Dynamics of Circular Motion

Dynamics of Circular Motion

Circular motion is not produced by a special force called “centripetal force.” Rather, circular motion occurs when the vector sum of the real forces acting on a body has an inward component. Newton’s second law then requires that inward component to equal the mass times the Normal, or centripetal, acceleration.

For motion along a circular path of radius R, the inward acceleration has magnitude

(1)

If the speed is changing, there can also be a tangential acceleration

(2)

The force equations therefore separate naturally into

(3)

and

(4)

The first equation controls the curvature of the path. The second controls how the speed changes along the path. This article develops these equations for flat turns, banked curves, vertical circles, contact forces, and conical pendulums.

1 The normal and tangential directions

For a particle constrained to a circular path, let e_t point tangent to the path in the direction of motion and let e_n point toward the center of curvature. The acceleration is

(5)

The inward term exists even when the speed is constant because the velocity vector changes direction.

For uniform circular motion, dv∕dt = 0, so

(6)

The acceleration is nonzero even though the speed is constant.

2 Centripetal force is not an additional force

The phrase centripetal force means the net inward force required by circular motion:

(7)

It is not another force to add to the free body diagram. The inward force may be supplied by Friction, Tension, gravity, a normal force, or a combination of real forces.

For example, on a flat road the inward force on a turning CAR can be static friction. For a ball on a string it can be tension. For a satellite it can be gravity. For a roller coaster it can be a combination of gravity and the normal force.

A useful procedure is:

1. Draw only the real forces.
2. Choose the inward normal direction.
3. Resolve the real forces along that direction.
4. Set their inward sum equal to mv²∕R.
5. Use a separate tangential equation if the speed is changing.

3 Flat turns and friction

Consider a car of mass m moving around a level circular road of radius R. The vertical forces balance,

(8)

so

(9)

The horizontal inward force is supplied by static friction.

The radial equation is

(10)

Static friction adjusts as needed up to

(11)

Therefore the no-slip condition is

(12)

The maximum speed without slipping is

(13)

Notice that the vehicle mass cancels in this ideal model.

4 Worked example 1: maximum speed on a flat curve

A car rounds a level curve of radius R = 65.0 m. The coefficient of static friction between the tires and the road is μ_s = 0.650. Find the maximum speed for which the car can follow the circular path without slipping.

At the limiting condition,

(14)

Thus

(15)

Substitution gives

(16)

Therefore

(17)

The static friction force is below its maximum value at any lower speed.

5 Banked curves without friction

A banked road tilts the normal force so that it has an inward horizontal component. At one particular speed, no friction is required.

If the road is banked by angle 𝜃, the vertical equation is

(18)

and the inward equation is

(19)

Dividing the equations eliminates N and m:

(20)

Equivalently, the frictionless design speed is

(21)

6 Worked example 2: frictionless banked turn

A highway curve has radius R = 120 m and is banked at 𝜃 = 12.0^∘. Find the speed for which no friction is required.

Using

(22)

we obtain

(23)

Thus

(24)

At other speeds, friction may be needed to maintain the same circular path.

7 Vertical circular motion

When a body moves in a vertical circle, gravity changes its radial contribution around the path. The radial equation must be written separately at each location.

For a body moving on the inside of a vertical circular track, at the bottom the inward direction is upward. Therefore

(25)

Hence

(26)

At the top, the inward direction is downward. Both gravity and the normal force point inward:

(27)

Thus

(28)

The normal force is a contact force and cannot pull the body toward the track. Therefore physical contact requires

(29)

At the top, the limiting condition N_t = 0 gives

(30)

so the minimum top speed for contact is

(31)

This result is purely local. Determining what speed is required at some other point to arrive at the top with this value requires an additional dynamical or energy analysis.

8 Worked example 3: apparent weight in a circular dip

A 70.0 kg rider passes through the bottom of a circular dip of radius R = 25.0 m at a speed of v = 18.0 m∕s. Find the normal force exerted by the seat on the rider.

At the bottom, inward is upward, so

(32)

Therefore

(33)

Substitution gives

(34)

Hence

(35)

The seat force exceeds the rider’s Weight because an additional upward net force is required to curve the motion upward.

9 Crests, contact loss, and apparent weight

At the top of a convex circular hill, the inward direction points downward. Gravity points inward while the normal force points outward. The radial equation is

(36)

Thus

(37)

As the speed increases, the normal force decreases. The limiting contact condition is N = 0, giving

(38)

If a mathematical calculation predicts N < 0, the assumed contact motion is no longer physically possible. The object loses contact instead.

10 Worked example 4: contact at the top of a loop

A small cart moves on the inside of a vertical circular track of radius R = 4.00 m. At the top of the loop its speed is v_t = 7.50 m∕s. Find the normal force per unit mass and determine whether contact is maintained.

