Amusement Park Physics!! Question..;) plz help?

Question

(This ride is a swing carousel)

An amusement parkride consists of a rotating circular platform 8.00m in diameter (radius = 4) from which 10 kg seats are suspended at the end of 2.50 m massless chains. When the system rotates, the chains make an angle (theta = 28.0 degrees) with the vertical.

here are the questions...

What is the speed of each child seat?
Find the tension in the chain when a 40 kg child sits in the seat (10kg + 40 kg)?

plz help

Thank you!

Answer 1

The radius of the circle r made by the seats is 4.00 m (the radius of the platform) plus (2.50 m) (sin 28 degrees).

Now apply Newton's Second Law to the seat in both the vertical and horizontal dimensions.

Vertical: the vertical forces are the contact force from the tension in the chain, magnitude T cos 28 degrees, and the force of gravity down, mg. Newton's Second Law is then T cos 28 = mg. Solve this for T.

Plug in m = 50 kg (10 kg for the seat, 40 kg for the child) into either equation to find T for the second part.

Horizontal: the only force in this dimension is the horizontal component of the chain contact force, T sin 28 degrees. This centripetal force causes the chair to move in a circle of radius r, so Newton's Second Law is T sin 28 = m v^2 / r. The only unknown is v, so solve for that value.