Ch. 10 - Rotational Motion

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 10 - Rotational Motion

Problem 102

Chapter 10, Problem 102

A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R₀ and mass M. It then has a circular hole of radius R₁ drilled into it (Fig. 10–80). The hole’s center is a distance h from the center of the disk. Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C. [Hint: Consider a solid disk and “subtract” the hole; use the parallel-axis theorem.]

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Start by recalling the formula for the moment of inertia of a solid uniform disk about its center. The moment of inertia is given by:

I = \frac{1}{2} M R^{2}

, where M is the mass of the disk and R is its radius.

Next, consider the disk with the hole. To find the moment of inertia of the disk with the hole, we can treat it as the moment of inertia of the full disk minus the moment of inertia of the removed hole. The mass of the removed hole is proportional to its area:

m = M * \frac{R^{2}}{1} / R^{0}

The moment of inertia of the removed hole about its own center is given by the same formula as for a solid disk:

I = \frac{1}{2} m R^{1}

. However, since the hole is off-center, we need to use the parallel-axis theorem to account for the distance h between the center of the hole and the center of the disk.

The parallel-axis theorem states that the moment of inertia about a new axis is the moment of inertia about the center of mass plus an additional term:

I = I cm + m h^{2}

. Here,

I cm

is the moment of inertia of the hole about its own center, and

m h^{2}

is the additional term due to the offset.

Finally, subtract the moment of inertia of the hole (including the parallel-axis term) from the moment of inertia of the full disk. The resulting expression will give the moment of inertia of the disk with the off-center hole about its center:

I = \frac{1}{2} M R^{0} - (\frac{1}{2} m R^{1} + m h^{2})

. Substitute the expressions for m and simplify as needed.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to rotational motion about an axis. It depends on the mass distribution relative to the axis of rotation. For a solid disk, the moment of inertia can be calculated using the formula I = (1/2)MR², where M is the mass and R is the radius. When a hole is drilled, the moment of inertia of the hole must be subtracted from that of the original disk.

Parallel-Axis Theorem

The parallel-axis theorem allows us to calculate the moment of inertia of an object about any axis parallel to an axis through its center of mass. It states that I = I_cm + Md², where I_cm is the moment of inertia about the center of mass, M is the mass of the object, and d is the distance between the two axes. This theorem is essential when dealing with off-center holes, as it helps in adjusting the moment of inertia accordingly.

Subtraction of Areas in Moment of Inertia

When calculating the moment of inertia for a disk with a hole, the approach involves treating the hole as a negative mass distribution. This means calculating the moment of inertia of the entire disk and then subtracting the moment of inertia of the hole. This method simplifies the problem by allowing us to use known formulas for standard shapes and apply the parallel-axis theorem to the hole's contribution.

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