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(II) A rotating merry-go-round makes one complete revolution in 4.0 s (Fig. 45). (a) What is the linear speed of a child seated 1.2 $\mathrm{m}$ from the center? (b) What is her acceleration (give components)?

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(a) 1.9 $\mathrm{m} / \mathrm{s}$(b) 3.0 $\mathrm{m} / \mathrm{s}^{2}$ towards the center

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Washington

Simon Fraser University

University of Sheffield

University of Winnipeg

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.

04:02

A rotating merry-go-round …

03:15

03:37

(II) A rotating merry-go-r…

01:45

02:55

A 5.0 -m-diameter merry-go…

02:15

Two children ride on the m…

01:46

(II) $(a)$ A grinding whee…

03:23

A child pushes a merry-go-…

05:36

A merry-go-round starts fr…

03:29

01:37

(II) (a) A grinding wheel …

01:48

03:12

04:19

A 3.0 -m-diameter merry-go…

So here the angular speed of the merry go round is going to be equal to two pi radiance. So one revolution for every 4.0 seconds. So we have a period of 4.0 seconds. This is giving us 1.57 radiance per second. And we know that four part a the velocity, the linear velocity would be equal to the Omega or the angular velocity times the radius. So this would be 1.57 radiance per second multiplied by the radius of 1.2 meters, and we find that the linear velocity ah would be equal to approximately 1.9 meters per second. For part B, the acceleration is radio and there isn't any tangential acceleration. So the radio acceleration would be equal to omega squared times are and this would be 1.57 radiance per second quantity squared times the radius of 1.2 meters. So the radio acceleration is found to be 3.0 meters per second and then this would be towards the center of the merry go round. That is the end of the solution. Thank you for watching

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