The “Giant Swing” ride consists of a vertical shaft with a number of horizontal arms attached at the top as shown. A rigid cable 5.0 m long connects a seat to the arm at a point 2.0 m from the center shaft. When the ride reaches its full speed, the

Question

angle made by the arm is 35.0o.
(i) What is the time that it takes for one full revolution at the maximum speed?
(ii) If a 60.0-kg person goes on the ride, and we ignore the mass and weight of the metal shaft, then what is the tension in the 5.00 m cable?

Answer 1

(i) We can find the maximum speed that the swing rotates by finding the horizontal component of motion. The radius of the circle traced by the seat is:

r = 2.0 m * sin(35°) = 1.14 m

Now we can find the linear velocity (v) using centripetal acceleration:

a = v^2 / r

For an object in circular motion, the centripetal acceleration is:

a = 4 * pi^2 * r / T^2

where T is the time period of one full revolution. Substituting the value of r, we get:

T^2 = 4 * pi^2 * (1.14 m) / a

Now, the centripetal acceleration can also be expressed using the angle (theta) and the radius (R) of the arm:

a = R * g * sin(theta)

We know R = 2.0 m, g = 9.81 m/s^2, and theta = 35°:

a = 2.0 m * 9.81 m/s^2 * sin(35°) = 11.36 m/s^2

Substituting the value of a into the equation for T, we get:

T^2 = 4 * pi^2 * (1.14 m) / 11.36 m/s^2

T^2 ≈ 3.996 s^2

Taking the square root of both sides, we find:

T = 2.0 s

So it takes 2.0 seconds for one full revolution at the maximum speed.

(ii) We can find the tension in the cable by considering the forces acting on the person. There are two forces acting on the person: the gravitational force (Fg) acting downward and the tension (T) acting along the cable.

We can divide the tension into two components: one acting horizontally (Tx) and another acting vertically (Ty). The sum of the forces in the vertical direction should be equal to the centripetal force, and the Ty should balance the person's weight.

Ty = Fg = m * g = 60 kg * 9.81 m/s^2 = 588.6 N

The centripetal force is given by:

Fc = m * a = 60 kg * 11.36 m/s^2 = 681.6 N

Now, the horizontal component of the force can be expressed using the angle:

Tx = Fc / cos(theta) = 681.6 N / cos(35°) = 831.75 N

The tension in the cable is the vector sum of the horizontal and vertical components of the force:

T = sqrt(Tx^2 + Ty^2) = sqrt(831.75 N^2 + 588.6 N^2) ≈ 1030 N

So the tension in the cable is approximately 1030 N.