What is the equation to calculate an angular momentum for a particle and for a rigid object?

Question

BlackTom AI · Accepted Answer

The question asks for the equations used to calculate angular momentum for a particle and for a rigid object. We will evaluate each option in turn, noting what is correct and what is incorrect.

Option a: 'p = mv for a particle, L = I ω for a rigid object' 
- For a particle, the linear momentum p = m v is not the angular momentum; angular momentum about a chosen origin is L = r × p, which equals m v r sinθ, not simply p. This makes the first part incorrect. 
- For a rigid body, L = I ω is correct provided the rotation is about a principal axis or the axis through the center of mass with the appropriate moment of inertia. However, since the particle part is wrong, the overall statement is incorrect.

Option b: 'L = mvr sin θ for a particle, L = I ω for a rigid object' 
- This matches the standard expressions: angular momentum of a point particle about a given origin is L = r × p = m r × v, which in magnitude is L = m v r sin θ, where θ is the angle between r and v. 
- For a rigid body (rotating rigid object), angular momentum about the axis of rotation is L = I ω, where I is the moment of inertia about that axis. This option correctly assigns the formulas to each case.

Option c: 'L = I ω for a particle, L = mvr sin θ for a rigid object' 
- This reverses the correct forms. A single particle does not have L = I ω unless you consider a continuous mass distribution and the appropriate parallel axis, but the standard point-particle formula is L = r × p = mvr sin θ, not I ω. The second part would be valid only if the rigid body’s rotation is described about a fixed axis, but the first part makes this option incorrect.

Option d: 'L = I ω for a particle, p = mv for a rigid object' 
- Again, the particle case is incorrect because a single particle’s angular momentum is not generally given by L = I ω; a point particle uses L = r × p = mvr sin θ. The second part, p = mv, is the linear momentum formula for a particle, but interpreting it as the rigid object’s momentum here is misleading since a rigid body can have both translational and rotational motion; this pairing is inconsistent with standard definitions of angular momentum for rigid bodies as L = I ω about the rotation axis. Therefore this option is incorrect.

To recap, option b correctly assigns L = mvr sin θ for a particle and L = I ω for a rigid object, while the other options mix incorrect pairings of angular momentum formulas with particle or rigid-body cases. The reasoning above highlights the standard definitions: point-particle angular momentum from r and p, and rigid-body angular momentum from I ω about the rotation axis.

What is the equation to calculate an angular momentum for a particle and for a rigid object?单项选择题

类似问题

Angular momentum is:

Why does an ice skater spin faster when she folds her arms close to her body?

What is an angular momentum and what is its SI unit?

Angular momentum is:

A child swings a 0.38 kg ball in a circle on a string that is 1.3 m long. If the ball makes 1.2 rev/s, what is the magnitude of the ball's angular momentum?

An ice skater performs a fast spin by pulling outstretched arms down and close to her body. What happens to her angular momentum with respect to the axis of rotation?

Angular momentum is:

How much angular momentum must be supplied by the hamstrings to bring a leg swinging at 20 rad/s to a stop, given that the leg’s moment of inertia is 0.7 kg.m2? NOTE: SI unit symbols must be used and round to two (2) significant figures. This question is worth two (2) marks.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单