Samples Physics Circular Motion in a Merry Go Round

# Circular Motion in a Merry Go Round

1006 words 4 page(s)

INTRODUCTION
Circular motion is where the motion of the object moves about an axis. When the axis of the motion is within the body, then the circular motion is referred to as the rotation. When the axis of motion is outside the bound, then the circular motion is referred to as revolution. Therefore, rotation and revolution are said to be the types of circular motion (Myers 125). Circular motion is experienced in almost all aspects of the university. The largest units of the universe, the solar system as well as the smallest units of matter that has property and chemical element, the atom, all experience circular motion within them.

Circular motion involves two types of speeds, the rotational speed and the tangential speed (Giancoli 213). The rotational speed is also referred to as the angular speed and indicates the rotations per unit time. The tangential speed is the speed at which an object revolving around an axis would fly at, was it to be released from the circular motion.

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The force involved is referred to as the centripetal force and is directed towards the center of rotation giving an impression that the object is being pulled to the center of rotation/revolution (Urone et al. 412). The force responsible for the circular motion is in some instances described by the outwards force which tends to pull the object in a tangential path of the circle. This kind of force is referred to as the tangential force. This research paper will therefore study the circular motion with special focus on the Merry go round, commonly referred to as the carousal.

FORMULA AND CALCULATIONS
The rotational speed is given by;:
v=2πR/T

Where the v is the rotational speed, R is the radius of the circle of motion, and T is the period (time required for a single revolution).
The acceleration of an object in a circular path is therefore:
a_rad=v^2/R=(4π^2 R)/T^2

From the Newton’s second law of motion expressed as;
F=ma
Then the centripetal force can be obtained by:
F_net=ma_rad=m v^2/R

The equation above applies for all uniform circular motion.

Considering a conical pendulum with the dimension as shown below:

As seen from the diagram, the angle β represents the angle of the deviation of the pendulum from the central axis. Since the bob has no vertical acceleration, the free body diagram can be drawn as:

In this case, T represents force F while θ represents the angle β. Therefore, the forces in the x-direction can be summed as:
∑▒F_x =F sin⁡β=ma_rad

The force in the y-direction is zero since the motion is only in the horizontal direction:
∑▒F_y =F cos⁡β+(-mg)=0

From equation (ii), we obtain:
F=mg/cos⁡β (iii)

Substituting (iii) in (i) gives:
mg/cos⁡β sin⁡β=ma_rad
a_rad=sin⁡β/cos⁡β g=g tan⁡β

Converting the acceleration equation to make period T the subject gives;
T=2π√(R/a_rad )
But a_rad=g tan⁡β

Hence:
T=2π√(R/(g tan⁡β ))

EXPERIENCE
In this case, we consider a Merry Go Round, also referred to as the Carousal. In a carousal, the metal bars rotate about the center while the people swinging revolve about the center. A swinger in carousal moves at a constant speed, however, the direction of speed keeps on changing after each point indicating that extra force is responsible for causing the change in direction (acceleration). Supposing there is a swinger seated in the carousal, his rotational speed (angular velocity) is given as ω, where
ω=2πR/T

A viewer below the carousal estimates the horizontal radius to be 10m and its period to be 9 seconds, then the rotational speed would be obtained as:
ω=(2π×10)/9=6.98 rad/sec

Assuming that by accident the swinger’s faulty seat flies off the carousal, then he would be fatally flying at a speed of:

Tangential Speed=radius×Angular velocity
=10m×6.98rad/sec=69.8m/s

Back to the normal case, the angular acceleration of the swinger in the carousal is given by:
a_rad=ω^2/R=〖6.98〗^2/10=4.872 m/s^2

Therefore, since the acceleration due to gravity is 9.81m/s2, then the angle of swinging will be obtained according to equation (iv) as:
a_rad=g tan⁡β
tan⁡β=a_rad/g=4.872/9.81=0.4966
β=tan^(-1) (0.4966)=〖26.4〗^0

The length of the arms of the pendulum can therefore be obtained as;
R=L sin⁡β
L=R/sin⁡β =10/sin⁡26.4 =22.48m

CONCLUSION
The Carousal is a perfect example of applying the study of circular motion. The carousal acts like a conical pendulum with the suspended mass being the swinger and the rope being the hoisting bars. As seen from this research paper, it is possible to estimate the swinging angle, the length of the suspending bars and hence the height off the ground (if the height of the carousal is known). All these measurements can be obtained by just estimating the horizontal radius of the circle made by the swinger and the time take for one revolution (period). In critical situations, such data can be crucial in determining the design of a carousal especially in built environment. The knowledge of the height of the carousal and the maximum speed of the driving motor can be used to estimate the length of the suspended arms to avoid hitting other objects when it hits maximum speed.

Circular motion can also be studied in many forms such as banking of road and sporting curves to avoid vehicles running off the track. The circular motions may also be studied by the slanting of the airplanes when they make circular turns in air. In further studies, circular motion is also used to explain why satellites are able to go round the earth in a geo-synchronous motion without falling to other orbits.

References
• “PhysicsLAB: Conical Pendulums.” PhysicsLAB: Conical Pendulums, http://dev.physicslab.org/ Accessed on 16th Dec 2016. Giancoli, Douglas C. Physics: principles with applications. Boston, Pearson, 2016.
• Urone, Paul P, Roger Hinrichs, Kim Dirks, and Manjula Sharma. College Physics. Houston, Texas: OpenStax College, Rice University, 2013.
• Young, Hugh D. et al. University Physics with Modern Physics. Harlow, Essex, Pearson, 2014.