What is the magnitude of acceleration for a minute hand in circular motion?

[dragonladies1]

Hi all,

I'm having a very difficult time with this problem. I know that the magnitude stays the same and it involves uniform circular motion, but I just can't seem to figure out how to begin the problem or what formula to use. I would really appreciate any kind of help.

A wall clock has a minute hand with a length of 0.47 m and an hour hand with a length of 0.24 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x-axis pointing to 3 o'clock and the positive y-axis pointing to 12 o'clock. What is the magnitude of the acceleration of the tip of the minute hand of the clock?

I also need to express it as a fraction of the magnitude of free-fall acceleration g.

Again thank you all very much, any kind of help would be so much help.

 

[berkeman]

Hi all,

I'm having a very difficult time with this problem. I know that the magnitude stays the same and it involves uniform circular motion, but I just can't seem to figure out how to begin the problem or what formula to use. I would really appreciate any kind of help.

A wall clock has a minute hand with a length of 0.47 m and an hour hand with a length of 0.24 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x-axis pointing to 3 o'clock and the positive y-axis pointing to 12 o'clock. What is the magnitude of the acceleration of the tip of the minute hand of the clock?

I also need to express it as a fraction of the magnitude of free-fall acceleration g.

Again thank you all very much, any kind of help would be so much help.

What is the equation for the centripetal acceleration of an object (the tip of the clock hand) as a function of radius and velocity?

 

[dragonladies1]

Centripetal acceleration's equation is but I've tried plugging the numbers into that equation and can't seem to get it right.

a= V^2/r

 

[berkeman]

Centripetal acceleration's equation is but I've tried plugging the numbers into that equation and can't seem to get it right.

a= V^2/r

There's another form of that equation in terms of omega. What is the omega of a minute hand on a clock?

Show us your work...

 

[dragonladies1]

I've never done anything with omega unfortunately. But right now I have this:

a= ((2*pi*.47)/3600)^2/.47

 

[berkeman]

I've never done anything with omega unfortunately. But right now I have this:

a= ((2*pi*.47)/3600)^2/.47

Be sure to show units in your equations -- it helps understandability and helps avoid mistakes.

Here's a page that should help you understand omega and uniform circular motion better:

http://en.wikipedia.org/wiki/Uniform_circular_motion

.

 

[dragonladies1]

Thank you very much for the help. I also just do not understand what the second statement, "express it as a fraction of the magnitude of free-fall acceleration g." is looking for.

 

[berkeman]

Thank you very much for the help. I also just do not understand what the second statement, "express it as a fraction of the magnitude of free-fall acceleration g." is looking for.

The magnitude of acceleration due to gravity at the surface of the Earth is generally called "g". Like you feel 1g while standing, and can feel higher g's when stunt flying, etc. Do you know the numerical value of 1g? It's in units of m/s^2.

 

[dragonladies1]

Yes, 9.81. So would it be:

3.86 E-4/9.81 ?

 

[dragonladies1]

Although, I do not believe my answer for the first portion of my problem is correct.

 

[berkeman]

Although, I do not believe my answer for the first portion of my problem is correct.

Yes, taking the ratio to 9.8m/s^2 would be the way to do the comparison. Post all of your work and I should be able to check it from home in a couple hours (or somebody else can).