U3AOS1 Topic 8: Vertical Circular Motion

Introduction

Vertical circular motion refers to the movement of an object in a circular path that is oriented vertically, such as a roller coaster loop or a spinning bucket of water. Unlike horizontal circular motion, which is generally governed by a constant centripetal force, vertical circular motion is influenced by gravitational forces, which vary with the position of the object in the circle. This introduces additional complexities to the analysis of the forces and accelerations involved.

Basic Concepts

  1. Centripetal Force:

    • In circular motion, centripetal force is the force that keeps the object moving in a circular path. It acts toward the center of the circle. For an object of mass mm moving with a velocity vv in a circle of radius RR, the centripetal force FcF_c is given by: Fc=mv2RF_c = \frac{mv^2}{R}
  2. Gravitational Force:

    • Gravity exerts a force downward with magnitude Fg=mgF_g = mg, where gg is the acceleration due to gravity (9.8 m/s2\approx 9.8\ \text{m/s}^2).

Forces in Vertical Circular Motion

In vertical circular motion, the forces acting on the object vary depending on its position in the circle. The net centripetal force required to keep the object moving in a circular path is the result of the combination of gravitational force and the normal force (or tension, depending on the context).

  1. At the Top of the Circle:

    • At the top of the circle, the gravitational force and the centripetal force both act toward the center of the circle. The centripetal force here is provided by the difference between the gravitational force and the normal force. Thus: Fc=mgNF_c = mg - N
    • Rearranging gives the normal force (or tension in a string) at the top of the circle: N=mgmv2RN = mg - \frac{mv^2}{R}
  2. At the Bottom of the Circle:

    • At the bottom of the circle, the gravitational force acts downward while the centripetal force acts upward, toward the center. The normal force or tension must overcome the gravitational force to provide the necessary centripetal force. Thus: N=mv2R+mgN = \frac{mv^2}{R} + mg
    • Here, the normal force (or tension) is larger due to the need to provide additional force to counteract gravity.
  3. At the Midpoint of the Circle:

    • At the midpoint of the vertical circle (if the object is a wheel or similar), the forces are balanced differently. The centripetal force must be provided by the combination of the forces acting perpendicular to the vertical axis. This situation often simplifies to only considering the centripetal force without the gravitational force being directly relevant to the calculations of normal force or tension.

Key Concepts and Equations

  1. Velocity for a Stable Circular Path:

    • For the object to complete the vertical circle without falling or breaking the string (if it is a pendulum or similar), it must have a minimum velocity at the top of the circle. This velocity ensures that the centripetal force is adequate to maintain circular motion. The critical condition at the top of the circle is: vmin=gRv_{\text{min}} = \sqrt{gR}
    • This is derived from setting the normal force to zero at the top of the circle: mg=mv2Rmg = \frac{mv^2}{R} v=gRv = \sqrt{gR}
  2. Energy Considerations:

    • Energy conservation is a useful tool in analyzing vertical circular motion. The total mechanical energy (kinetic plus potential) remains constant if only conservative forces are involved. At any point in the motion: E=12mv2+mghE = \frac{1}{2}mv^2 + mg h
    • Here, hh is the height above the lowest point of the circle. This equation can be used to find the speed at different points in the circle based on the height and known values of initial conditions.

Applications and Examples

  1. Roller Coasters:

    • Roller coasters are classic examples of vertical circular motion. At the top of a loop, the coaster must maintain a minimum speed to prevent falling out of the track. Engineers design roller coasters with enough initial speed to ensure safety and comfort. The forces at the top and bottom of the loops are carefully calculated to ensure the structural integrity of the coaster and the safety of the riders.
  2. Pendulums:

    • A simple pendulum exhibits vertical circular motion if its amplitude is large enough. At the highest point of the swing, the pendulum's speed is lowest, and the tension in the string is minimum, whereas at the lowest point, the tension is maximum due to the addition of the gravitational force.
  3. Bucket of Water:

    • A common demonstration of vertical circular motion involves swinging a bucket of water in a vertical circle. To keep the water from falling out at the top of the circle, the bucket must be swung fast enough so that the centripetal force due to the bucket's motion is sufficient to provide the necessary force to counteract gravity.

Example 1
A roller coaster car of mass 500 kg500\ \text{kg} is at the top of a vertical loop with a radius of 10 m10\ \text{m}. What is the minimum speed required at the top of the loop to ensure the car does not fall off?

At the top of the loop, the centripetal force required to keep the car in motion is provided by the gravitational force. Therefore: Fc=mgF_c = mg mv2R=mg\frac{mv^2}{R} = mg v2=gRv^2 = gR v=gRv = \sqrt{gR} Using g=9.8 m/s2g = 9.8\ \text{m/s}^2 and R=10 mR = 10\ \text{m}: v=9.8×10=98=9.90 m/sv = \sqrt{9.8 \times 10} = \sqrt{98} = 9.90\ \text{m/s}

So, the minimum speed required at the top of the loop is 9.90 m/s9.90\ \text{m/s}.

Example 2
 A 2 kg2\ \text{kg} object is moving in a vertical circle with a radius of 5 m5\ \text{m} at a speed of 12 m/s12\ \text{m/s}. Calculate the tension in the string at the bottom of the loop.
At the bottom of the loop, the tension TT in the string must provide the centripetal force and support the weight of the object. So: T=mv2R+mgT = \frac{mv^2}{R} + mg Substitute m=2 kgm = 2\ \text{kg}, v=12 m/sv = 12\ \text{m/s}, R=5 mR = 5\ \text{m}, and g=9.8 m/s2g = 9.8\ \text{m/s}^2: T=2×1225+2×9.8T = \frac{2 \times 12^2}{5} + 2 \times 9.8 T=2885+19.6T = \frac{288}{5} + 19.6 T=57.6+19.6=77.2 NT = 57.6 + 19.6 = 77.2\ \text{N}
Exercise &&1&& (&&1&& Question)

A 1 kg1\ \text{kg} object is moving in a vertical circle with a radius of 8 m8\ \text{m}. The tension in the string at the top of the loop is 5 N5\ \text{N}. Calculate the speed of the object at the top of the loop.

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Exercise &&2&& (&&1&& Question)

A 3 kg3\ \text{kg} object moves in a vertical circle of radius 6 m6\ \text{m} with a speed of 15 m/s15\ \text{m/s}. Find the tension in the string at the top of the loop.

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Exercise &&3&& (&&1&& Question)

A projectile is launched with a speed of 20 m/s20\ \text{m/s} and follows a vertical circular path with a radius of 15 m15\ \text{m}. Calculate the maximum height the projectile reaches above the lowest point.

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