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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 3 / Problem 79P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: Cycloid. A particle moves in the xy-plane. Its

Chapter 3, Problem 79P

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QUESTION:

Problem 79P

Cycloid. A particle moves in the xy-plane. Its coordinates are given as functions of time by

x(t) = R(ωt − sin ωt) y(t) = R(1 − cos ωt)

Where R and ω are constants. (a) Sketch the trajectory of the particle. (This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space is called a cycloid.) (b) Determine the velocity components and the acceleration components of the particle at any time t. (c) At what times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.

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QUESTION:

Problem 79P

Cycloid. A particle moves in the xy-plane. Its coordinates are given as functions of time by

x(t) = R(ωt − sin ωt) y(t) = R(1 − cos ωt)

Where R and ω are constants. (a) Sketch the trajectory of the particle. (This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space is called a cycloid.) (b) Determine the velocity components and the acceleration components of the particle at any time t. (c) At what times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.

ANSWER:

Solution 79P

Given that,

$x(t)=R(\omega{}t-sin\omega{}t)$

$y(t)=R(1-cos\omega{}t)$

(a) A rough sketch of the trajectory of a cycloid is shown below.

(b) To calculate velocity, we have to differentiate the $x(t)$ and $y(t)$ expressions.

Differentiating $x(t)=R(\omega{}t-sin\omega{}t)$ with respect to $t$ .

$\frac{dx}{dt}=R(\omega{}-\omega{}cos\omega{}t)$

$\frac{dx}{dt}=R\omega{}(1-cos\omega{}t)$

$v(x)=R\omega{}(1-cos\omega{}t)$ …..(1)

This is the x-component of velocity.

The x-component of acceleration can be calculated by deriving this equation (1).

$\frac{dv(x)}{dt}=R\omega{}(0-\omega{}(-sin\omega{}t))$

$\frac{dv(x)}{dt}=R\omega{}{\ }^{2}sin\omega{}t$

$a(x)=R\omega{}{\ }^{2}sin\omega{}t$ …..(2)

This is the x-component of acceleration.

Now, $y(t)=R(1-cos\omega{$

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