The Normal Force in Circular Motion

Expert reviewed • 22 November 2024 • 9 minute read

Understanding the Normal Force in Circular Motion

The Normal force is the perpendicular force exerted by a surface on an object in contact with it. In circular motion, this force often plays a pivotal role in providing the necessary centripetal force to maintain the circular path.

What are the key Concepts:

  • Centripetal Force: The force directed towards the centre of the circular path, responsible for keeping an object in circular motion.
  • Normal Force: The force perpendicular to the surface of contact.
  • Friction: The force resisting relative motion between surfaces in contact.

Horizontal Circular Motion

Case Study: The Rotor Ride

The rotor ride provides an excellent example of horizontal circular motion involving normal force.

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Forces at Play:

  • Centripetal Force: Provided by the normal force from the wall.
  • Normal Force: Exerted by the wall on the rider, directed towards the centre.
  • Friction: Static friction between the rider and the wall, supporting the rider’s weight.

Mathematical Analysis:

  • Normal Force (Centripetal Force):
N=mv2rN = \frac{mv^2}{r}

Where:

  • NN is the normal force
  • mm is the mass of the rider
  • vv is the velocity
  • rr is the radius of the rotor
  • Static Friction:
fs=μN=μmv2rf_s = \mu N = \mu \frac{mv^2}{r}

Where,

  • μ\mu is the coefficient of static friction
  • Minimum Velocity to Stay Suspended: When fs = mgfs = mg, we get:
μmv2r=mgv=grμ\mu \frac{mv^2}{r} = mg \\v = \sqrt{\frac{gr}{\mu}}

Practice Question 1

A rotor ride has a radius of 5 m. If the coefficient of static friction between the wall and a rider is 0.6, calculate the minimum angular velocity required to keep the rider suspended.

v=grμ=(9.8)(5)0.6=9.06m/sv=\sqrt{\frac{gr}{\mu}} \\= \sqrt{\frac{(9.8)(5)}{0.6}} \\= 9.06 m/s

Thus, angular velocity is:

ω=vr=9.065=1.81rad/s\omega = \frac{v}{r} \\= \frac{9.06}{5} \\= 1.81 rad/s

Vertical Circular Motion

Vertical circular motion, such as in a loop-the-loop roller coaster, presents a more complex scenario where the normal force changes in both magnitude and direction.

Forces at Different Points:

  • At the Bottom of the Loop:
N=mv2r+mgN = \frac{mv^2}{r} + mg
  • At the Top of the Loop:
N=mv2r−mgN = \frac{mv^2}{r} - mg

Critical Velocity:

The minimum velocity required at the top of the loop to maintain contact is when N = 0:

mv2r−mg=0vmin=rg\frac{mv^2}{r} - mg = 0 \\v_{min} = \sqrt{rg}

Practice Question 2

A roller coaster car enters a vertical loop with a radius of 20 m. Calculate:

  • The normal force at the bottom if the car’s mass is 1000 kg and its speed is 25 m/s.
  • The minimum speed required at the top to maintain contact.
  • At the bottom:
N=mv2r+mg=1000(25)220+1000(9.8)=41,300NN = \frac{mv^2}{r} + mg \\= \frac{1000(25)^2}{20} + 1000(9.8) \\= 41,300 N
  • Minimum speed at top:
vmin=rg=20(9.8)=14m/sv_{min} = \sqrt{rg} \\= \sqrt{20(9.8)} \\= 14 m/s

Banked Curves

Banked curves are designed to help vehicles navigate turns at higher speeds by providing some of the necessary centripetal force through the normal force.

Force Analysis:

On a banked curve with angle θ:

  • Normal Force:
N = mgcos θN = mgcos θ
  • Centripetal Force:
Fc = Nsin θ + fF_c = Nsin θ + f

Where,

  • ff is the friction force

Furthermore, for an ideally banked curve (meaning no friction needed):

tan⁡θ=v2rg\tan \theta = \frac{v^2}{rg}

Practice Question 3

A car travels around a circular turn of radius 100 m on a road banked at an angle of 10°. If the coefficient of friction between the tires and the road is 0.1, calculate the maximum speed the car can travel without slipping.

Using:

v2r=g(tan⁡θ+μssec⁡θ):v=rg(tan⁡θ+μssec⁡θ)\frac{v^2}{r} = g(\tan \theta + \mu_s \sec \theta): v = \sqrt{rg(\tan \theta + \mu_s \sec \theta)}

We can calculate the velocity:

v=100(9.8)(tan⁡10°+0.1sec⁡10°)=21.7m/s\\v = \sqrt{100(9.8)(\tan 10° + 0.1 \sec 10°)} = 21.7 m/s

Key Takeaways

  • Normal force plays a crucial role in providing centripetal force in circular motion.
  • The magnitude and direction of normal force change based on the position in vertical circular motion.
  • Banked curves utilise normal force to assist in providing centripetal force, reducing reliance on friction.
  • Understanding the interplay of forces in circular motion is crucial for analysing real-world scenarios like amusement park rides and vehicle dynamics.

By mastering these concepts, you’ll be well-equipped to analyse and solve a wide range of circular motion problems involving normal forces.

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