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10.5 Calculating moments of inertia  (Page 5/9)

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Angular velocity of a pendulum

A pendulum in the shape of a rod ( [link] ) is released from rest at an angle of 30 ° . It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point?

Figure shows a pendulum in the form of a rod with a mass of 300 grams and length of 30 centimeters. Pendulum is released from rest at an angle of 30 degrees.
A pendulum in the form of a rod is released from rest at an angle of 30 ° .

Strategy

Use conservation of energy to solve the problem. At the point of release, the pendulum has gravitational potential energy, which is determined from the height of the center of mass above its lowest point in the swing. At the bottom of the swing, all of the gravitational potential energy is converted into rotational kinetic energy.

Solution

The change in potential energy is equal to the change in rotational kinetic energy, Δ U + Δ K = 0 .

At the top of the swing: U = m g h cm = m g L 2 ( cos θ ) . At the bottom of the swing, U = m g L 2 .

At the top of the swing, the rotational kinetic energy is K = 0 . At the bottom of the swing, K = 1 2 I ω 2 . Therefore:

Δ U + Δ K = 0 ⇒ ( m g L 2 ( 1 − cos θ ) − 0 ) + ( 0 − 1 2 I ω 2 ) = 0

or

1 2 I ω 2 = m g L 2 ( 1 − cos θ ) .

Solving for ω , we have

ω = m g L I ( 1 − cos θ ) = m g L 1 / 3 m L 2 ( 1 − cos θ ) = g 3 L ( 1 − cos θ ) .

Inserting numerical values, we have

ω = 9.8 m / s 2 3 0.3 m ( 1 − cos 30 ) = 3.6 rad / s .

Significance

Note that the angular velocity of the pendulum does not depend on its mass.

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Summary

  • Moments of inertia can be found by summing or integrating over every ‘piece of mass’ that makes up an object, multiplied by the square of the distance of each ‘piece of mass’ to the axis. In integral form the moment of inertia is I = ∫ r 2 d m .
  • Moment of inertia is larger when an object’s mass is farther from the axis of rotation.
  • It is possible to find the moment of inertia of an object about a new axis of rotation once it is known for a parallel axis. This is called the parallel axis theorem given by I parallel-axis = I center of mass + m d 2 , where d is the distance from the initial axis to the parallel axis.
  • Moment of inertia for a compound object is simply the sum of the moments of inertia for each individual object that makes up the compound object.

Conceptual questions

If a child walks toward the center of a merry-go-round, does the moment of inertia increase or decrease?

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A discus thrower rotates with a discus in his hand before letting it go. (a) How does his moment of inertia change after releasing the discus? (b) What would be a good approximation to use in calculating the moment of inertia of the discus thrower and discus?

a. It decreases. b. The arms could be approximated with rods and the discus with a disk. The torso is near the axis of rotation so it doesn’t contribute much to the moment of inertia.

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Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why?

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The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is m L 2 / 3 . Why is this moment of inertia greater than it would be if you spun a point mass m at the location of the center of mass of the rod (at L /2) (that would be m L 2 / 4 )?

Because the moment of inertia varies as the square of the distance to the axis of rotation. The mass of the rod located at distances greater than L /2 would provide the larger contribution to make its moment of inertia greater than the point mass at L /2.

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Why is the moment of inertia of a hoop that has a mass M and a radius R greater than the moment of inertia of a disk that has the same mass and radius?

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Problems

While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg-m 2 and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter’s shoe if it is 1.05 m from the hip joint?

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Using the parallel axis theorem, what is the moment of inertia of the rod of mass m about the axis shown below?

Figure shows a rod that rotates around the axis that passes through it at 1/6 of length from one end and 5/6 of length from the opposite end.

I = 7 36 m L 2

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Find the moment of inertia of the rod in the previous problem by direct integration.

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A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of 60 ° with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

Figure shows a rod that is released from rest at an angle of 60 degrees with respect to the horizontal.

v = 7.14 m / s .

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A pendulum consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm (see the following figure). If the pendulum is released from rest at an angle of 30 ° , what is the angular velocity at the lowest point?

Figure shows a pendulum that consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm. The pendulum is released from rest at an angle of 30 degrees.
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A solid sphere of radius 10 cm is allowed to rotate freely about an axis. The sphere is given a sharp blow so that its center of mass starts from the position shown in the following figure with speed 15 cm/s. What is the maximum angle that the diameter makes with the vertical?

Left figure shows a solid sphere of radius 10 cm that first rotates freely about an axis and then received a sharp blow in its center of mass. Right figure is the image of the same sphere after the blow. An angle that the diameter makes with the vertical is marked as theta.

θ = 10.2 °

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Calculate the moment of inertia by direct integration of a thin rod of mass M and length L about an axis through the rod at L /3, as shown below. Check your answer with the parallel-axis theorem.

Figure shows a rod that rotates around the axis that passes through it at 1/3 of length from one end and 2/3 of length from the opposite end.
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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