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5.2Rotational equilibrium and Newton’s second law

Before discussing rotational equilibrium and mechanics, we need to introduce torque the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the axis of rotation:
τ = Fd or τ = Fr sin θ.

Torque

Torque has to do with the ability of a force to produce rotation.

Figure 5.5 shows a rigid body that is free to rotate about the axis indicated. A force F is applied at a point Q a distance r from the axis.

Figure 5.5: The line through the vector representing the force is called the line of action of the force. Torque is the product of the magnitude of the force times the perpendicular distance between the axis and the line of action of the force.

The green dotted line is the extension of the force and is called the line of action of the force. The perpendicular distance between the line of force and the axis is d.

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KEY POINT

We define the torque τ of the force F about the axis to be the product of the force times the perpendicular distance between the line of action of the force and the axis: τ = F d.

Since the distance d is given by d = r sin θ we also have τ = Fr sin θ, where θ is the angle between the force and the vector from the axis to the point where the force is applied (Figure 5.6).

The unit of torque is the N m. (Although this combination of units is equivalent to the joule, we never express torque in joules.)

Figure 5.6: a The force tends to rotate the body about the axis in the clockwise direction. This is because the force has a torque about that axis. b A wrench turning a screw is a real-life application of torque. A smaller force further from the axis has the same turning effect as a larger force closer to the axis.

Notice that if the force is directed through the axis, the torque is zero (Figure 5.7). The force cannot produce a rotation in this case.

Figure 5.7: These two forces pass through the axis and so have zero torque about that axis.

CHECK YOURSELF 1

Look at Figure 5.8. In which diagram, a–d, is the torque of the force the greatest? The rod is 1 m long.

Figure 5.8: For check yourself question 1.

Worked example 5.5

Find the torque of the forces in the three diagrams of Figure 5.9. The length of the rod is 4 m.

Figure 5.9: For worked example 5.5.

Answer

a τ = Fr sin θ = 20 × 4 × sin 90° = 80 N m.

b τ = Fr sin θ = 20 × 4 × sin 30° = 40 N m.

c τ = Fr sin θ = 20 × 4 × sin 0° = 0. This force is directed through the axis; it cannot turn the body and has zero torque.

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Torque is essential in discussing situations of equilibrium. For the case of point particles, equilibrium means that the net force is zero. For a rigid body we have to distinguish between translational equilibrium the net force of the body is zero, when the net force of the body is zero, and rotational equilibrium the net torque on the body is zero, when the net torque of the body is zero. We will discuss equilibrium further in the next section.

In translational equilibrium, the centre of mass of the body (the point where all the mass is assumed to be located) remains at rest or moves in a straight line at constant speed.

Equilibrium

A simple example of equilibrium is a see-saw, as shown in Figure 5.10. The rod can rotate about its pivot—the point of support. How can we find the force F 2 required for equilibrium? We do not want the rod to move or to rotate. We must apply the conditions for translational and rotational equilibrium.

Figure 5.10: A see-saw in equilibrium.

Translational equilibrium says that the net force is zero. Apart from the two forces F 1 and F 2 we also have the normal force N on the rod from the point of support. (We are ignoring the mass of the rod.) Equilibrium demands that

F 1+F 2=N

Now, for rotational equilibrium, the net torque must be zero. The normal force does not have a torque about the axis. Force F 1 tends to rotate the body in the counterclockwise direction; force F 2 tends to rotate the body in the clockwise direction. The two forces have opposite torques. We must set these torques equal to each other to get a zero net torque. This gives

F 1×1.2=F 2×0.8

240×1.2=F 2×0.8

F 2=360 N

From the translational equilibrium condition we can now find

N=600 N.

Worked example 5.6

Figure 5.11a shows a uniform ladder of length 5.0 m resting against a vertical, frictionless wall. The other end of the ladder rests on the floor where a frictional force f prevents the ladder from slipping. The weight of the ladder is 350 N.

Calculate the minimum coefficient of static friction between the ladder and the floor so that the ladder does not slip.

The ladder makes an angle of 60° with the floor.

