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Problem 115 Easy Difficulty

A small clamp of mass $m_{B}$ is attached at $B$ to a hoop of mass $m_{h}$ Knowing that the system is released from rest and rolls without sliding, derive an expression for the angular acceleration of the hoop in terms of $m_{B}, m_{h}, r,$ and $\theta$

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Physics 101 Mechanics

Vector Mechanics for Engineers: Statics and Dynamics

Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

Section 2

Constrained Plane Motion

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Video Transcript

Hello friends, as in the figure, the small length of mass is a mass of clamp v attached to the hoop of mass m h. S here, m h is given mass of fop of radius r. The system is released from rest, so omega naught is 0, have roles without sliding or rolling. Without saliding we have to find angular acceleration alp. Let us see here suppose it's an acceleration alais in this direction, to acceleration of a will be towards left acceleration of, b or b with respect to a at angle. Thein. This angle is given theta, so we can write acceleration of a to b r into alpha and acceleration of b, with respect to a r into alpha at in the direction at angle, theta down to the origin, acceleration of the acceleration of a towards y plus acceleration Of p, with respect to a down to the horizontal biangletnow, we will draw 3 diagram f b. Tits weight will act downward. This is b point here. Its beak will act in this direction. This is a weight of h m h into a a m b into r into al. You can measure this distance from herethis is r, and this is r cos. Theta momentum interet will be m h into r square. This will be i into alpha so moment of coupon at c this in here it will be normal reaction. There will be a friction summation of mc. If you point taking clockwise to be positive, we can write bait of b into our sine theta is equal to. I into alpha a a and 2 r m b r alpha into r plus r cos theta plus mb into r alpha sine of theta into r sine theta plus cos theta r, plus r cos cetait of p will be mb into g r, sine theta high Momentum inertia will be m h into r square into acceleration of a r into a m b r alpha or you can take comma 2 r square 1 plus cos theta same here. You may take just simplify. Take all the common m r square cos: theta 1 plus cos theta, so solving the above equation for alpha alpha. You will get m b g sin of theta upon 2 r m h, plus mb 1 plus cos theta pi. This is the angular acceleration of the hoop. That'S all for it, thanks for watching it.

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Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

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Problem

A $4-$ lb bar is attached to a 10 -lb uniform cyl…

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Related Courses

Vector Mechanics for Engineers: Statics and Dynamics

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Top Physics 101 Mechanics Educators

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Watch More Solved Questions in Chapter 16

Video Transcript

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Christina Krawiec

Farnaz Mohseni

Meghan Miholics

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

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Christina Krawiec

Farnaz Mohseni

Meghan Miholics

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Christina Krawiec

Farnaz Mohseni

Meghan Miholics

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics