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Rotational motion and angular momentum

73 Dynamics of Rotational Motion: Rotational Inertia

Learning Objectives

  • Understand the relationship between force, mass, and acceleration.
  • Explore the turning effect of force, or torque.
  • Examine analogies between force and torque, mass and moment of inertia, and linear and angular acceleration.

If you’ve ever spun a bike wheel or pushed a merry-go-round, you know that force is required to change rotational motion. As shown in Figure 73.1, applying a force to a rotating object causes angular acceleration. This mirrors how a net force causes linear acceleration in translational motion. The farther from the pivot point the force is applied, the more effective it is at causing rotation. Similarly, more massive or distributed systems resist changes in rotation—just as heavier objects resist linear acceleration.

The given figure shows a bike tire being pulled by a hand with a force F backward indicated by a red horizontal arrow that produces an angular acceleration alpha indicated by a curved yellow arrow in counter-clockwise direction.
Figure 73.1 A force is required to spin the bike wheel. The greater the force, the greater the angular acceleration. If the force is applied closer to the center (the axle), the angular acceleration is smaller. A more massive wheel also responds with less angular acceleration.

To develop a precise mathematical relationship, consider a force [latex]F[/latex] applied to a point mass [latex]m[/latex] located a distance [latex]r[/latex] from the pivot point, as in Figure 73.2. The resulting linear acceleration is:

[latex]a = \frac{F}{m}[/latex]

Rewriting Newton’s second law, [latex]F = ma[/latex], and substituting [latex]a = r\alpha[/latex], we get:

[latex]F = mr\alpha[/latex]

Torque [latex]\tau[/latex] is defined as the turning effect of a force applied at a distance from a pivot:

[latex]\tau = rF[/latex]

Multiplying both sides of the previous equation by [latex]r[/latex] yields:

[latex]\tau = mr^2\alpha[/latex]

This equation is the rotational analog of Newton’s second law, where torque plays the role of force, angular acceleration [latex]\alpha[/latex] replaces linear acceleration, and [latex]mr^2[/latex] represents a rotational equivalent of mass called the moment of inertia.

The given figure shows an object of mass m, kept on a horizontal frictionless table, attached to a pivot point, which is in the center of the table, by a cord that supplies centripetal force. A force F is applied to the object perpendicular to the radius r, which is indicated by a red arrow tangential to the circle, causing the object to move in counterclockwise direcion.
Figure 73.2 A force [latex]F[/latex] applied perpendicular to the radius [latex]r[/latex] causes the mass to accelerate in a circle about the pivot point.

Making Connections: Rotational Motion Dynamics

Rotational dynamics closely parallels translational dynamics. Just as Newton’s second law relates force, mass, and acceleration, the rotational form relates torque, moment of inertia, and angular acceleration.

Rotational Inertia and Moment of Inertia

We extend the concept of rotational inertia to more complex objects by summing the individual contributions of all their mass elements. For a collection of point masses, the moment of inertia [latex]I[/latex] is:

[latex]I = \sum mr^2[/latex]

In this context, [latex]I[/latex] is the rotational equivalent of mass, and its value depends on both the mass and how it is distributed relative to the axis of rotation. For example, the moment of inertia of a hoop with mass [latex]M[/latex] and radius [latex]R[/latex] is:

[latex]I = MR^2[/latex]

The general rotational form of Newton’s second law is:

[latex]\tau_{\text{net}} = I\alpha[/latex]

Solving for angular acceleration:

[latex]\alpha = \frac{\tau_{\text{net}}}{I}[/latex]

This relationship tells us that the greater the torque, the greater the angular acceleration, and the larger the moment of inertia, the more resistance there is to changes in rotational motion.

For biological and mechanical systems—like rotating limbs or prosthetic joints—the moment of inertia is critical to understanding motion control and energy efficiency. Distributing mass closer to the pivot decreases rotational resistance, which is why figure skaters pull their arms inward to spin faster.

Take-Home Experiment

Cut a 10 cm radius cardboard circle and label it like a clock face. Mount it on a wall so it can rotate freely. Place a lump of poster putty at 3 o’clock and observe if it rotates the wheel. Try different putty masses or move the putty closer to the center. How does changing the moment of inertia affect the system’s motion?

