# Homework Help: Displacement vector on a watch?

1. Aug 28, 2011

### skysunsand

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

The minute hand on a watch is 2.0cm in length. What is the displacement vector of the tip of the minute hand from 8:00 am to 8:20 am?
It wants it in x and y components.

2. Relevant equations

I found this equation online- i(L sin 2pi(T/12)) + j ((L cos 2pi (T/12))

where L is length, T is time in hours
3. The attempt at a solution

Tried plugging my numbers into the equation- 2 for L, and 1/3 for T, since it's 20 minutes. And then each "side" (sin or cos) would be the components of the vector. But not so, something is obviously not correct. Help please? Physics is terribly discouraging. :(

2. Aug 28, 2011

### PeterO

I highlighted part of your OP.
To the rest of us there not something obviously incorrect, as you have not shown even your calculated answers, let alone your working [what you were substituting in where] ??
It may have been as simple as having your calculator set to radians not degrees, or vice-versa.

3. Aug 28, 2011

### rl.bhat

Position vector at 8.00 am is V1 = L( 0 + j)
Position vector at 8.20 am is V2 = i(L sin 2pi(T/12)) + j ((L cos 2pi (T/12))
So the displacement vector Vd = V2- V1

4. Aug 28, 2011

### skysunsand

Oh, my fault. Allow me to further clarify-

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

The minute hand on a watch is 2.0cm in length. What is the displacement vector of the tip of the minute hand from 8:00 am to 8:20 am?
It wants it in x and y components.

2. Relevant equations

I found this equation online- i(L sin 2pi(T/12)) + j ((L cos 2pi (T/12))

where L is length, T is time in hours
3. The attempt at a solution

Tried plugging my numbers into the equation- 2 for L, and 1/3 for T, since it's 20 minutes. And then each "side" (sin or cos) would be the components of the vector.

i (2 sin 2pi ((1/3)/12) + j((2 cos 2pi ((1/3)/12)) and put it into my calculator to get 2.00 as my final answer for the whole thing. My calculator is set correctly, that I do know.
Then in thinking each of them were split, for the sin side I got 0.005 and for the cos side I got 1.99.

5. Aug 28, 2011

### skysunsand

at 8:00 am, plugged in 2 for L to get V1=2j

Then at 8:20 am, i (2 sin 2pi ((1/3)/12) + j ((2 cos 2pi ((1/3)/12)), to have V2 = .005i + 1.99j.

So V2-V1 would be (.005i+1.99j)-(2j) to get the answer of the coordinates (.005,-.001), right?

6. Aug 28, 2011

### PeterO

Looks more like the tip of a 2cm hour hand!

Try substituting T = 1 and see if the hand is back where it started, because a minute hand certainly is.

7. Aug 28, 2011

### skysunsand

at 8:00 am, plugged in 2 for L to get V1=2j

Then at 8:20 am, i (2 sin 2pi ((1/3)/12) + j ((2 cos 2pi ((1/3)/12)), to have V2 = .005i + 1.99j.

So V2-V1 would be (.005i+1.99j)-(2j) to get the answer of the coordinates (.005,-.001), right?

For T=1

at 8am, 2 for L, 1 for T=
V1=2j

V2= i(2sin2pi(1/12)) + j(2cos2pi(1/12)) = 0.016 i + 1.999j

Then V2-V1 = 0.016i + 1.999j -2j still gets me (0.016i-.001j)

8. Aug 28, 2011

### PeterO

You said your calculator was set to the appropriate radians or degrees.
Perhaps you don't realise whether the following is in radians or degrees
2sin2pi(1/12)
I'll give you a hint - the value of 2sin2pi(1/12) is 1.0

9. Aug 28, 2011

### skysunsand

Oh. Well then it was a calculator problem...

Then I should go back to plugging in 1/3 for my T and then I should have the correct answer, right?

at 8:00 am, plugged in 2 for L to get V1=2j

Then at 8:20 am, i (2 sin 2pi ((1/3)/12) + j ((2 cos 2pi ((1/3)/12)), to have V2 = .347i + 1.96j.

So V2-V1 would be (.347i + 1.96j) - 2j to get the answer of the coordinates (.347, -.04)

10. Aug 28, 2011

### PeterO

Those numbers are still wrong???

I think the formula is not the formula for the end of a minute hand.

use that formula to find out the displacement from 8:00 until 9:00 - because the answer had better be 0

EDIT: and not approximately zero; exactly zero

11. Aug 28, 2011

### PeterO

at 8:20, the j component of the tip of the minute hand is negative for a start ???

I reckon it is 1.7i - j, or more accurately sqrt(3)i - j

Last edited: Aug 28, 2011
12. Aug 28, 2011

### skysunsand

13. Aug 28, 2011

### PeterO

even you subtracted 2j from your (incorrect) displacement vector. why did you do that?

14. Aug 28, 2011

### rl.bhat

In the problem the displacement vector of the tip of the minute hand from 8:00 am to 8:20 am is required.
If OA be the position vector at 8.00 am and OB be the position vector at 8.20 am, then the displacement vector is AB = OB - OA.

15. Aug 28, 2011

### PeterO

Now that you know the solution, and how it was found: I offer the following:

Once a "random" formula you found on-line resulted in an incorrect answer you should have been shown something.

Either this is not the formula you thought it was [you should now know it wasn't: this formula was for an hour hand], or you don't know how to use the formula.

How you should have solved this problem.

Draw a clock face with two minute hands, one at 8:00 and one for 8:20 [don't bother drawing the hour hands - and in case you are wondering this hour-hand-less clock would show the same thing if you were drawing 10:00 and 10:20]

What is the angle between the two hands ?

Imagine x-y axes drawn over the clock fact. The co-ordinates of the end of the 8:00 hand are (0,2)

To get to 8:20, we rotate 120 degrees.
The new x-coordinate is simple 2*sin120 = 1.732
The new y-coordinate is 2*cos120 = -1

to get from (0,2) to (1.732,-1) you move 1.732 right (in the x direction) and 3 units down (-3 in the y - direction)

with the 120o angle involved, you don't even need sine and cosine functions, these values relate to the 60-30 or 2,1,root(3) triangle. The hypotenuse [hand] is even two units long so the x and y coordinates are simply 1 and root(3)

16. Aug 29, 2011

### perucho

That's correct Peter!

Origin at watch's center.

y axis at 8:00 ------> vector V_1 = 0i + 2j.

x axis at 8:15 ----> 15 min. = (1/4)60 min ----> (1/4)360 deg. = 90 deg.

20 min = (1/3)60 min. -----> (1/3)360 deg. = 120 deg.

angle among 8:15 and 8:20 = 120 - 90 = 30 deg (below x).

vector V_2 = 2(cos30i - sin30j) ----> V_2 = sqrt{3}i - j (2 cancel).

By definition,

dV = V_2 - V_1 = (sqrt{3}i - j ) - ( 0i + 2j).