Figure 5.11: For worked example 5.6.

Answer

It is useful to first draw in the forces acting, as shown in Figure 5.11b.

Translational equilibrium demands that

N=f

R=W

Since W = 350 N, we find R = 350 N right away.

We must now apply the condition for rotational equilibrium. But what is the axis of rotation? As stated in the exam tip box, we can take anything we wish for the rotation axis. It is convenient to choose an axis such that as many forces as possible pass through. In this way their torques will be zero and we will not have to deal with them. So choosing the point where the ladder touches the floor, we have to find the torque of W and of N. The torque of N is N × b = N × 5.0 × sin 60°. The torque of W is W × a = W × 2.5 × cos 60°. The two torques are opposite so we set them equal to each other:

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N×5.0×sin 60°=W×2.5×cos 60°N=W×2.5×cos 60°5.0×sin 60°=101 N

This implies that f = 101 N. Now f max = μ s R and so μ s=f max R=101 350≈0.29.

Exam tip

To solve a rotational equilibrium problem:

●demand that the net force is zero

●demand that the net torque is zero

●you may take torques about any axis, not just the actual axis of rotation.

Moment of inertia

Consider a rigid body that is free to rotate about some axis. Imagine splitting the body into tiny bits, each of some mass and distance from the axis. Figure 5.12 shows three of these bits—but you have to imagine the entire rigid body is covered by bits like these.

Figure 5.12: To calculate the moment of inertia we imagine splitting the entire body into tiny bits.

Each bit has mass m i, and its distance from the axis is r i. We multiply the mass of each bit by the distance of the bit to the axis squared and sum up for all the bits making up the body; that is, we form the quantity

I=∑m i r i 2

which we call the moment of inertia a measure of the distribution of mass of an extended body about an axis of rotation: I=∑m i r i 2 of the body about the given axis.

Consider a particle of mass m that is free to rotate about an axis a distance R from the particle, as shown in Figure 5.13a.

The moment of inertia of the particle about the given axis is

I=∑m i r i 2

But here there is just one ‘bit’ making up the body (the particle itself) and so the sum just has one term:

I=m R 2

For two particles of the same mass (Figure 5.13b) the moment of inertia is

I=∑m i r i 2=m R 2+m R 2=2 m R 2

Notice that the moment of inertia depends on the axis we choose. The same body will have different moments of inertia for different axes of rotation.

Figure 5.13: a A single particle rotating about a fixed axis. b Two particles moving on the same circle.

Consider a ring of radius R, as shown in Figure 5.14. We break up the ring into small bits of mass m i. In this case, each bit has the same distance from the axis, so

I=∑m i r i 2=∑m i R 2=R 2∑m i=M R 2

where M = ∑m i is the total mass of the ring.

Figure 5.14: A ring rotating about a fixed axis.

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It is not possible to find the moment of inertia of other arrangements as easily as in the previous cases. A few examples are shown with the axis of rotation indicated in Figure 5.15.

Figure 5.15: Moments of inertia of a few common shapes about the axes shown.

So we see that the moment of inertia is a quantity that depends on the mass of the body and how this mass is distributed around the rotation axis. The closer the mass is to the axis, the smaller the moment of inertia. The unit of moment of inertia is kg m2.

CHECK YOURSELF 2

Look at Figure 5.16. About which axis (dashed line) is the moment of inertia of the uniform equilateral triangle the greatest?

Figure 5.16: For check yourself question 2.

But why do we need to define such a quantity? Consider a body that rotates about some axis with angular speed ω and again break up the body into tiny bits, each of mass m i, as in Figure 5.17.

Figure 5.17: The kinetic energy of each bit of the body makes the total kinetic energy of the entire body.

The kinetic energy of rotational motion E K=1 2 I ω 2 of the body is the sum of the kinetic energies of all the bits:

E K=1 2 m 1 v 1 2+1 2 m 2 v 2 2+1 2 m 3 v 3 2+⋯=∑1 2 m i v i 2

Now, each bit is at a different distance r i from the axis of rotation, but all bits have the same angular velocity ω.

