A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o

Answers

Answer 1

Answer:

Answer:

1) I_ pendulum = 2.3159 kg m², 2) I_pendulum = 24.683 kg m²

Explanation:

In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum

Sphere

They indicate the density of the sphere roh = 37800 kg / m³ and its radius

r = 5 cm = 0.05 m

we use the definition of density

ρ = M / V

M = ρ V

the volume of a sphere is

V = 4/3 π r³

we substitute

M = ρ 4/3 π r³

we calculate

M = 37800 4/3 π 0.05³

M = 19,792 kg

Bar

the density is ρ = 32800 kg / m³ and its dimensions are 1 m,

0.8 cm = 0.0008 m and 4cm = 0.04 m

The volume of the bar is

V = l w h

m = ρ l w h

we calculate

m = 32800 (1 0.008 0.04)

m = 10.496 kg

Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is

r_cm = 1 / M (M r_sphere + m r_bar)

M = 19,792 + 10,496 = 30,288 kg

r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))

r_cm = 1 / 30,288 (5,248 + 20,7816)

r_cm = 0.859 m

This is the center of mass of the pendulum.

1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:

Sphere I = 2/5 M r2

Bar I = 1/12 m L2

parallel axes theorem

I = I_cm + m D²

where m is the mass of the body and D is the distance from the body to the axis of rotation

Sphere

m = 19,792 ka

the distance D is

D = 1.05 -0.85

D = 0.2 m

we calculate

I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168

I_sphere = 0.811472 kg m²

Bar

m = 10.496 kg

distance D

D = 0.85 - 0.5

D = 0.35 m

I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576

I_bar = 1.5044 kg m²

The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts

I_pendulum = I_sphere + I_bar

I_pendulum = 0.811472 +1.5044

I_ pendulum = 2.3159 kg m²

this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m

2) The moment is requested with respect to the pivot point at r = 0 m

Sphere

D = 1.05 m

I_sphere = 2/5 M r2 + M D2

I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82

I_sphere = 21.84 kg m²

Bar

D = 0.5 m

I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624

I_bar = 2,84266 kg m 2

The pendulum moment of inertia is

I_pendulum = 21.84 +2.843

I_pendulum = 24.683 kg m²

This moment of inertia is about the turning point at r = 0 m

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Answers

Answer:

P = 31.83 W

Explanation:

Our data are,

Magnitude of the force F = 26 N

Radius of the circular path r = 0.26 m

The angle between force and handle $\theta = 0$ °

Time t = 2 s

We know that the formula to find the velocity is given by

Velocity $v = (2\pi r)/(t)$

$v= (2\pi r)/(t)$

$v=(2 \pi 0.26)/(2)$

$v= 0.8168m/s$

We know also that the formula to find the power is given by,

$P = F*v$

$P = (26)(0.8168)$

$P = 31.83 W$

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?

Answers

To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.

By definition we know that velocity is defined as the change of position due to time, therefore

$V = (d)/(t)$

Where,

d = Distance

t = Time

Speed can also be expressed in vector form through its components $V_x$ and $V_y$

In the case of the horizontal component X, we have to

$V_x = (d)/(t)$

Here d means the horizontal displacement, then

$t = (d)/(V_x)$

$t = (67)/(V_x)$

At the same time we have that the vertical component of the velocity is

$V_y = gt$

Here,

g = Gravity

Therefore using the relation previously found we have that

$V_y = g (67)/(V_x)$

The relationship between the two velocities and the angle can be expressed through the Tangent, therefore

$tan\theta = (V_y)/(V_x)$

$tan \theta = (g (67)/(V_x) )/(V_x)$

$tan 3 = (9.8(67)/(V_x) )/(V_x)$

$tan 3 = (9.8*67)/(V_x^2)$

$V_x^2 = (9.8*67)/(tan 3)$

$V_x= \sqrt{ (9.8*67)/(tan 3)}$

$V_x = 111.93m/s \hat{i}$

This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,

Then $V_y,$

$V_y = g (67)/(V_x)$

$V_y = 9.8*(67)/(111.93)$

$V_y = 5.866m/s\hat{j}$

$|\vec{V}| = √(111.93^2+5.866^2)$

$|\vec{V}| = 112.084m/s$

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The valve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.25 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer. Use data from the tables.

Answers

Answer:

Answer

The Final Quality of teh R-134a in the container is 0.5056

The Total Heat transfer is $Q_(in) = 22.62 KJ$

Explanation:

Explanation is in the following attachments

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answers

Answer:

a) W₁ = - 127 J, b) W₂ = 148.18 J, c) $v_(f)$ = 3.43 m/s and d) $v_(f)$ = 3.43 m / s

Explanation:

The work is given the equation

W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

W-T = m a

T = W -m a

T = mg -mg/7

T = mg 6/7

T = 3.6 9.8 6/7

T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

W₁ = F d cos θ

W₁ = 30.24 4.2 cos 180

W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

W₂ = (mg) d cos 0º

W₂ = 3.6 9.8 4.2

W₂ = 148.18 J

c) kinetic energy

K = ½ m v²

Let's calculate speed with kinematics

$v_(f)$ ² = vo² + 2 a y

v₀ = 0

a = g / 7

$v_(f)$ ² = 2g / 7 y

$v_(f)$ = √ (2 9.8 4.2 / 7)

$v_(f)$ = 3.43 m/s

We calculate

K = ½ 3.6 3.43²

K = 21.18 J

d) the speed of the block and we calculate it in the previous part

$v_(f)$ = 3.43 m / s

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acceleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above

Answers

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the position must also be changing.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

Final answer:

In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.

Explanation:

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.

The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.

The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.