Rotational and Linear Example

A mass m is placed on a rod of length r and negligible mass, and constrained to rotate about a fixed axis. If the mass is released from a horizontal orientation, it can be described either in terms of force and accleration with Newton's second law for linear motion, or as a pure rotation about the axis with Newton's second law for rotation. This provides a setting for comparing linear and rotational quantities for the same system. This process leads to the expression for the moment of inertia of a point mass.

Moment of Inertia: Rod

For a uniform rod with negligible thickness, the moment of inertia about its center of mass is

For mass M = kg and length L = m, the moment of inertia is I = kg m² Show 显示 对于质量 M = kg 和长度 L = m，转动惯量是 I = kg m² 中文译文中的待填/计算数值依次对应：1：m 2：L 3：I。实际数值以上方原输入框为准。

The moment of inertia about the end of the rod can be calculated directly or obtained from the center of mass expression by use of the Parallel axis theorem.

The moment of inertia about the end of the rod is I = kg m². Show 显示 关于杆的端点的转动惯量是 I = kg m²。 中文译文中的待填/计算数值依次对应：1：Ie。实际数值以上方原输入框为准。

If the thickness is not negligible, then the expression for I of a cylinder about its end can be used.

如果厚度不可忽略，那么可以使用关于圆柱体端面的转动惯量表达式。

利用平行轴定理，可以利用质心表达式计算出杆的端点处的转动惯量，或者直接计算得到。

Moment of Inertia: Rod

Calculating the moment of inertia of a rod about its center of mass is a good example of the need for calculus to deal with the properties of continuous mass distributions. The moment of inertia of a point mass is given by I = mr², but the rod would have to be considered to be an infinite number of point masses, and each must be multiplied by the square of its distance from the axis. The resulting infinite sum is called an integral. The general form for the moment of inertia is:

When the mass element dm is expressed in terms of a length element dr along the rod and the sum taken over the entire length, the integral takes the form:

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Rod Moment Calculation

The moment of inertia calculation for a uniform rod involves expressing any mass element in terms of a distance element dr along the rod. To perform the integral, it is necessary to express eveything in the integral in terms of one variable, in this case the length variable r. Since the total length L has mass M, then M/L is the proportion of mass to length and the mass element can be expressed as shown. Integrating from -L/2 to +L/2 from the center includes the entire rod. The integral is of polynomial type: 均匀杆的转动惯量计算涉及将任意质量元表示为沿杆的长度元素dr。为了进行积分，必须将积分中的所有量表示为一个变量，即长度变量r。由于总长度L有质量M，因此M/L是质量与长度的比例，质量元可以如图所示表示。从中心积分从-L/2到+L/2包括整个杆。积分是多项式类型的：

Rod Moment About End

Once the moment of inertia of an object about its center of mass has been determined, the moment about any other axis can be determined by the Parallel axis theorem: 一旦确定了物体关于其质心的转动惯量，就可以利用平行轴定理确定其关于任何其他轴的转动惯量：

In this case that becomes

This can be confirmed by direct integration