Since v i = ω r i we deduce that

E K=1 2 m 1 ω 2 r 1 2+1 2 m 2 ω 2 r 2 2+1 2 m 3 ω 2 r 3 2+…=1 2(∑m i r i 2)ω 2

This is similar to the formula for kinetic energy of a point particle, but here velocity is replaced by angular velocity and mass is replaced by the quantity I=∑m i r i 2.

This is precisely the quantity we defined as the moment of inertia of the body.

Exam tip

The moment of inertia is to rotational motion what mass is to linear motion.

So, for a body rotating about some axis, the kinetic energy is

E K=1 2 I ω 2

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where I is the moment of inertia about the rotation axis and ω is the angular velocity of the body about the (same) rotation axis.

Worked example 5.7

A ring (moment of inertia MR 2) of mass 0.45 kg and radius 0.20 m rotates with angular velocity 5.2 rad s–1. Calculate the kinetic energy of the ring.

Answer

The moment of inertia is

I=M R 2=0.45×0.20 2=1.8×10−2 kg m 2

The kinetic energy is therefore

E K=1 2 I ω 2=1 2×1.8×10−2×5.2 2=0.24 J

Rolling without slipping

When you ride a bicycle, the wheels rotate and at the same time the bicycle rolls forward.

So, consider a wheel of radius R that just rotates about its axis as in Figure 5.18a. All points on the wheel have the same angular velocity. In particular, points on the circumference have the same linear speed v = ω R.

Now stop the rotation and imagine making the wheel just slide with velocity u along a horizontal floor. Every point of the wheel has a velocity u to the right.

Now imagine that the wheel moves forward as it rotates. Every point on the circumference of the wheel now has two motions:

●the motion due to the rotation

●the motion due to the translation of the wheel forward.

This means that the top point has linear velocity v = u + ω R and the bottom point a linear velocity v = u − ω R.

Figure 5.18: a A rotating disc: every point of the disc has a different linear velocity. b A sliding disc: every point on the disc has the same velocity. c A disc that rolls without slipping: the point of contact has zero velocity.

Rolling without slipping means that the centre of mass of the body has moved forward a distance of 2πR in a time equal to the period of revolution T (see Figure 5.19). (R is the distance from the centre of mass to the point of contact with the ground.)

So, u=2 π R T.

Figure 5.19: In one period, the centre of mass moves forward a distance 2πR.

Since ω=2 π T, it follows that u = ω R; in other words, for rolling without slipping the linear speed of the centre of mass is given by ω R.

This means that the point of contact with the ground has net velocity zero (see Figure 5.20).

Figure 5.20: For rolling without slipping, the contact point has net velocity zero.

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If we have a sphere of radius R and we apply a force on it, the centre of mass of the sphere will start to slide in the direction of the force and may also rotate.

If the force is applied

●through the centre of mass, the ball will slide but will not rotate (there is no torque to cause rotation)

●anywhere else, it will cause sliding as well as rotation.

Because angular acceleration is the rate of change of angular velocity, the condition of rolling without slipping may also be expressed in terms of angular acceleration: a = α R.

Kinetic energy of a body that rolls without slipping

We saw that a rotating body has kinetic energy

E K=1 2 I ω 2

If the body rolls, then we have to include the kinetic energy due to the translational motion of the centre of mass. The total kinetic energy is then

E K=1 2 M v 2+1 2 I ω 2

where v is the speed of the centre of mass.

If the centre of mass of the body is at a height h from some horizontal level then the gravitational potential energy is Mgh, and so the total mechanical energy is

E T=1 2 I ω 2+1 2 M v 2+M g h

In the absence of resistance forces the total energy is conserved.

If we have rolling without slipping v and ω are related by v = ω R.

Worked example 5.8

A cylinder of mass M and radius R (moment of inertia I=1 2 M R 2) begins to roll without slipping from rest down an inclined plane (Figure 5.21).

Calculate the linear speed of the cylinder when it reaches level ground a vertical distance h lower.

Figure 5.21: For worked example 5.8.

Answer

The initial energy is Mgh.