Problem-Solving Strategy: Rotational Dynamics

  1. Identify that torque and rotational motion are present.
  2. Define the system of interest.
  3. Draw a free-body diagram showing all external forces.
  4. Apply the rotational analog of Newton’s second law: [latex]\tau_{\text{net}} = I\alpha[/latex]
  5. Check your result for physical reasonableness (direction, magnitude, units).

Making Connections

In static systems like in Figure 73.3, where the object does not rotate, the net torque is zero. But in dynamic systems, net torque produces angular acceleration—just as net force causes linear acceleration.

Illustrations of ten different objects accompanied by their rotational inertias.
Figure 73.3 The moment of inertia depends not just on total mass, but on how that mass is distributed relative to the axis of rotation. This figure summarizes rotational inertias for several geometric shapes.

69.1 Calculating the Effect of Mass Distribution on a Merry-Go-Round

Consider the father pushing a playground merry-go-round in Figure 73.4. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

The given figure shows a man pushing a merry-go-round by a force F, indicated by a red arrow which is perpendicular to the radius r, of the merry-go-round, such that it moves in counter-clockwise direction.
Figure 73.4 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque.

Strategy

Angular acceleration is given directly by the expression [latex]\alpha =\frac{\text{net τ}}{I}[/latex]:

[latex]\alpha =\frac{\tau }{I}.[/latex]

To solve for [latex]\alpha[/latex], we must first calculate the torque [latex]\tau[/latex] (which is the same in both cases) and moment of inertia [latex]I[/latex] (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that

[latex]\mathrm{\tau }=\mathrm{rF}\phantom{\rule{0.25em}{0ex}}\text{sin θ}=\left(\text{1.50 m}\right)\left(\text{250 N}\right)=\text{375 N}\cdot \text{m.}[/latex]

Solution for (a)

The moment of inertia of a solid disk about this axis is given in Figure 73.3 to be

[latex]\frac{1}{2}{\text{MR}}^{2}\text{,}[/latex]

where [latex]M=\text{50.0 kg}[/latex] and [latex]R=\text{1.50 m}[/latex], so that

[latex]I=(0\text{.}500)(\text{50.0 kg})(\text{1.50 m})^{2}=\text{56.25 kg}\cdot {\text{m}}^{2}\text{.}[/latex]

Now, after we substitute the known values, we find the angular acceleration to be

[latex]\alpha =\frac{\tau }{I}=\frac{\text{375 N}\phantom{\rule{0.25em}{0ex}}\text{⋅}\phantom{\rule{0.25em}{0ex}}\text{m}}{\text{56.25 kg}\cdot {\text{m}}^{2}}=\text{6.67}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}[/latex]

Solution for (b)

We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia [latex]I[/latex], we first find the child’s moment of inertia [latex]{I}_{\text{c}}[/latex] by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

[latex]{I}_{\text{c}}={\mathrm{MR}}^{2}=\left(\text{18.0 kg}\right)\left(\text{1.25 m}\right)^{2}=\text{28.13 kg}\cdot {\text{m}}^{2}\text{.}[/latex]

The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of [latex]I[/latex]:

[latex]I=\text{28.13 kg}\cdot {\text{m}}^{2}+\text{56.25 kg}\cdot {\text{m}}^{2}=\text{84.38 kg}\cdot {\text{m}}^{2}.[/latex]

Substituting known values into the equation for [latex]\alpha[/latex] gives

[latex]\alpha =\frac{\text{τ}}{I}=\frac{\text{375 N}\cdot \text{m}}{\text{84.38 kg}\cdot {\text{m}}^{2}}=\text{4.44}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}[/latex]

Discussion

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.

Check Your Understanding

Torque is the analog of force, and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the number of atoms in an object. Are torque and moment of inertia similarly simple?

No. Torque depends on three factors: the magnitude of the force, the direction of the force, and the point of application (distance from the axis). Moment of inertia depends not only on the amount of mass, but also on how that mass is distributed relative to the axis of rotation. So, while the analogies are conceptually valid, these rotational quantities are more complex and involve additional parameters.

