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The shutter shown was formed by removing one quarter of a disk of 0.75 -in. radius and is used to interrupt a beam of light emanating from a lens at $C .$ Knowing that the shutter weighs 0.125 lb and rotates at the constant rate of 24 cycles per second, determine the magnitude of the force exerted by the shutter on the shaft at $A$

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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

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

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.

01:49

The shutter shown was form…

06:46

The circular disk has a we…

04:48

A 6 -in-diameter hole is c…

01:16

A Newton's rings appa…

02:58

The uniform semicircular r…

02:13

So here we want to find you're the central oId of a circular sector, and first we know that here are Equalling point 75 inches and so we can actually find be centrally for the circular sector. This would be to our sign of Alfa divided by three Alfa. This would then be equaling 22 multiplied by 220.75 inches, multiplied by sine of three pie over four. This would be divided by three multiplied by three pie over four, and we find that then the century is equaling 2.15 005 inches. Then your acceleration Equalling our omega squared Regan say Omega equaling, then 24. No, it's blood by two pi Equalling, then 150 0.8 radiance her second. We want to then use Newton's second law, and we can say that our equaling M times ace of an Equalling and our omega squared simply to substitute in for the linear acceleration. So this is the equal to then 0.1 to £5. This would be divided by 32 0.2 feet, her second squared. This would be multiplied by 0.1500 five inches, divided by 12 inches per foot and then multiplied by 150.8 radiance per second. Quantity squared and we find that they are four stars equaling 1.1 04 pounds. So this would be and then if we were to say here we can say this would be essentially the force on the shaft pointing towards the centre of this 3/4 circle here so again be £1.104. So that would be the force. The magnitude of the force on the shaft. That is the end of the solution. Thank you for watching.

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