Displacement vector on a watch?

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The discussion revolves around calculating the displacement vector of a watch's minute hand from 8:00 am to 8:20 am, with the minute hand being 2.0 cm long. The initial formula used for the calculation was incorrect, as it was meant for an hour hand rather than a minute hand. The correct approach involves determining the position vectors at both times, with the angle of rotation being 120 degrees for the 20 minutes elapsed. The final displacement vector is derived by subtracting the initial position from the final position, resulting in coordinates of approximately (1.73, -3.00). The importance of correctly identifying the formula and understanding the geometry of the problem was emphasized throughout the discussion.
skysunsand
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Homework Statement



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.

Homework 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

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. :(
 
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skysunsand said:

Homework Statement



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.

Homework 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

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. :(

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.
 
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
 
PeterO said:
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.

Oh, my fault. Allow me to further clarify-

1. Homework Statement

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. Homework 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.
 
rl.bhat said:
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


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?
 
skysunsand said:
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?

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.
 
PeterO said:
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.

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)
 
skysunsand said:
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)

You said your calculator was set to the appropriate radians or degrees.
Perhaps you don't realize 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
 
PeterO said:
You said your calculator was set to the appropriate radians or degrees.
Perhaps you don't realize 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

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
skysunsand said:
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)

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 0EDIT: and not approximately zero; exactly zero
 
  • #11
skysunsand said:
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)

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:
  • #12
The answer was 1.73,-3.00. Why??
 
  • #13
skysunsand said:
The answer was 1.73,-3.00. Why??

even you subtracted 2j from your (incorrect) displacement vector. why did you do that?
 
  • #14
PeterO said:
even you subtracted 2j from your (incorrect) displacement vector. why did you do that?

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
skysunsand said:

Homework Statement



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.

Homework 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

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. :(


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
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).

And the answer is:

dV = sqrt{3}i - 3j.
 

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