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Show that in the case of an unbalanced disk, the equation derived in Prob. 16.92 is valid only when the mass center $G$, the geometric center $O,$ and the instantaneous center $C$ happen to lie in a straight line.

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Physics 101 Mechanics

Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

Section 2

Constrained Plane Motion

Motion Along a Straight Line

Cornell University

University of Washington

McMaster University

Lectures

04:34

In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

07:57

In mathematics, a position is a point in space. The concept is abstracted from physical space, in which a position is a location given by the coordinates of a point. In physics, the term is used to describe a family of quantities which describe the configuration of a physical system in a given state. The term is also used to describe the set of possible configurations of a system.

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Hello: everyone in this problem. We have to prove that, in case of unbalanced disc, the equation i c is equal to m r. Squared plus, i would be valid in case center of mass geometrical center and instantaneous center, like along the same line. Let us see here suppose this is a disk o is the geometrical center c is instant in a center, and g is center of mass. So this is position. Factor of center of mass is with respect to instantaneous center of rotation. It is rotating with angular velocity, omega and angular acceleration alpha in block by direction, so acceleration a can be defined as acceleration of c plus. This is acceleration a c, and this is the acceleration of g with respect to c acceleration of g. With respect to c, it will be the sum of normal reaction and tangential acceleration, so tangential acceleration is alpha, cross r g vector and normal is omega. Cross cross of r dc vector since omega is perpendicular to r g c. So we can right acceleration a will. Be a c plus alpha cross r g c, vector minus omega square r g c, vector equation: 1 petrothe, free body, diagram of the disk. So at g point it's the weight will act here it having mass into acceleration, and this is i into alphand this vector. You know r g vector position vector of center of mass from distant anent. So if we find moment of couple in a moment of force about c point having the value hive into alpha r d c cross m a substituting, the value acceleration of g already, we have calculated by using equation 1- was m. A c plus alpha cross r g c vector minus omega square r g c vector so this would be written as momentum force about ce point can be written as i into l, r g c cross m, a c vector, plus m r g c, vector crss. Alpha r d c, minus m, omega square r d, cross r d, r c cross r, t c 0 say this is equation, but 2, now r c vector cross r g c is 0 and alpha is perpendicular to r g vector. So we can say r g c cross m alpha cross r g c to be equal m r. Squared g c into l 4 substituting this value in equation 2. We can write summation of m c to be equal to hi, alpha plus m r square g c, alpha plus hermit cross m a c. This is i plus m r squared d into alpha plus r dc vector cross m. A c vector since i c is equal to i plus g, so equation 3 can be written as we can write moment of force about c point. I c into alpha r g c: cross equation: 3 rereverses 2 summission of mc is ecall to i c into alpha 1 d, f, r g c and a c r cho linear, hence g point o point c must be along the same line. That is the answer for this problem. Thanks for watchin.

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