The energy on the horizontal level is

E K=1 2 M v 2+1 2 I ω 2.

Equating the two energies gives

1 2 M v 2+1 2 I ω 2=M g h

The moment of inertia of the cylinder is

I=1 2 M R 2

Because the cylinder rolls without slipping, v = ω R or ω=v D

Therefore, 1 2 M v 2+1 2(1 2 M R 2)v 2 R 2=M g h 1 2 v 2+1 4 v 2=g h 3 v 2 4=g h

The linear speed is

v=4 g h 3

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Exam tip

Notice that, if the body were a point particle, the answer for the speed would be v=2 g h, greater than the answer to worked example 5.8.

Newton’s second law applied to rotational motion

When we solve problems in mechanics with rigid bodies we must apply Newton’s second law to the centre of mass of the body

F net=M a

That is, in exactly the same way we would treat a point particle.

But we must also apply Newton’s second law to the rotational motion of the body; it can be shown that this law becomes

τ net=I α

(This is expected because torque plays the role of force and moment of inertia plays the role of mass.)

Exam tip

Moment inertia is to rotational motion what mass is to linear motion.

Torque is to rotational motion what force is to linear motion.

theory of knowledge

Once the correspondence between linear and angular quantities is established, the formulae relating the angular quantities can be guessed from the corresponding formulae for linear quantities. The formalism is the same, and so the formulae are the same. We have seen similar connections in physics before, for example between electricity and gravitation.

We begin with an example of a sphere that rolls without slipping down an inclined plane that makes an angle θ to the horizontal. The sphere has mass M and radius R. We want to find the linear acceleration of the sphere as it comes down the plane.

Figure 5.22 shows the forces on the sphere. Notice right away that a frictional force must be present. Without one, the sphere would just slide down the plane and would not roll. (The problem would then be identical to a problem with a point mass.)

Figure 5.22: A sphere that rolls without slipping down an inclined plane. There has to be a frictional force to provide the torque for turning.

The net force on the sphere down the plane is

M g sin θ−f

So

F net =M a

Mg sin θ − f = Ma (Newton’s second law for translational motion)

(We also have that N = Mg cos θ.)

The net torque about a horizontal axis through the centre of mass is fR.

N and Mg have zero torques about this axis.

The moment of inertia of a sphere about its axis is I=2 5 M R 2.

So

τ=I α

f R=2 5 M R 2 α (Newton’s second law for rotational motion)

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We assume rolling without slipping and so

α=a R

And our two equations become

{M g sin θ−f=M a f R=2 5 M R 2 a R

These simplify to

{M g sin θ−f=M a f=2 5 M a

So

M g sin θ−2 5 M a=M a⇒a=5 g sin θ 7

(For a body other than a sphere, the result would be a=g sin θ 1+I M R 2, where I is the moment of inertia about the centre of mass.)

Exam tip

Notice that if the body were a point particle the answer for the acceleration would be a = g sin θ and is larger than that for the sphere (or any extended body).

Let us now calculate the speed of the ball after its centre of mass is lowered by a vertical distance h. The distance travelled down the plane is s and h = s sin θ (see Figure 5.23).

Figure 5.23: Relation between the vertical distance and the distance along the plane.

v 2=2 a s=2×5 g sin θ 7×s=10 g h 7

We will get the same result if we apply conservation of energy:

1 2 M v 2+1 2 I ω 2=M g h 1 2 M v 2+1 2(2 5 M R 2)v 2 R 2=M g h 1 2 v 2+1 5 v 2=g h 7 v 2 10=g h v 2=10 g h 7

At this point you may be wondering why we have been able to use energy conservation when there is a frictional force present. The answer is that the point of contact where the frictional force acts is not sliding; it is a point that is instantaneously at rest (as we saw earlier), and so the frictional force does not do any work and does not transfer any energy into thermal energy. Friction forces would do work, if we had rolling with slipping; but we will not consider such cases in this course.

CHECK YOURSELF 3

Figure 5.24 shows four diagrams; in each diagram, two identical, uniform spheres are joined by a rod of negligible mass. In which diagram, a–d, will the angular acceleration be the greatest? The applied forces are all the same and the distances are centre-to-centre distances.

