Problem

(II) A bicycle wheel of diameter 65 $\mathrm{cm}$…

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Problem 56 Hard Difficulty

(II) If a mass equal to half the mass of the wheel in Problem 55 is placed at the free end of the axle, what will be the precession rate now? Treat the extra mass as insignificant in size.

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Problem 55

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Video Transcript

to solve this problem with Lose Equation 11.13 C for the positional angular velocity. It will be given by MGR over Iomega that AJ is now the center off moss off the combination combination off the wheel on the additional mast that this be the displaced on the axis of proficient motions. The moss it is half off. The moss off the wheel is pleased on the access off petition, so it does not contribute. D'oh! The total moment off inertia. It's a moment of inertia due to this mask about the axis of rotation in Cedo. Therefore, the total moment off inertia will still be saying on will be equal dough that off the real Ah, it is. Am I scared? As if being the one is a circle now? But we only this expression now. So let's forget about that. And let's just call the moment of inertia, the total moment of inertia equal dough, the moment of inertia after real. I. Now the addition off the moss does not change the center off Moss, but it does change the total mass that is now equal dough months off the wheel. Unless the mass that is placed on the access off rotation, which is, um, overto so the total mass is now three m over. Now let's find the sent it off, boss. So I will not be equal to. So we will use this formula for the same bet off, Boss. So m one is Thomas off the real. So that is him, x one, which is the distance off the real from the access of tradition. So that is l over to that ideas, okay. And do is the small lamb on X two for that is just and it is lent off the exits. And total mass is three m overto m one. Bless him dough as we found here. So this gives the center of mass to the equal toe do over three times. Now, we can use the seclusion to find that issue off the new positional velocity over the original or old one. New loss time. See, dimes. I knew. So this is basically our new now. Oh, large Iomega on Dhe III and Omega staying the same. So yet I have our Disick wish in now. Modest. Simplify this a cushion in Burt. So these stones cancel a we now em neurologist The total mass now is three m over too. On our on our new is this distance The news ended off months and in old is just the moss off the wheel before the small moss was added at the center. So this will be going toe just end on DDE the center off. Months off, this will be half off as we rode over here This gifts the ratio off both the positional velocities processional angular velocity is to be equally do don't. Now we can use the answer from Problem 55 to substitute for this old positional velocity that we got to be equal to eat radiant for a second in problem 55 those wars, the new position and velocity will be two times eight and so this would be 16 radian per second.

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Physics for Scientists and Engineers with Modern Physics

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Problem

(II) A bicycle wheel of diameter 65 $\mathrm{cm}$…

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Related Courses

Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 11

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Andy Chen

Zachary Mitchell

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Andy Chen

Zachary Mitchell

Jared Enns

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Andy Chen

Zachary Mitchell

Jared Enns

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics