What is the centripetal acceleration of the end of the rod?

[ilkjester]

Homework Statement

According to the Guinness Book of world records. the highest rotary speed ever attained was 2010 m/s (4500 mph). the rotating rod was 15.3 cm (6 in.) long. assume that the speed quoted is that of the end of the rod.
a) What is the centripetal acceleration of the end of the rod?
b) if you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on its own.

Homework Equations

ac=v^2/r
fnet=m(ac)

The Attempt at a Solution

ac=2010m/s^2/7.56
ac=528117 that just seems ridiculous for an answer.

fnet=1.0g(528117)
fnet=528117

 

[learningphysics]

You didn't convert to m. Also, are you sure the rod is rotating about the center and not the end?

I think you're supposed to use r = 0.153m

 

[ilkjester]

You didn't convert to m. Also, are you sure the rod is rotating about the center and not the end?

I think you're supposed to use r = 0.153m

ok so you think i should use r = 0.153m. So just wondering but what would be the difference between rotating about the center and not the end?

 

[ilkjester]

ok so you think i should use r = 0.153m. So just wondering but what would be the difference between rotating about the center and not the end?

ok so i got ac=2010^2/0.153
ac=41312.970

 

[learningphysics]

ok so you think i should use r = 0.153m. So just wondering but what would be the difference between rotating about the center and not the end?

I'm actually not really sure if they want you to use r = 0.153m or r = 0.0765m.

Just got the feeling they meant the distance from the pivot to the end of the rod is 0.153m... but I really don't know.

 

[learningphysics]

ok so i got ac=2010^2/0.153
ac=41312.970

how did you get that? I'm getting 2.64*10^7 m/s^2.

 

[ilkjester]

Thats alright I can do them with both that way i can't go wrong thank you again. But do you mind showing me how you converted them.

 

[ilkjester]

how did you get that? I'm getting 2.64*10^7 m/s^2.

Im not sure how i got it. Now I get the 2.64*10^7 m/s^2. I must have just hit some button on accident on the calc.

 

[learningphysics]

Thats alright I can do them with both that way i can't go wrong thank you again. But do you mind showing me how you converted them.

convert them? you mean the cm to m, g to kg?

 

[ilkjester]

convert them? you mean the cm to m, g to kg?

Yeah I should probably learn how to do it.

 

[learningphysics]

Yeah I should probably learn how to do it.

no prob.

$$1cm = 10^{-2} m$$

$$1g = 10^{-3} kg$$

I can get the inverse relations... multiply both sides of the first equation by 100. second by 1000.

so:

$$100cm = 1 m$$

$$1000g = 1 kg$$

so I can convert one way or the other...

so $$15.3cm = 15.3*(10^{-2}m) = 0.153m$$

what's really handy for converting units is the "factor label" method:

http://en.wikipedia.org/wiki/Units_conversion_by_factor-label

 

[ilkjester]

ok so the teacher said that we use r=0.153m so to find the centripetal acceleration. ac=v^2/r so to find velocity i do distance/time so would it be 0.153m/2010m/s. would that be how i find velocity. because i also have a circular motion equation for velocity which is v=delta r/ delta time

 

[learningphysics]

ok so the teacher said that we use r=0.153m so to find the centripetal acceleration. ac=v^2/r so to find velocity i do distance/time so would it be 0.153m/2010m/s. would that be how i find velocity. because i also have a circular motion equation for velocity which is v=delta r/ delta time

I don't understand. We already got the answer. You said you got the same answer as me 2.64*10^7 in post#8.

 

[ilkjester]

Yeah I remember now but yeah we used the whole diameter instead of the circumfrance.

 

[learningphysics]

Yeah I remember now but yeah we used the whole diameter instead of the circumfrance.

Hmmm... ac = v^2/r. Your teacher said we use r = 0.153 right?

ac = 2010^2/0.153 = 2.64*10^7 m/s^2

 

[ilkjester]

yep that's what i got thanks