Question

The figure shows a photo of a swing ride at an amusement park. The structure consists of a horizontal, rotating, circular platform of diameter D from which seats of mass m are suspended at the end of massless chains of length d. When the system rotates at constant speed, the chains swing outward and make an angle with the vertical. Consider such a ride with the following parameters: D = 7.50 m, d = 2.50 m, m = 9.0 kg, and = 29.09. Stuart Gregory/Photographer's Choice/Getty Images (a) What is the speed of each seat (in m/s)? m/s (b) Draw a diagram of forces acting on a 40.0-kg child riding in a seat. Choose File No file chosen (c) Find the tension in the chain (in N). N What If? The chains holding each seat can withstand a maximum total tension of 700 N. (d) What is the angle that the chains make with the vertical (in degrees) when they have this tension? (e) What is the maximum angular speed (in rad/s) with which the ride can rotate? rad/s

Answer 1

(a) In x direction we have

Tsin$\theta$ = mv² / r ------------- (1)

In y direction ,we have

Tcos$\theta$ = mg -------------- (2)

Here, r is the total radius and is given as

r = (D/2) + dsin$\theta$

r = 7.50 /2 + 2.5sin29

r = 4.962 m

solving (1) and (2) , we get

v = sqrt ( grtan$\theta$)

v = sqrt ( 9.8 * 4.962 * tan 29)

v = 5.2 m/s

--------------------------------------------------------------

(b) Free body diagram will be

where T is the tension in chain and mg is the weight of child acting downwards.

--------------------------------------------------------------------------------------

(c) Let M be the mass of child then

Tcos$\theta$ -  mg - Mg = 0

Here m is mass of seat

so,

Tcos$\theta$ - 9*9.8 - 40*9.8 = 0

T cos 29 - 88.2 - 392 = 0

T = 549 N

----------------------------------------------------------

(d) The question did not make it clear if the tension of 700 N is with the child or without the child, I am assuming with the child.

Here, we need to use same equation as part (c) .

Tcos$\theta$ -  mg - Mg = 0

T = 700 N (given)

700 * cos $\theta$ - 88.2 - 392 = 0

cos $\theta$ = 0.686

$\theta$ = 46.68 degree

-------------------------------------------------------------