Figure 5.24: For check yourself question 3.

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Worked example 5.9

A block of mass m is attached to a string that goes over a cylindrical pulley (Figure 5.25).

When the block is released the pulley begins to turn as the block falls.

The mass of the pulley is M, and its radius is R. Calculate the acceleration of the block.

Figure 5.25: For worked example 5.9.

Answer

Applying the second law to the motion of the hanging mass,

m g−T=m a

Applying the second law to the rotation of the pulley gives

T R=I α

The angular and the linear accelerations are related by a = α R. So the two equations become

{T R=I a R m g−T=m a

Solve the first equation for T:
T=I a R 2=1 2 M R 2 a R 2=1 2 M a

and substitute in the second equation to get

m g−1 2 M a=m a m g=a(m+M 2)a=m g m+M 2

Worked example 5.10

A snooker ball of mass M and radius R is hit horizontally at a point a distance d above the centre of mass (Figure 5.26). Determine d such that the ball rolls without slipping.

Figure 5.26: For worked example 5.10.

Answer

The torque on the ball about an axis through the centre of mass is Fd.

Therefore

F d=2 5 M R 2 α

The net force on the ball is F, and so

F=M a

The ball does not slip, and so the angular and the linear accelerations are related by a = α R.

This means that we have two equations:

{F d=2 5 M R 2 a R F=M a

Dividing side by side to get rid of F we get

d=2 5 R

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Work and power

Consider a constant force F that is applied on a rigid body for a time ∆t. Let this force move its point of application by a distance ∆s, rotating the body by an angle ∆θ in the process. The force does work W = F ∆s. But ∆s = R ∆θ, and so the work is also given by

W=F R Δ θ

The torque of the force is τ = FR and so

W=τ Δ θ

The familiar result from linear mechanics that the work done by the net force equals the change in kinetic energy holds here as well in the form that work done by the net torque is the change of rotational kinetic energy.

The power developed by the force is

P=W Δ t=τ Δ θ Δ t

P=τ ω

Linear motion | Rotational motion
F = m a | τ = I α
W = F ∆s | W = τΔθ
P = F v | P = τ ω

The formulae are direct analogues of the corresponding quantities in linear motion.

CHECK YOURSELF 4

A point particle slides down an inclined plane. A cylinder and a ring have the same radius and roll without slipping down the same inclined plane. All three objects have the same mass. All three start from rest at the same point on the inclined plane. Which arrives at the base of the inclined plane first and which arrives last?

Worked example 5.11

A disc of moment of inertia 25 kg m2 rotating at 320 revolutions per minute must be stopped in 12 s.

aCalculate:

ithe work needed to stop it

iithe average power developed in stopping it.

bDetermine:

ithe torque that stopped the disc (assuming it is constant).

iithe angle by which the body turned while stopping.

Answer

a iThe initial kinetic energy of the rotating disc is E K=1 2 I ω 2.

We must find ω:ω=320×2 π 60=33.5 rad s−1.

Hence E K=1 2×25×33.5 2=1.40×10 4≈1.4×10 4 J.

This is the work required.

iiThe average power is, therefore, P=1.40×10 4 12=1.17×10 3≈1.2×10 3 W.

b iWe can find the torque from τ=I α=25×33.5 12=69.8≈70 N m.

But we can also use P = τ ω.

The average power involves the average angular speed, and so P¯=τ ω¯.

Since ω¯=33.5 2=16.75 rad s−1, the torque stopping the disc is τ=1.17×10 3 16.75=69.8≈70 N m.

iiFrom W = τΔθ, we find Δ θ=W τ=1.40×10 4 69.85≈200 rad.

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TEST YOUR UNDERSTANDING

6A uniform rod of weight 450 N and length 5.0 m is supported at its ends (Figure 5.27). A block of weight 120 N is placed at a distance of 2.0 m from the left end. Calculate the force at each support.

Figure 5.27: For question 6.