At the top,

(39)

Divide by m:

(40)

Numerically,

(41)

Therefore

(42)

Because N > 0, contact is maintained. The minimum top speed would be

(43)

so 7.50 m∕s is safely above the local contact threshold.

11 Conical pendulum

A conical pendulum is a mass attached to a string of length L that moves in a horizontal circle while the string makes a constant angle 𝜃 with the vertical.

The vertical force balance is

(44)

The horizontal inward equation is

(45)

Dividing gives

(46)

Since

(47)

we may also write the angular-speed relation

(48)

The tension is

(49)

12 Worked example 5: conical pendulum

A 0.500 kg mass is attached to a 1.20 m string and moves as a conical pendulum with the string at 30.0^∘ from the vertical. Find the radius of the horizontal circle, the tension, the speed, and the period.

The circular radius is

(50)

The tension is

(51)

From

(52)

we obtain

(53)

The period is circumference divided by speed:

(54)

Thus

(55)

13 When speed changes around a circle

Circular motion does not require constant speed. If a tangential component of net force exists, then

(56)

At the same instant, the radial equation remains

(57)

The two equations answer different questions. The tangential force changes the magnitude of the velocity. The inward force changes its direction.

The total acceleration magnitude is

(58)

Likewise, if the net force has perpendicular tangential and normal components,

(59)

14 Common mistakes

• Drawing a separate “centripetal force” in addition to the real forces.
• Setting one particular force equal to mv²∕R without first finding the net inward force.
• Assuming circular motion means constant speed.
• Forgetting that the inward direction changes continuously around the path.
• Using N = mg in a vertical circle, dip, hill, or banked turn without checking the force equations.
• Assuming the normal force can become negative. A passive contact force can push but cannot pull.
• Using kinetic friction for a car that is rolling without slipping around a curve. The relevant tire-road force is ordinarily static friction.
• Reversing the force directions at the top of an inside vertical loop. Both gravity and the inward normal force point toward the center there.
• Treating the bank angle equation as valid for every speed. The frictionless relation selects one design speed for a specified R and 𝜃.

15 Practice problems

Use g = 9.81 m∕s² unless otherwise stated.

1. A particle moves at 8.00 m∕s on a circular path of radius 5.00 m. Find its inward acceleration.
2. A 2.50 kg object moves at 6.00 m∕s in a circle of radius 3.00 m. Find the required net inward force.
3. A 1200 kg car rounds a level curve of radius 80.0 m at 18.0 m∕s. Find the static friction force required and the minimum coefficient of static friction.
4. A level road has μ_s = 0.500 and curve radius 50.0 m. Find the maximum no-slip speed.
5. A frictionless banked curve has radius 90.0 m and bank angle 10.0^∘. Find its design speed.
6. A frictionless curve is designed for v = 25.0 m∕s at radius R = 180 m. Find the bank angle.
7. A 65.0 kg passenger moves through the bottom of a circular dip of radius 30.0 m at 20.0 m∕s. Find the normal force.
8. A car passes over the top of a convex hill of radius 45.0 m at 15.0 m∕s. Express the normal force as a fraction of the car’s weight.
9. For the hill in Problem 8, find the speed at which the car would just lose contact with the road.
10. A 0.750 kg cart moves on the inside of a vertical loop of radius 2.50 m. At the top its speed is 6.00 m∕s. Find the normal force.
11. A conical pendulum has length 0.800 m and angle 25.0^∘ from the vertical. Find the angular speed and period.
12. At an instant, a 3.00 kg particle moves on a circle of radius 4.00 m with speed 5.00 m∕s. Its speed is increasing at 2.00 m∕s². Find the normal force component, the tangential force component, and the magnitude of the net force.

16 Answers

1. 12.8 m∕s² inward.
2. 30.0 N inward.
3. 4.86 × 10³ N inward; μ _s ≥ 0.413.
4. 15.7 m∕s.
5. 12.5 m∕s.
6. 𝜃 = 19.5^∘.
7. 1.50 × 10³ N.
8. N∕W = 1 − v²∕(Rg) = 0.490.
9. 21.0 m∕s.
10. 3.44 N toward the center.
11. ω = 3.68 rad∕s; P = 1.71 s.
12. F_n = 18.8 N, F_t = 6.00 N, F_net = 19.7 N.

17 Summary

Circular dynamics is Newton’s second law resolved into directions adapted to the path. The normal equation is

(60)

while changing speed is governed by

(61)

The term mv²∕R is not a new force. It is the required net inward force. Identifying which real forces provide that inward component is the central step in circular-motion dynamics.

For a level friction-limited turn,

(62)

For a frictionless banked curve,

(63)

At the top of an inside vertical loop, maintaining contact requires

(64)

These results all come from the same principle: draw the real forces, choose the inward direction, and apply Newton’s second law.

References