Question

PART ONE:
An amusement park ride consists of a rotating
circular platform 8.78 m in diameter from
which 10 kg seats are suspended at the end
of 3.54 m massless chains. When the system
rotates, the chains make an angle of 40.7◦ with the vertical.
The acceleration of gravity is 9.8 m/s^2.
What is the speed of each seat?
Answer in units of m/s.

PART TWO
If a child of mass 49.1 kg sits in a seat, what is
the tension in the chain (for the same angle)?
Answer in units of N.

Answer 1

Answer:

7.51 m/s

764 N

Explanation:

Part One:

Draw a free body diagram.  There are two forces on the seat: tension force T along the chain, and weight force mg pulling down.

Sum of forces in the y direction:

∑F = ma

T cos θ − mg = 0

T = mg / cos θ

Sum of forces in the centripetal direction:

∑F = ma

T sin θ = m v² / r

(mg / cos θ) sin θ = m v² / r

mg tan θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

First, find the radius of the path:

r = 8.78 m / 2 + 3.54 m sin 40.7°

r = 6.70 m

v = √(9.8 m/s² × 6.70 m × tan 40.7°)

v = 7.51 m/s

Part Two:

T = mg / cos θ

T = (10 kg + 49.1 kg) (9.8 m/s²) / cos 40.7°

T = 764 N