tension T; find the vertical and horizontal components in terms of sines and cosines.

What must the vertical forces add to?
What must the horizontal forces add to?

11. Dec 18, 2005

### yosup231

ok...

well i think what i said was right then for the weight and then for the tension would it be the vertical tension equals the Ft times cos(angle) and then the horizontal tension equals the ft times sin angle???

12. Dec 18, 2005

### yosup231

??

what should i do?

13. Dec 18, 2005

### yosup231

confused

If anyone can help me, it would be greatly appreciated.

Last edited: Dec 18, 2005
14. Dec 22, 2005

### joex444

uniform circular motion, so the plane doesn't descend. Therefore, vertical acceleration is 0. So, your weight is all vertical, and since acceleration vertical has to be 0, the Ty = -mg. Now, on the string there is a vertical and, well, let's say horizontal (perhaps tangential is a better term?), components of it. You already solved for Ty, and if you knew the angle the plane makes with the axis of rotation you could find Tx, from which you can find Ax, and since $$a=\frac{v^2}{R}$$ the speed of the plane. Hoping this helps...