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3 years ago

15

A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on

the hoop to stop it

1 answer:

Ede4ka [16]3 years ago

4 0

Answer:

The work done by the hoop is equal to 5.529 Joules.

Explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

$K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

I is the moment of inertia, $I=mr^2$

Since, $\omega=\dfrac{v}{r}$

$K_i=mv^2$

$K_i=96\times (0.24)^2=5.529\ J$

Finally it stops, so the final energy of the hoop will be, $K_f=0$

The work done by the hoop is equal to the change in kinetic energy as :

$W=K_f-K_i$

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.

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Answer:

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Explanation:

1) Notation and info given

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2) Solution to the problem

So we can use a model to describe the density as function of the radius

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So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

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We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

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Answer:

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3 years ago