Mechanics - circular motion/tension?

Question

Another problem which I can't solve...now in 3 dimensions.

A mass of 1kg is fastened by a string of length 1m to a point 0.5m above a smooth horizontal table and is describing a circle on the table with uniform angular speed of 1 revolution in 2 seconds [I take this to mean omega = pi]. Find the force exerted on the table and the tension in the string.

I could do this easily if the string was ON the table, but the third dimension makes it too hard. Help is greatly appreciated! Thank you...and please try not to assume I know a lot about mechanics.

Answer 1

The mass is continually accelerating in the direction of the centre of the circle at a rate of r*omega^2, where r is the radius of the circle described.

Now imagine a two-dimensional, side-on view. The geometry of the situation dictates, by Pythagoras' Theorem, that the radius of the circle is SQRT(3)/2 m. Consequently, the linear acceleration of the mass is SQRT(3)/2 *pi^2 ms^-2.

This centripetal acceleration is provided by the horizontal component of the tension in the string. This component is SQRT(3)/2 *pi^2 kgms^-2.

The horizontal and vertical components of tension are in similar proportion to the horizontal and vertical distances between the mass and the apex of the string's path. Hence, the total tension in the string (mass 1 kg) must be pi^2 kgms^-2.

The vertical component of this tension is 0.5*pi^2 kgms^-2. This vertical tension helps counteract the weight of the mass, which is 10 kgms^2. The resistance of the table surface acts to balance the remainder of the weight,  (10-0.5*pi^2) kgms^2, which is about 5 kgms^2. A downward force of equal strength acts on the table, exerted by the mass.