"Cat on a fan" physics analysis.

[MDJH]

So basically I posted the following analysis as a comment on this cat video on youtube.

 

"At about 2 minutes in, the fan circles 4 times in 5 seconds. That's 0.8 cycles per second.

 

If we assume the cat's spinning an average distance of a metre from the centre of the fan, that means it's moving at 2.5 metres per second. (v=2*pi*f*r)

 

Also, the circular acceleration would be (v*v)/r, which would mean a velocity change of 6.25 metres per second every second.

 

And so, to the cat, the pull of the rope probably feels like two thirds of the pull of gravity."

 

What say you, Science Forums? Is my analysis valid?

Edited February 21, 2011 by MDJH

[John Cuthber]

If there are two cats at equal radii on the fan and the fan speed is increased slowly which cat falls off first?

[swansont]

So basically I posted the following analysis as a comment on this cat video on youtube.

 

"At about 2 minutes in, the fan circles 4 times in 5 seconds. That's 0.8 cycles per second.

 

If we assume the cat's spinning an average distance of a metre from the centre of the fan, that means it's moving at 2.5 metres per second. (v=2*pi*f*r)

 

Also, the circular acceleration would be (v*v)/r, which would mean a velocity change of 6.25 metres per second every second.

 

And so, to the cat, the pull of the rope probably feels like two thirds of the pull of gravity."

 

What say you, Science Forums? Is my analysis valid?

 

2.5 m/s is true at half a meter. At one meter, it's 5 m/s

 

What's the link for the video?

[MDJH]

2.5 m/s is true at half a meter. At one meter, it's 5 m/s

 

What's the link for the video?

 

I meant to link to it in the original post.

 

Yeah, I multiplied f by pi and by r, but forgot to multiply by 2. That's a pretty big mistake.

 

I suppose I should remove and replace the comment, then.

[lemur]

If there are two cats at equal radii on the fan and the fan speed is increased slowly which cat falls off first?

The one with the most accurate ability to estimate a tangent line to the canary cage hanging in the corner and let go at the right moment.