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2.3Circular motion

Circular motion is a common type of motion which includes as a very special case the motion of planets around the sun that is approximately circular. As we will see, circular motion requires the presence of a force directed towards the centre of the circle. In the case of planets going around the sun a new force had to be discovered to accommodate for this fact: the force of gravitation.

Circular motion and angular speed

Consider the object in Figure 2.56, which rotates on a circle of radius r in the counterclockwise direction with constant speed v.

Figure 2.56: An object moving on a circle of radius r.

Let T be the time taken to complete one full revolution (or orbit). We call T the period the time it takes for an orbiting object to make one complete revolution; the time for one full oscillation or one full wave of the motion. Since the speed is constant and the object covers a distance of 2πr in a time of T seconds, it follows that

v=2 π r T

We may also note that the object sweeps out an angle of 2π radians in a time equal to the period, so we define the angular speed the rate of change of the angle swept by an object as it moves along a circular path: ω=2 π T=2 π f of the object (denoted by ω) by (Figure 2.57)

angular speed, ω= angle swept time taken =Δ θ Δ t

For a complete revolution, Δθ = 2π and Δt = T so that we also have

ω=2 π T

Figure 2.57: As the object rotates around the circle it sweeps out an angle measured from some arbitrary reference line.

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The units of angular speed are radians per second or just s−1.

We may define a quantity called the frequency, f the number of complete revolutions or full waves made in one second; f=1 T of the motion. This will be useful also in the context of waves later on. Frequency is the number of full revolutions per second. Since we make one full revolution in the course of one period T, the number in one second is f=1 T.

So, in terms of frequency we also have that ω = 2π f.

Comparing v=2 π r T with ω=2 π T w we see that v = r ω

This is the relation between the linear speed v and the angular speed ω.

Summarising all the formulae:

v=r ω=2 π r T=2 π r f.

theory of knowledge

The photograph in Figure 2.58 shows stars tracing arcs. Most of the ancients would explain this observation by saying that the stars rotate around the fixed earth. Today we explain this by saying that the earth rotates about its axis. Which view is correct, and how do we know? In 1851 the French physicist Leon Foucault constructed a long pendulum with a heavy bob at the end. When set into oscillation the plane of oscillation rotates slowly clockwise (in the northern hemisphere). The simple explanation for this effect is the rotation of the earth about its axis. The same reason that makes the Foucault pendulum precess, that is, change its plane of oscillation, is responsible for the patterns of wind and ocean currents on earth.

It is quite remarkable how very simple observations reveal something deep about the world around us. In the same category we have Eratosthenes’ ingeniously simple method for the measurement of the radius of the earth and Olbers’ observation about the darkness of the night sky leading to the abandonment of the Newtonian view of an infinite universe.

Figure 2.58: As the earth rotates about its axis, stars appear to trace circles in the night sky.

Worked Example 2.14

Calculate:

athe angular speed of the earth as it rotates around the sun

bthe angle swept by the earth in 30 days

cthe linear speed of the earth.

(The radius of the orbit is about 1.5 × 1011 m.)

Answer

aThe earth completes one full revolution in approximately 365 days. We use ω=2 π T for angular speed:

ω=2 π 365×24×60×60 ω=1.99×10−7≈2.0×10−7 s−1

b θ=30 365×2 π=0.52 radians or approximately 30°.

cWe need the relation between angular and linear speed v = ω r:

v=1.99×10−7×1.5×10 11=29580=3.0×10 4 m s−1

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Centripetal acceleration

It is important to note right away that in circular motion even if the linear speed is constant the velocity is not: it is changing because it is changing direction. Since the velocity changes, we have acceleration. What follows is a derivation of the expression for acceleration in circular motion. You may want to skip the derivation and go directly to the result just before the end of this section.

Look at Figure 2.59 which shows the velocity of a particle at two points P and Q. We are making use of the fact that the velocity vector has to be tangent to the circle.

Figure 2.59: a The velocity vector changes direction as the particle moves from P to Q. b The change in the velocity vector from P to Q is given by the red arrow. c The distance travelled and the length of the arc are the same if the angle Δθ is very small.

The acceleration is defined as

a→=Δ ν→Δ t

where Δ ν→ is a vector. Thus, we have acceleration every time the velocity vector changes. This vector will change if:

●its magnitude changes

●or the direction changes

●or both magnitude and direction change.

For motion in a circle with constant speed, it is the direction of the velocity vector that changes. We must thus find the difference Δ v→=v→Q−v→P. (Note that the magnitude of the velocity vectors at P and Q are the same—they are equal to the constant speed v of the moving particle.)

