How Does the Hour Hand's Velocity and Acceleration Change from Noon to 6pm?

[negation]

Homework Statement

What are a) average velocity and b) average acceleration of the tip of the 2.4cm long hour hand clock in the interval from noon to 6pm. Use unit vector to express, wth the x-axis pointing towards 3 an y-axis towards noon.

Homework Equations

The Attempt at a Solution

V = deposition/dt

P2 = -2.4j
P1= 2.5j
P2-p1 = -4.8
-4.8e-3/21600= -2.22e-6j

b) can I have a leg up?
I know a = dv/dt but I can't build a mental geometrical set up.

 

[voko]

Check your answer for (a). You seem off by a degree of magnitude.

For (b), you need initial and final velocities. What are they?

 

[negation]

Check your answer for (a). You seem off by a degree of magnitude.

For (b), you need initial and final velocities. What are they?

A) it should be -4.8e-2/21600= -2.22e-6

B)
I'll attempt it procedurally:
Vi = -2.22e-6 sin 90 j
Vf -2.22e-6 sin (-90) j

 

[voko]

How did you obtain the initial and final velocities?

 

[negation]

How did you obtain the initial and final velocities?

At 12 noon, the minute hand is at pi/2
Since this is purely vertical; vi = -2.22e-6 sin pi/2
At 6pm, the minute hand is at -pi/2
This is also purely vertical; hence, vf = -2.22e-6 sin(-pi/2)

 

[voko]

Does the tip of the hour hand move vertically at 12 and 6? And where does the magnitude of velocity come from?

 

[negation]

Does the tip of the hour hand move vertically at 12 and 6? And where does the magnitude of velocity come from?

It comes from part (a).
I presume it does since in moving from 12 to 6, the unit vector goes from +J to -J

 

[voko]

Part (a) is about the average velocity about these two positions. Part (b) requires instantaneous velocities at these two positions.

 

[negation]

Part (a) is about the average velocity about these two positions. Part (b) requires instantaneous velocities at these two positions.

I believe my part(a) is correct. The answer sheet confirms it.
As for part(b), I do not know how to go about setting up the question.
I know a = dv/dt but I do not know how should I find vf and viEdit: Do I assume the velocity is the angular velocity or linear velocity?

 

[negation]

a = dv/dt
v = ds/ dt
ds = rΘ/dt=rω
v = rω/dt
∴a = d(rw)/dt

 

[Chestermiller]

Check out again what Voko said in post #8. What is the instantaneous velocity vector at 6 pm (in terms of the unit vector in the x- direction)? What is the instantaneous velocity vector at noon? What is the difference between these two vectors?

Chet

 

[negation]

Check out again what Voko said in post #8. What is the instantaneous velocity vector at 6 pm (in terms of the unit vector in the x- direction)? What is the instantaneous velocity vector at noon? What is the difference between these two vectors?

Chet

At 6pm: (0,-2.22e-6j)
At 12pm (0,2.22e-6j)

 

[voko]

Edit: Do I assume the velocity is the angular velocity or linear velocity?

Part (b) needs linear velocity. But it is related to angular velocity. Again, what are the directions of the instantaneous velocity of the tip of the hour hand at 12 and 6? Image for a second that the hour hand rotates at a visible pace. Over a very short arc at 12 and 6, where is it going?

 

[negation]

Part (b) needs linear velocity. But it is related to angular velocity. Again, what are the directions of the instantaneous velocity of the tip of the hour hand at 12 and 6? Image for a second that the hour hand rotates at a visible pace. Over a very short arc at 12 and 6, where is it going?

Δv is headed towards the center
v on the other hand is perpendicular to the hour hand.

 

[voko]

$v_i$ and $v_f$ are indeed perpendicular to the hour hand. But what are their magnitudes and signs?

 

[Chestermiller]

At 6pm: (0,-2.22e-6j)
At 12pm (0,2.22e-6j)

These are not the instantaneous velocities at those times. The instantaneous velocities of the tip of the hour hand at those times multiply i, not j. The angular velocity is 2π radians divided by 43200 seconds.

 

[negation]

These are not the instantaneous velocities at those times. The instantaneous velocities of the tip of the hour hand at those times multiply i, not j. The angular velocity is 2π radians divided by 43200 seconds.

Why i and not j?
Why 43200s? If the hand moves from 12 to 6, then effectively, the change in radians is pi; therefore, v = rω
r = 2.4cm A ω = pi/6(60x60) = pi/21600s
v = 2.4cm x pi/6(60x60)

Edit: I think you might be right with instantaneous velocity having an i direction.
At 12pm, the instantaneous velocity has a +i unit vector; at 6pm, it has a -i unit vector.
But my contention against 43200s and theta = 2pi stands.

 

[negation]

Is instantaneous velocity at 12pm = (-2.22j,0) and at 6pm = (2.22j,0) ?

 

[negation]

$v_i$ and $v_f$ are indeed perpendicular to the hour hand. But what are their magnitudes and signs?

(2.22e-6i,0) and (-2.22e-6i,0) ?

 

[Chestermiller]

Why i and not j?
Why 43200s? If the hand moves from 12 to 6, then effectively, the change in radians is pi; therefore, v = rω
r = 2.4cm A ω = pi/6(60x60) = pi/21600s
v = 2.4cm x pi/6(60x60)

Edit: I think you might be right with instantaneous velocity having an i direction.
At 12pm, the instantaneous velocity has a +i unit vector; at 6pm, it has a -i unit vector.
But my contention against 43200s and theta = 2pi stands.

2π divided by 43200 is the same thing as π divided by 21600.

 

[Chestermiller]

(2.22e-6i,0) and (-2.22e-6i,0) ?

Good. So what's the average acceleration?

 

[negation]

Good. So what's the average acceleration?

