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Taylor CMCH 1 P 41

The document provides a solution to a physics problem involving an astronaut twirling a mass on a string in a gravity-free environment. The problem involves writing Newton's second law in polar coordinates and finding the tension in the string. The solution derives an expression for the tension as mRω^2, where m is the mass, R is the string length, and ω is the angular velocity.

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38 views2 pages

Taylor CMCH 1 P 41

The document provides a solution to a physics problem involving an astronaut twirling a mass on a string in a gravity-free environment. The problem involves writing Newton's second law in polar coordinates and finding the tension in the string. The solution derives an expression for the tension as mRω^2, where m is the mass, R is the string length, and ω is the angular velocity.

Uploaded by

noemiefontaine55
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Taylor Classical Mechanics - Problem 1.

41 Page 1 of 2

Problem 1.41
An astronaut in gravity-free space is twirling a mass m on the end of a string of length R in a
circle, with constant angular velocity ω. Write down Newton’s second law (1.48) in polar
coordinates and find the tension in the string.

Solution

Start by drawing a free-body diagram of the mass. There’s only a tensile force acting on the
mass, which points toward the center of the circle.

Newton’s second law states that the sum of the forces on the mass is equal to its mass times
acceleration.
X


 Fr = mar


X X
F = ma ⇒ Fϕ = maϕ



 X
Fz = maz

The tension acts in the negative r-direction.




−T = mar


0 = maϕ



 0 = ma
z

Divide both sides of each equation by m.


T


− = ar
 m

 0 = aϕ



0 = az

Substitute the formulas for acceleration in cylindrical coordinates.


 2  2
d r dϕ T
−r =−




 dt2 dt m



 2
d ϕ dr dϕ
r 2 +2 =0


 dt dt dt


d2 z



 =0
dt2

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Taylor Classical Mechanics - Problem 1.41 Page 2 of 2

Because the mass moves in a circle, r = R, dr/dt = 0, and d2 r/dt2 = 0.


  2
 dϕ T

 0 − R =−
dt m






 2  
d ϕ dϕ
R
 dt2 + 2(0) =0

 dt


d2 z



=0


dt2
And since the mass moves with a constant angular velocity, dϕ/dt = ω and d2 ϕ/dt2 = 0.

T

 0 − R(ω)2 = −
m





R(0) + 2(0)(ω) = 0


d2 z



=0


dt2
Therefore, multiplying both sides of the first equation by −m,

T = mRω 2 .

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