The magnitude of the vector Δ v→ can be found from simple trigonometry. Look at the small triangle in Figure 2.59b. The triangle is isosceles, and two sides have length v, the speed of the particle. The third side is |Δ v→|. This is approximately an arc (Figure 2.59c) in a circle of radius v and subtending an angle Δθ (in radians). Hence Δv = vΔθ. Therefore, using a=Δ v Δ t the acceleration has a magnitude given by

a=v Δ θ Δ t=v ω

But ω=v r and so the acceleration is

a=v×v r=v 2 r

This gives us the magnitude of the acceleration vector for motion around a circle of radius r with constant speed v. As we see, the magnitude of the acceleration vector is constant if v is constant. But what about its direction? As Δt gets smaller and smaller, the angle Δθ gets smaller and smaller, which means that the vector Δ v→, which is in the direction of acceleration, becomes perpendicular to v→. This means that the acceleration vector is normal to the circle and directed toward the centre of the circle. It is a centripetal acceleration the rate of change of velocity of a body in circular motion; given as a=Δ v Δ t=v 2 r=r ω 2=v ω and is directed towards the centre of the circle (see Figure 2.60).

A body moving along a circle of radius r with speed v experiences centripetal acceleration that has magnitude a c=v 2 r and is directed toward the centre of the circle.

Figure 2.60: The centripetal acceleration vector is normal to the velocity vector.

We can find many equivalent expressions for the centripetal acceleration as follows. Using v=2 π r T we have that

a c=4 π 2 r 2 r T 2=4 π 2 r T 2

Using v = r ω, gives a c = ω 2 r.

Since ω = 2π f we also have a c = 4π 2 r f 2. So we can use any one of

a c=v 2 r or a c=ω 2 r or a c=4 π 2 r T 2 or a c=4 π 2 r f 2

depending on what is convenient.

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Check yourself 8

The earth has a mass of 6 × 1024 kg and completes one revolution around the sun in a circular orbit of radius 1.5 × 1011 m. What is the gravitational force between the earth and the sun?

Worked Example 2.15

aA particle moves along a circle of radius 2.0 m with constant angular speed 2.1 s−1. Determine the centripetal acceleration of the particle.

bThe radius of the earth is R = 6.4 × 106 m. Determine the centripetal acceleration due to the spinning earth experienced by someone on the equator.

cA mass moves in a circle with constant speed in a counterclockwise direction, as in Figure 2.61a. Determine the direction of the velocity change when the mass moves from A to B.

Figure 2.61: Diagram to find the change in velocity.

Answer

aWe remember that a c = ω 2 r and so a c = 2.12 × 2.0. Hence a c ≈ 8.8 m s−2.

bA mass on the equator covers a distance of 2πR in a time T = 1 day. Thus v=2×π×6.4×10 6 24×60×60=4.65×10 2 m s−1 and so a=(4.65×10 2)2 6.4×10 6=3.4×10−2 m s−2. This is quite small compared with the acceleration of free fall, and we are not aware of it in daily life.

cThe velocity at A is vertical and at B it points to the left. The change in the velocity vector is v→B−v→A, and this difference of vectors is directed as shown in Figure 2.61b.

Centripetal forces

If we know that a body moves in a circle, then we know at once that a net force must be acting on the body, since it moves with acceleration. If the speed is constant, the direction of the acceleration is toward the centre of the circle and therefore that is also the direction of the net force. It is a centripetal force the force directed toward the centre of a circular path; given as F=m v 2 r=m ω 2 r=m v ω. Consider a car that moves on a circular level road of radius r with constant speed v. Friction between the wheels and the road provides the necessary force directed toward the centre of the circle that enables the car to take the turn. (See Figure 2.62.)

Figure 2.62: A car will skid outward (i.e., will cover a circle of larger radius) if the friction force is not large enough.

Check yourself 9

A pendulum swings back and forth. The instant the string is vertical, is the pendulum bob in equilibrium?

Worked Example 2.16

aThe coefficient of static friction between the tyres of a car of mass 1100 kg and dry asphalt is about 0.70. Determine the maximum speed with which a car can take a circular turn of radius 95 m.

bIn wet conditions the coefficient of friction is reduced to 0.40. Predict the safe maximum speed now.

Answer

aThe maximum frictional force is given by f max = μ s N where the normal force is N. Hence f max = 0.70 × 1100 × 9.81 = 7554 ≈ 7.6 × 103 N. This frictional force provides the centripetal force for the car and so

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m v 2 r=f

v=f r m

v=7554×95 1100

v≈26 m s−1

bWe may repeat part a or better notice that m v 2 r=μ, m g⇒v=μ s g r. So the mass is not relevant, and the new speed is v=26×0.40 0.70≈20 m s−1.

Exam tip

It is very important that you understand worked example 2.17.

Worked Example 2.17

A particle is tied to a string and moves with constant speed in a horizontal circle. The string is tied to the ceiling. Draw the forces on the particle.

Answer

The common mistake here is to put a horizontal force pointing toward the centre and call it the centripetal force. When you are asked to find forces on a body, the list of forces that are available include the weight, normal forces (if the body touches another body), friction (if there is friction), tension (if there are strings or springs), resistance forces (if the body moves in air or a fluid), electric forces (if electric charges are involved), etc. Nowhere in this list is there an entry for a centripetal force.

Think of the word centripetal as simply an adjective that describes forces already acting on the body, not as a new force. In this example, the only forces on the mass are the weight and the tension. If we decompose the tension into horizontal and vertical components, we see that the weight is equal and opposite to the vertical component of the tension. This means that the only force left on the body is the horizontal component of the tension. We may now call this force the centripetal force. But this is not a new force. It is simply the component of a force that is already acting on the body. (See Figure 2.63.)

