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3 years ago

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A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

ttom of the ramp. (a) What is its (constant) acceleration down the ramp? (b) What is its angular acceleration?

1 answer:

RoseWind [281]3 years ago

3 0

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

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Answer:

6.91 Hz

Explanation:

Volume of the cube

= (1.5 x 10⁻² )³m³

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mass of the cube

= 3.375 x 10⁻⁶ x 8920 ( 8.92 g / cm³ = 8920 kg/m³ )

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K x2.65 x 10⁻² = 1.5

k = .566 x 10² N / m

Now frequency of oscillation for spring - mass system is given by

n = $\frac{1}{2\pi} \sqrt{\frac{k}{m} }$

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n = 6.91 Hz

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3 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

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We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

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7 0

3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

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Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

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puteri [66]

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