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.