7A uniform rod of mass 30 kg and length 4.0 m is supported from its left end and at a point a distance of 0.80 m from the middle as shown in Figure 5.28.

Figure 5.28: For question 7.

Calculate the largest distance x that a boy of mass 40 kg can walk to without tipping the rod over.

8A uniform rod of length 2.0 m and weight 58 N is supported horizontally by a cable attached to a vertical wall (Figure 5.29).

Figure 5.29: For question 8.

Calculate:

athe tension in the cable

bthe magnitude and direction of the force exerted by the wall on the rod.

9A body of moment of inertia 0.12 kg m2 is uniformly accelerated from rest to an angular velocity of 24 rad s–1 in 6.0 s. What is the torque that acted on the body?

10A torque of 28 N m acts on a body of moment of inertia 3.2 kg m2 for 4.0 s. The body is initially at rest.

aWhat is the final angular velocity?

bWhat is the change in the rotational kinetic energy of the body?

11A uniform disc has radius R. A circular disc of radius r is cut away leaving a disc with a hole. The mass of the disc with the hole is M. Show that the moment of inertia of this solid about the axis shown is 1 2 M(R 2+r 2). (You do not need calculus to do this. This is an above exam level question, just for fun.)

Figure 5.30: For question 11.

12A cylinder of mass 5.0 kg and radius 0.20 m is attached to an axle through its centre of mass and parallel to its axis. A constant force of 6.5 N acts on the surface of the cylinder at right angles to the axle. Find the angular speed of the cylinder after 5.0 s.

13A point particle, a sphere, a cylinder and a ring all have mass M. The solid bodies have the same radius R. The four bodies are released from the same height on an inclined plane. Determine the order, from least to greatest, of the speed when they reach level ground. (Assume rolling without slipping.)

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14A disc of mass 12 kg and radius 0.35 m is spinning with angular velocity 45 rad s–1.

aDetermine how many revolutions per minute (rpm) the disc is making.

bA force is applied to the rim of the disc in a direction that is tangential to the disc’s circumference. Determine the magnitude of this force if the disc is to stop spinning after 4.0 s.

cCalculate the work done by the force in stopping the disc.

dWhat is the average power developed by the force?

eCalculate the number of revolutions made before stopping.

15A rod of length L = 1.20 m is supported by a fixed support at its left end and is held horizontal (Figure 5.31). The moment of inertia about the axis is given by 1 3 M L 2.

Figure 5.31: For question 15.

The rod is released from rest.

aCalculate the initial angular acceleration of the rod.

bState and explain whether the angular acceleration is constant in magnitude as the rod rotates.

cCalculate the angular velocity of the rod as it moves past the vertical position.

16A force F is applied horizontally to a cylinder of mass M and radius R as shown in Figure 5.32.

Figure 5.32: For question 16.

The force is applied at a point a distance d directly above the centre of mass of the cylinder. Determine d in terms of R such that the cylinder rolls without slipping. (The moment of inertia of a cylinder about its axis is 1 2 M R 2.)

17A block of mass m hangs from the end of a string that goes over a pulley and is connected to another block of mass M that rests on a horizontal table (Figure 5.33).

Figure 5.33: For question 17.

aAssuming that the pulley is massless, calculate the acceleration of each block when the smaller block is released. There are no frictional forces.

bThe pulley is now assumed to have mass M and radius R. The small block is again released. The pulley turns as the blocks move. Calculate the acceleration of each block. (Hint: the tensions in each string are different.)

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18The moment of inertia of a cylinder of mass M and radius R is 1 2 M R 2. A cylinder rolls without slipping on a horizontal surface with angular speed ω.

aShow that the kinetic energy of the cylinder is E K=3 4 M ω 2 R 2.

A cylinder of mass 4.0 kg and radius 0.20 m rolls without slipping on a horizontal surface with angular speed 5.0 rad s−1. The cylinder then descends a vertical distance of 0.80 m along a ramp that makes an angle of 30° to the horizontal.

bWhat is the final angular speed of the cylinder?

cWhat is the angular acceleration of the cylinder on the ramp?

dWhat is the net torque on the cylinder when on the ramp?