[(2.22e-6i,0)-(-2.22e-6i,0)]/21600s

 

[negation]

2π divided by 43200 is the same thing as π divided by 21600.

It is but the mathematical reasoning comes to me as very intuitively strange

 

[Chestermiller]

It is but the mathematical reasoning comes to me as very intuitively strange

My reasoning was that the hour hand makes one complete revolution (2π radians) every half-a-day. Please tell me why that seems strange.

Chet

 

[Chestermiller]

[(2.22e-6i,0)-(-2.22e-6i,0)]/21600s

Yeah, but, c'mon, let's see the final vector.

Chet

 

[negation]

Yeah, but, c'mon, let's see the final vector.

Chet

I'm getting 2.05e-10 which apparently does not correspond to the book's answer

 

[negation]

My reasoning was that the hour hand makes one complete revolution (2π radians) every half-a-day. Please tell me why that seems strange.

Chet

Because while 2pi/43200 = pi/21600 is tautological, one would intuitively take pi/21600 givens the question. But it's no biggie either way

 

[negation]

It should be:

vi = (-2.22e-6i,0) because at 12 position, acceleration is in the -j direction and velocity in the +i direction
vf = (0,2.22e-6j) because at 6 position, acceleration is in the -i direction and velocity in the -j direction
The magnitude works out to be 3.2e-6 and this corresponds to the book.
However, I cannot make sense of the unit vector.

 

[Chestermiller]

I'm getting 2.05e-10 which apparently does not correspond to the book's answer

This is because your instantaneous velocities are wrong for noon and 6 pm. You had the correct results for the instantaneous velocities in post #17: ωr=(0.00015)x2.4=0.000349 cm/sec =3.49E-7 m/sec. So the instantaneous velocities are 3.49E-7i at noon and -3.49E-7i at 6 pm. So the average acceleration is -3.23E-11i m/sec². This compares with the exact instantaneous velocity at 3pm (the half-way time) of -5.08E-11i.

Chet

 

[negation]

This is because your instantaneous velocities are wrong for noon and 6 pm. You had the correct results for the instantaneous velocities in post #17: ωr=(0.00015)x2.4=0.000349 cm/sec =3.49E-7 m/sec. So the instantaneous velocities are 3.49E-7i at noon and -3.49E-7i at 6 pm. So the average acceleration is -3.23E-11i m/sec². This compares with the exact instantaneous velocity at 3pm (the half-way time) of -5.08E-11i.

Chet

I did an update prior to this current post. I got the right magnitude. However, I cannot make sense of the unit vector.

 

[Chestermiller]

I did an update prior to this current post. I got the right magnitude. However, I cannot make sense of the unit vector.

If the book didn't match my result in post #29, then the book must be wrong. Now for the direction of the unit vector. In what direction is the tip of the hour hand moving at noon (i or j)? In what direction is the tip of the hour hand moving at 6 pm (-i or -j)? (j is the up-down direction, and i is the right-left direction).

 

[negation]

If the book didn't match my result in post #29, then the book must be wrong. Now for the direction of the unit vector. In what direction is the tip of the hour hand moving at noon (i or j)? In what direction is the tip of the hour hand moving at 6 pm (-i or -j)? (j is the up-down direction, and i is the right-left direction).

It struck me while I was in the toilet I made a blunder.

At 12 position: (3.5e-6,0)
Velocity is in the +i direction, j is irrelevant here since its magnitude is 0.
At 6 position: (-3.5e-6i,0)
Velocity is in the -i direction.
p(6) - p(12) = (-7e-6,0)/21600

what I got now is -3.24 e-10. This exactly coincides with the book.

Now, it would be helpful if you could shed some light on the conceptual difference between
v = -2.22e-6 in part(a) and v = rω = 3.5e-6
While I under v = rω implies linear velocity, I do not understand why it is not mathematically valid to use v = -2.22e-6 instead of 3.5e-6

 

[Chestermiller]

It struck me while I was in the toilet I made a blunder.

At 12 position: (3.5e-6,0)
Velocity is in the +i direction, j is irrelevant here since its magnitude is 0.
At 6 position: (-3.5e-6i,0)
Velocity is in the -i direction.
p(6) - p(12) = (-7e-6,0)/21600

what I got now is -3.24 e-10. This exactly coincides with the book.

Now, it would be helpful if you could shed some light on the conceptual difference between
v = -2.22e-6 in part(a) and v = rω = 3.5e-6
While I under v = rω implies linear velocity, I do not understand why it is not mathematically valid to use v = -2.22e-6 instead of 3.5e-6

These are very good questions. The velocity is changing with time (its direction is changing), so the average velocity between the initial time and the final time is only an approximation to the instantaneous velocity at some point between the initial time and the final time. The time at which the average velocity best approximates the instantaneous velocity is at the half-way point, 3 pm. At 3 pm, the instantaneous velocity is -3.5E-6j, while the average velocity is -2.22E-6 j. So the average velocity gives the correct direction for the instantaneous velocity at the half way point, and it is only about 30% lower in magnitude. This is pretty good, considering the huge time interval and the huge change in the displacement vectors over this time interval.

The same approach applies to the acceleration. The average acceleration is -3.23E-11i over the 6 hour interval between noon and 6 pm, while the instantaneous acceleration a at 3pm is -5.08E-11i. So the average acceleration gets the direction of the instantaneous acceleration at the half-way point correct, and it's only about 40% lower in magnitude. This is again pretty good, considering the huge time interval and the huge change in the instantaneous velocity vectors over this time interval.

What we have been dealing with here are finite difference approximation formulas to the instantaneous velocity and instantaneous acceleration from finite changes in displacement and velocity over a finite time interval. The finite difference formula is second order accurate at the half-way point.