# Hope someone can help

• Nov 12th 2009, 02:50 AM
srobrien
Hope someone can help
I have this od chestnut to solve, I have done as much as what I think is right but I need some assistance:

a) A body of mass m kg is attached to a point by string of length
1.25 m. If the mass is rotating in a horizontal circle 0.75 m below the
point of attachment, calculate its angular velocity.

So far I have:

$\displaystyle r = \sqrt{1.25^2 - 0.75^2}$
$\displaystyle r = 1$

$\displaystyle \tan\theta = \frac{\omega^2r}{g}$

$\displaystyle \tan\theta = \frac{1}{0.75}$

$\displaystyle therefore:$
$\displaystyle 53.13 = \frac{\omega^2}{9.81}$

$\displaystyle \omega = 22.83$

Have I totally done this wrong, also were do I need to go from here?

Thanks
• Nov 12th 2009, 03:13 AM
Gusbob
From your own (correct) equations:

$\displaystyle \tan \theta = \frac{\omega^2r}{g}$

$\displaystyle \tan \theta = \frac{1}{0.75}$

Equating the two is

$\displaystyle \frac{\omega^2r}{g} = \frac{4}{3}$

There is where you went wrong. I think you equated $\displaystyle \theta$ to $\displaystyle \tan \theta$ instead of $\displaystyle \tan \theta$ = $\displaystyle \tan \theta$
• Nov 12th 2009, 03:26 AM
srobrien
Thanks

so w = 3.63 ms-1 ?

R
• Nov 12th 2009, 04:18 AM
srobrien
Ok so part 2:

(b) If the mass rotates on a table, calculate the force on the table when
the speed of rotation is 25 rpm and the mass is 6 kg

we have:

$\displaystyle T\cos\theta = mr\omega^2$

$\displaystyle T = \frac{6*1*25^2}{\cos\theta}$

Which angle is $\displaystyle \theta$?
• Nov 12th 2009, 06:17 AM
Gusbob
The angle $\displaystyle \theta$ is shown in the following picture. You can find the value from your equation in the first part: $\displaystyle \tan \theta = \frac{1}{0.75}$

http://farside.ph.utexas.edu/teachin...es/img1085.png

Also, you need to convert 25 the rpm into radians per second.

I also got the same answer as you did on the first part.
• Nov 12th 2009, 06:27 AM
srobrien
thanks ive got it.