Section Summary

  • The farther a force is applied from the pivot point, the greater the resulting angular acceleration. Angular acceleration is inversely proportional to the moment of inertia.
  • When a force [latex]F[/latex] is applied to a point mass [latex]m[/latex] at a distance [latex]r[/latex] from the pivot (perpendicular to [latex]r[/latex]), the linear acceleration is:
    [latex]a = \frac{F}{m}[/latex]

    Using [latex]a = r\alpha[/latex], we substitute to obtain:

    [latex]F = mr\alpha[/latex]
  • Torque is the turning effect of a force. When the force is perpendicular to the radius, torque is:
    [latex]\tau = rF[/latex]

    Substituting from above:

    [latex]\tau = mr^2\alpha[/latex]
  • The moment of inertia [latex]I[/latex] for a system of point masses is:
    [latex]I = \sum mr^2[/latex]

    It represents the resistance to angular acceleration and depends on both mass and its distribution relative to the axis.

  • The general rotational form of Newton’s second law is:
    [latex]\tau = I\alpha[/latex]

    Or, solving for angular acceleration:

    [latex]\alpha = \frac{\tau_{\text{net}}}{I}[/latex]

Conceptual Questions

  1. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is [latex]{\mathit{ML}}^{2}\text{/3}[/latex]. Why is this moment of inertia greater than it would be if you spun a point mass [latex]M[/latex] at the location of the center of mass of the rod (at [latex]L/2[/latex]
    )? (That would be [latex]{\mathit{ML}}^{2}\text{/4}[/latex].)
  2. Why is the moment of inertia of a hoop that has a mass [latex]M[/latex] and a radius [latex]R[/latex] greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass [latex]M[/latex] and a radius [latex]R[/latex] greater than that of a solid sphere that has the same mass and radius?
  3. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque.
  4. While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?
    The given figure shows a racing bicycle leaning on a door.
    Figure 73.5 The image shows a side view of a racing bicycle. Can you see evidence in the design of the wheels on this racing bicycle that their moment of inertia has been purposely reduced? (credit: Jesús Rodriguez)
  5. A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why?

Problems & Exercises

  1. This problem considers additional aspects of example 69.1 Calculating the Effect of Mass Distribution on a Merry-Go-Round. (a) How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them?
  2. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.
  3. The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of [latex]\text{2.00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of [latex]\text{120}\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}[/latex]. What is the moment of inertia of the boxer’s forearm?
  4. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of [latex]30.00 rad/{\text{s}}^{2}[/latex] and her lower leg has a moment of inertia of [latex]\text{0.750 kg}\cdot {\text{m}}^{2}[/latex]. What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm?
  5. Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk).(a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if  there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
  6. Consider the 12.0 kg motorcycle wheel shown in Figure 73.6. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?
    The given figure shows the rear wheel of a motorcycle. A force F is indicated by a red arrow pointing leftward at a distance r from its center. Two arrows representing radii R-one and R-two are also indicated. A curved yellow arrow indicates an acceleration alpha and a curved blue arrow indicates an angular velocity omega, both in counter-clockwise direction.
    Figure 73.6 A motorcycle wheel has a moment of inertia approximately that of an annular ring.
  7. Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of [latex]4.00×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.)
  8. An automobile engine can produce 200 N ∙ m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.
  9. Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length [latex]\left(I={\mathrm{M\ell }}^{ 2}/3\right)[/latex] , prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is [latex]I={\mathrm{M\ell }}^{ 2}/12[/latex]. You will find the graphics in Figure 73.3 useful in visualizing these rotations.
  10. Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a 500-N ∙ m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is [latex]0.050\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}[/latex]. (a) What time is required for her to exactly reverse her spin? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent
  11. Unreasonable Results An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent?

Glossary

torque
the turning effectiveness of a force
rotational inertia
resistance to change of rotation. The more rotational inertia an object has, the harder it is to rotate
moment of inertia
mass times the square of perpendicular distance from the rotation axis; for a point mass, it is [latex]I={\text{mr}}^{2}[/latex] and, because any object can be built up from a collection of point masses, this relationship is the basis for all other moments of inertia
definition

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