Kinematics of a Watch's Second Hand: Calculating Speed and Velocity

[mattstjean]

Homework Statement

A watch has a second hand 2.0cm long.
a) compute the speed of the tip of the second hand.
b) What is the velocity of the tip of the second hand at 0.0s and 15s

Homework Equations

No Idea...

The Attempt at a Solution

I drew a clock...but I don't even know where to start, any tips would be great.

 

[Kruum]

All you need to know is that the clock is circle and circles have a certain number of degrees or radians. From there you can determine the length the hand moves every second. Also s=rθ, ω=∆θ/∆t and v=rω.

 

[tiny-tim]

Hi mattstjean!

How long does it take the second hand to go round once?

How far does the second hand move in that time?

 

[mattstjean]

Hi mattstjean!

How long does it take the second hand to go round once?

How far does the second hand move in that time?

Thanks. It seems the simplest questions give me the hardest trouble. I always tend to over-complicate something that ends up being a lot more straight forward.

 

[abhikesbhat]

Im not sure but the answer to the first one is pi/15 cm/sec

The second one is (2qrt2)/15. I don't understand which way the velocity goes. Is it the way from the starting point? If it is then it is SouthEast?

 

[mattstjean]

Well, it's not as simple as I thought ... I think I got the first part right. I did:

r= 2cm (because the second hand is 2cm long and it is ask about from the tip)

c=[tex]\pi\ [/tex] r 2
c = 12.5663706 cm

V = c/t
V = 12.5663706cm / 60s
V = 0.209439510 s

So I said the speed of the tip of the second hand is 0.21 cm/s. Does that sound right?

As for the second part. I don't really know how I should go about finding the velocity. I'm thinking it might be 0.21cm/s [E] because it's right at the top.

Any tips for the velocity would be nice. I don't understand what the last guy tried to say.

 

[mattstjean]

Im not sure but the answer to the first one is pi/15 cm/sec

The second one is (2qrt2)/15. I don't understand which way the velocity goes. Is it the way from the starting point? If it is then it is SouthEast?

I'd rather be steered in the right direction with some hints and tips rather than given the answer. I'll die on my test without knowing how to arrive on the answer. Could you tell me how you get to that answer for the second one? And what does (2qrt2)/15 mean...

 

[mattstjean]

For the second one I'm thinking that the velocity at 0s would be 0.21 cm/s [E] because it's at the very top and at 15s I think it's 0.21cm/s because it's at the very right.

I was also asked to compute the change in velocity between 0.0 and 15 seconds and the average vector acceleration between 0.0 and 15 seconds.

For the change in velocity I did:

r= [tex]\sqrt{}0.21^2+0.21^2[/tex]
r=0.3cm/s

tan[tex]\theta[/tex] = 0.21/0.21
[tex]\theta[/tex] = 45degrees

so the change in velocity would be 0.3cm/s [45 degrees E of N]

I don't really know what to do for acceleration...I'm assuming it is 0 because of the constant magnitude of the velocity ...

Any help would be greatly appreciated,
Matt

 

[abhikesbhat]

I'd rather be steered in the right direction with some hints and tips rather than given the answer. I'll die on my test without knowing how to arrive on the answer. Could you tell me how you get to that answer for the second one? And what does (2qrt2)/15 mean...

i meant (2sqrt2)/15. OK what we know is that the average speed is distance/time. We can easily set the time to 60 seconds. The distance the second hand travels is the circumference because it makes a revolution every 60 seconds. To find the circumference we know radius is 2. And you hopefully know circumference formula. So we get 4pi/60 and that simplifies to pi/15 which is about .21 cm/sec.

 

[Office_Shredder]

matt, your velocities look good, except:

Considering your velocity went from pointing east to pointing south, I would imagine your change in velocity vector (taking the south and subtracting the east) would give you something pointing in the south-west direction.

To get the average acceleration, just divide the change in velocity by the time that it takes to change