Figure 2.63: For worked example 2.17.

Worked Example 2.18

A mass m is tied to a string and made to move in a vertical circle of radius R with constant speed v (Figure 2.64).

aDetermine the tension in the string at the lowest and highest points of the circle.

bCalculate the minimum speed so that the string never goes slack.

Answer

aThe forces are as shown. At the lowest point, the net force is T 1 − mg and so

T 1−m g=m v 2 r

giving

T 1=m v 2 p+m g

At the highest point, the net force is mg + T 2 and so

T 2=m v 2 r−m g

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bThe string goes slack when the tension in the string becomes zero at the top. So we need v 2 > gr.

Figure 2.64: For worked example 2.18.

It is a common mistake in circular motion problems to include a force pushing the body away from the centre of the circle: a centrifugal force. It is important to stress that no such force exists. A body in circular motion cannot be in equilibrium, and so no force pushing away from the centre is required.

nature of science

Simple deductions

Newton’s second law of mechanics implies that when a body moves with acceleration a net force of magnitude ma must be acting on the body and must have a direction that is the same as that of the acceleration vector. As we saw, circular motion involves an acceleration directed toward the centre of the circle. This means that the observation of the (approximate) circular motion of planets implies the existence of a force. Newton used this fact to deduce the existence of the gravitational force: the same force that is responsible for the falling of objects toward the surface of the earth is responsible for the motion of planets around the sun.

TEST YOUR UNDERSTANDING

32 aCalculate the angular speed and linear speed of a particle that completes a 3.50 m radius circle in 1.24 s.

bDetermine the frequency of the motion.

33Calculate the centripetal acceleration of a body that moves in a circle of radius 2.45 m making 3.5 revolutions per second.

34A mass moves on a circular path of radius 2.0 m at constant speed 4.0 m s−1 (see Figure 2.65).

aCalculate the magnitude and direction of the average acceleration during a quarter of a revolution (from A to B).

bCalculate the centripetal acceleration of the mass.

Figure 2.65: For question 34.

35An astronaut rotates at the end of a horizontal test machine whose arm has a length of 10.0 m, as shown in Figure 2.66. The acceleration she experiences must not exceed 5g (take g = 10 m s–2). Determine the maximum number of revolutions per minute of the arm.

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Figure 2.66: For question 35.

36A horizontal disc of radius R rotates about a vertical axis through its centre, Figure 2.67.

Figure 2.67: For question 36.

P and Q are two points on the disc. Q is on the circumference of the disc, and P is at a distance of R/2 from the centre. Calculate the ratios:

a ω p ω Q

b v p v Q

c a p a Q

37A body of mass 1.00 kg is tied to a string and rotates on a horizontal, frictionless table.

aThe length of the string is 40.0 cm, and the speed of revolution is 2.0 m s−1. Calculate the tension in the string.

bThe string breaks when the tension exceeds 20.0 N. Determine the largest speed at which the mass can rotate.

cThe breaking tension of the string is 20.0 N, but you want the mass to rotate at 4.00 m s−1. Determine the shortest length of string that can be used.

38Estimate the length of the day if the centripetal acceleration at the equator due to the spinning earth were equal to the acceleration of free fall (g = 9.8 m s–2, earth radius = 6400 km.)

39A neutron star has a radius of 50.0 km and completes one revolution every 25 ms.

aCalculate the centripetal acceleration experienced at the equator of the star.

bThe acceleration of free fall at the surface of the star is 8.0 × 1010 m s–2. State and explain whether a probe that landed on the star can stay on the surface or whether it will be thrown off.

40The earth (mass = 6.0 × 1024 kg) rotates around the sun in an orbit that is approximately circular, with a radius of 1.5 × 1011 m.

aEstimate the orbital speed of the earth around the sun.

bDetermine the centripetal acceleration experienced by the earth.

cDeduce the magnitude of the gravitational force exerted on the sun by the earth.

41A plane travelling at a speed of 180 m s−1 along a horizontal circle makes an angle of θ = 35° to the horizontal. The lift force L is acting in the direction shown in Figure 2.68.

Figure 2.68: For question 41.

Calculate the radius of the circle.

42A cylinder of radius 5.0 m rotates about its vertical axis. A girl stands inside the cylinder with her back touching the inside wall of the cylinder. The floor is suddenly lowered, but the girl stays ‘glued’ to the wall. The coefficient of friction between the girl and the wall is 0.60.

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aDraw a free-body diagram of the forces on the girl.

bDetermine the minimum number of revolutions per minute for which the girl does not slip down the wall.

43A horizontal disc has a hole through its centre. A string passes through the hole and connects a mass m on top of the disc to a bigger mass M that hangs below the disc. The smaller mass is rotating on the disc in a circle of radius r. Determine the speed of m such that the big mass stands still. (See Figure 2.69.)

Figure 2.69: For question 43.

44The ball in Figure 2.70 is attached to a rotating pole with two strings. The ball has a mass of 0.250 kg and rotates in a horizontal circle at a speed of 8.0 m s−1. Determine the tension in each string.

Figure 2.70: For question 44.