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(II) Four equal masses $M$ are spaced at equal intervals, $\ell$ along a horizontal straight rod whose mass can be ignored. The system is to be rotated about a vertical axis passing through the mass at the left end of the rod and perpendicular to it. (a) What is the moment of inertia of the system about this axis? $(b)$ What minimum force, applied to the farthest mass, will impart an angular acceleration $\alpha ?$ (c) What is the direction of this force?

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(a) 14 $\mathrm{Ml}^{2}$(b) $\frac{14}{3} M \ell \alpha$(c) The force must be perpendicular to the rod connecting the masses, and perpendicular to the axisof rotation. An appropriate direction is shown in the diagram.

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

University of Washington

Hope College

McMaster University

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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but essentially be the diagram of the system. And in order to find, we know that we're gonna treat each masses a point particle. So essentially, we know that the first master, the first mass, is at the of axis of rotation This staff lines essentially representing the axis of rotation. And so, for part A to find the total moment of inertia, this would simply be the second mass ml squared, plus the third mass m times to l squared plus the third mass, Uh, M times three l quantity squared. And so this is equaling 14 m. L squared. This would be the total moment of inertia for part A. Now, for part B, they want us to find the perpendicular force. We know that the torque is equaling the moment of inertia, times the angular acceleration. And this would be equaling the perpendicular force, which we have labeled here multiplied by our and so the perpendicular force would be equaling the moment of inertia times the angular acceleration divided by our This is gonna be equaling two, 14 times. Um l squared times, Alfa, and then here the total Essentially the total radius of this would be three times l And so this would be equaling 2 14 m. L Alfa over three. This would be your answer to part B for the magnitude of the perpendicular force and for part C, we know that the force must be We can say that the force must be perpendicular to the rod, connecting the masses and perpendicular to the access of rotation. That is the end of the solution.

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