What is the correct angle of displacement for a clock arm?

[rudransh verma]

Homework Statement

The minute hand of a wall clock measures 12cm from its tip to axis. The magnitude and angle of the displacement vector of tip are to be calculated for three intervals. What are the A) magnitude and B) angle from quarter after the hour to half past, C) for next half hour and D) for the hour after that?

Relevant Equations

Delta r= r2-r1

I calculated the angle as 225 whereas answer says something else.

 

[berkeman]

Please show your work, and please re-post your image rightside-up and scanned at higher quality so we can read it. Thank you.

 

[rudransh verma]

Please show your work, and please re-post your image rightside-up and scanned at higher quality so we can read it. Thank you.

Edited

 

[Steve4Physics]

I calculated the angle as 225 whereas answer says something else.

Assuming:
- the +x-axis runs from the dial’s centre in the direction of 3 o’clock (for a 12 hour clock);
- angles are measured anticlockwise relative to the +x axis;
then I agree: the angle of displacement of the tip from quarter past the hour to half past the hour is 225º.

What is the ‘official’ answer?

Also, note that your homework statement in Post #1, has 4 parts (labelled A to D). But the image of the question from the textbook shows 6 parts (labelled a to f).

 

[rudransh verma]

What is the ‘official’ answer?

You can follow in image numbering.

 

[Steve4Physics]

Here are some comments on the ‘official’ answers in the Post #5 attachment. Just my opinions though.

The units for the magnitudes of displacements should be cm. The units are missing (a very basic mistake!).

a) $12\sqrt 2$ should be $12\sqrt 2$ cm.

b) 135º only makes sense if the angle is measured anticlockwise from the 12 o'clock direction.

c) $24$ should be $24$ cm.

d) If the angles are measured as for part b) (above), then the answer should be 0º, not 180º.

e) Answer ‘e’ is mislabelled ‘f’ but is correct.

f) Answer ‘f’ is mislabelled ‘g’. Since the displacement has zero magnitude we should really say ‘θ is undefined’, not θ = 0.

What text-book/source are these questions/answers from?

 

[rudransh verma]

135º only makes sense if the angle is measured anticlockwise from the 12 o'clock direction.

This is the fig. The angle should be -135.
Also it is coming -45. Is it right? Because we take - angle when measuring clockwise.

Also this 180 degree doesn’t make sense. Because we measure angles from +x axis

 

[rudransh verma]

What text-book/source are these questions/answers from?

Principles of physics Resnik. 2D motion

 

[rudransh verma]

@berkeman can you clear my doubt in post 7

 

[haruspex]

135º only makes sense if the angle is measured anticlockwise from the 12 o'clock direction.

Or if measured clockwise from 3 o'clock.

d) If the angles are measured as for part b) (above), then the answer should be 0º, not 180º.

I make it 90° if measured anticlockwise from 3 o'clock. The given answer seems to be the rotational angle undergone by the hand, not the orientation of the displacement vector.

 

[rudransh verma]

Or if measured clockwise from 3 o'clock.

Every angle of the displacement will be measured from 3 or +x axis. @Steve4Physics is walking on different path.

I make it 90° if measured anticlockwise from 3 o'clock.

For the first sweep the angle should be -135. For second and third it should be 90.
But the answer says different.
The answer given are measuring the sweep angle , the angle between the arms (well there are not two arms, you understand) but we have to find the angle of the displacement vector.

 

[rudransh verma]

The given answer seems to be the rotational angle undergone by the hand, not the orientation of the displacement vector.

Yeah!
What will be the displacement vector magnitude for the next half hour(6oclock to 12 o clock). I don’t understand the answer. There is no triangle formation here.

 

[haruspex]

Yeah!
What will be the displacement vector magnitude for the next half hour(6oclock to 12 o clock). I don’t understand the answer. There is no triangle formation here.

There is a degenerate triangle, having sides 12, 12, 24. But that is not the reason they've written it that way. What seems to have been intended is $|\vec r|=\sqrt{\vec r.\vec r}=\sqrt{24\hat y .24\hat y }=\sqrt{24^2}$.

 

[rudransh verma]

There is a degenerate triangle, having sides 12, 12, 24.

what is a degenerate triangle?

What seems to have been intended is |r→|=r→.r→=24y^.24y^=242.

I don't understand the math

 

[haruspex]

what is a degenerate triangle?

One in which the long side equals the sum of the other two. It collapses to a straight line.

 

[rudransh verma]

One in which the long side equals the sum of the other two. It collapses to a straight line.

Ok. what is it with the square root of dot product? What are you doing there?

 

[haruspex]

Ok. what is it with the square root of dot product? What are you doing there?

That is how the modulus of a vector is defined: $|\vec r|=\sqrt{\vec r.\vec r}$.

 

[rudransh verma]

That is how the modulus of a vector is defined: $|\vec r|=\sqrt{\vec r.\vec r}$.

Modulus is defined as r=under root(r1^2 +r2^2).
In which book it is defined like this?

 

[haruspex]

Modulus is defined as r=under root(r1^2 +r2^2).
In which book it is defined like this?

https://en.m.wikipedia.org/wiki/Vector_space#Normed_vector_spaces_and_inner_product_spaces

 

[rudransh verma]

https://en.m.wikipedia.org/wiki/Vector_space#Normed_vector_spaces_and_inner_product_spaces

Seems higher grade stuff.

 

[haruspex]

Seems higher grade stuff.

The entire question seems to be part of an introduction to vectors, and this definition of the modulus is fairly basic in that context.
The Wikipedia discussion may seem higher grade because it makes the distinction between vector spaces in general (which need not have anything corresponding to a dot product) and inner product spaces.

 

[hmmm27]

Is there a definitive answer on the angles ? I get 135, 0 and not-applicable, based on the assumption that a "displacement vector" angle is an absolute to the coordinate system, and "12" on a clock-face seems a good place to have a 0/360.

 

[haruspex]

I get 135, 0 and not-applicable

Where are you measuring the angles from? Isn't anticlockwise from the 3 o'clock position standard?

 

[hmmm27]

Where are you measuring the angles from? Isn't anticlockwise from the 3 o'clock position standard?

The question doesn't mention Cartesian xy coordinate system. 2d polar, clockwise from noon, seemed appropriate for a clock.

For that matter, given precedent of WWII air-battle movies, I'd prefer to say 7:30, 12:00 and n/a.

Wonder where the OP sits on the second and third parts of the question.

 

[haruspex]

The question doesn't mention Cartesian xy coordinate system.

It doesn’t, true, but the preceding three questions all use Cartesian coordinates.

 

[rudransh verma]

Is there a definitive answer on the angles ? I get 135, 0 and not-applicable, based on the assumption that a "displacement vector"

I got -135, 90, 90 from x axis

 

[haruspex]

As

I got -135, 90, 90 from x axis

has been pointed out a couple of times, the angle for the final case is indeterminate since the vector has zero magnitude.