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3 years ago

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A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man

y radians did the disk turn while stopping ? C. how many revolutions?

1 answer:

Ne4ueva [31]3 years ago

6 0

Answer:

A. α = - 1.047 rad/s²

B. θ = 14.1 rad

C. θ = 2.24 rev

Explanation:

A.

We can use the first equation of motion to find the acceleration:

$\omega_f = \omega_i + \alpha t$

where,

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s

t = time = 3 s

α = angular acceleration = ?

Therefore,

$0\ rad/s = 3.14\ rad/s + \alpha(3\ s)$

<u>α = - 1.047 rad/s²</u>

B.

We can use the second equation of motion to find the angular distance:

$\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2$

<u>θ = 14.1 rad</u>

C.

θ = (14.1 rad)(1 rev/2π rad)

<u>θ = 2.24 rev</u>

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Answer:

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Part a

For this case we have the following differential equation:

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$A = e^{kt} e^c = C e^{kt}$

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$A_o = C$

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$ln (\frac{1}{2}) = kt$

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$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

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$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

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$t = T \frac{ln(8)}{ln(2)}$

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Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

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3 years ago