
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

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 $

Thanks
so w = 3.63 ms1 ?
R

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 $\displaystyle \theta$?

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.
