Physics, 28.01.2020 07:31 triddi666

Atorque of 35.6 n · m is applied to an initially motionless wheel which rotates around a fixed axis. this torque is the result of a directed force combined with a friction force. as a result of the applied torque the angular speed of the wheel increases from 0 to 10.1 rad/s. after 5.90 s the directed force is removed, and the wheel comes to rest 59.0 s later. (a) what is the wheel's moment of inertia (in kg · m2)? kg · m2 (b) what is the magnitude of the torque caused by friction (in n · m)? n · m (c) from the time the directed force is initially applied, how many revolutions does the wheel go through? revolutions

Answers: 2

Show answers

Answers

Answer from: jaidalynkimora

(a) The moment of inertia of wheel is $20.81 \;\rm kg-m^{2}$ .

(b) The magnitude of torque due to friction is $-3.558 \;\rm N-m^{2}$ .

Given data:

Magnitude of torque on wheel is, $T = 35.6 \;\rm N-m$ .

Initial angular speed during first half is, $\omega_{1}= 0\;\rm rad/s$ .

Final angular speed during the first half is, $\omega_{2}= 10.1\;\rm rad/s$ .

Time interval during the first half is, $t=5.90 \;\rm s$ .

Time interval during the second half is, $t'=59.0 \;\rm s$ .

(a)

Use the expression of torque to calculate the moment of inertia (I) as,

$T = I \times \alpha$

Here, $\alpha$ is the angular acceleration of wheel. And its value is,

$\alpha =\dfrac{\omega_{2}-\omega_{1}}{t}\\\alpha =\dfrac{10.1-0}{5.90}\\\alpha =1.71 \;\rm rad/s^{2}$

Then, moment of inertia is,

$35.6 = I \times 1.71\\I = 20.81 \;\rm kg-m^{2}$

Thus, the moment of inertia of wheel is $20.81 \;\rm kg-m^{2}$ .

(b)

The angular acceleration due to friction is,

$\alpha'=\dfrac{\omega'_{2}-\omega_{2}}{t'}\\\alpha'=\dfrac{0-10.1}{59.0}\\\alpha'=-0.171 \;\rm rad/s^{2}$

Then torque due to friction is,

$T'= I \times \alpha'\\T'= 20.81 \times (-0.171)\\T'=-3.558 \;\rm N-m^{2}$

Thus, the magnitude of torque due to friction is $-3.558 \;\rm N-m^{2}$ .

(c)

For first half, apply the third rotational equation of motion to obtain angular distance as,

$\omega^{2}_{2}=\omega^{2}_{1}+2 \alpha \theta\\10.1^{2}=0^{2}+2 \times 1.71 \times \theta\\\theta = 29.82 \;\rm rad$

For second half, again apply the third rotational equation of motion to obtain angular distance as,

$\omega'^{2}_{2}=\omega^{2}_{2}+2 \alpha'\theta'\\0^{2}=10.1^{2}+2 \times (-0.171) \times \theta'\\\theta' = 298.27 \;\rm rad$

Total angular distance is,

$\theta_{t}=\theta + \theta'\\\theta_{t}=29.82 + 298.27\\\theta_{t}=328.09 \;\rm rad$

And number of revolution during the entire journey is,

$N=\dfrac{\theta_{t}}{2 \pi} \\N=\dfrac{328.09}{2 \pi} \\N \approx 52$

Thus, the number of revolutions made by wheel during its entire journey is 52.

Learn more about rotational motion here:

link

Answer from: ubdental

(a) 20.8 kg m^2

First of all, we can find the angular acceleration of the wheel when both the directed force and the friction force are acting on it:

$\alpha=\frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 10.1 rad/s$ is the final angular velocity

$\omega_i=0$ is the initial angular velocity

t = 5.90 s is the time taken

Substituting,

$\alpha=\frac{10.1 rad/s-0}{5.90 s}=1.71 rad/s^2$

Now we can find the moment of inertia of the wheel by using the equivalent of Newton's second law for rotational motions:

$\tau = I\alpha$ (1)

where

$\tau=35.6 Nm$ is the torque applied

I is the moment of inertia

$\alpha=1.71 rad/s^2$ is the angular acceleration

Solving the equation for I,

$I=\frac{\tau}{\alpha}=\frac{35.6 Nm}{1.71 rad/s^2}=20.8 kg m^2$

(b) -3.5 Nm

In the second part, the directed force is removed, and only the friction force acts on the wheel. The wheel comes to rest after t=59.0 s, so the angular acceleration in this part is

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{0-(10.1 rad/s)}{59.0 s}=-0.17 rad/s^2$

And it is negative since it is a deceleration. The moment of inertia of the wheel has not changed, so we can still use eq.(1) to find the torque caused by the friction only:

$\tau=I \alpha=(20.8 kg m^2)(-0.17 rad/s^2)=-3.5 Nm$

The total angular displacement covered by the wheel in the first part of the motion is given by:

$\omega_f^2 - \omega_i^2 = 2 \alpha_1 \theta_1$

where

$\omega_f = 10.1 rad/s\\\omega_i = 0\\\alpha_1 = 1.71 rad/s^2$

Solving for $\theta_1$ ,

$\theta_1 = \frac{\omega_f^2-\omega_i^2}{2\alpha_1}=\frac{(10.1 rad/s)^2-0}{2(1.71 rad/s^2)}=29.8 rad$

The total angular displacement covered by the wheel in the second part of the motion is given by:

$\omega_f^2 - \omega_i^2 = 2 \alpha_2 \theta_2$

where

$\omega_f = 0 rad/s\\\omega_i = 10.1 rad/s\\\alpha_2 = -0.17 rad/s^2$

Solving for $\theta_2$ ,

$\theta_2 = \frac{\omega_f^2-\omega_i^2}{2\alpha_2}=\frac{0-(10.1 rad/s)^2}{2(-0.17 rad/s^2)}=300.0 rad$

So the total angular displacement in radians is

$\theta=\theta_1+\theta_2=29.8 rad+300.0 rad=329.8 rad$

And since $1 rev = 2\pi rad$

the angle convered in revolutions is

$\theta=\frac{329.8 rad}{2\pi rad/rev}=52.5 rev$

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

Answer from: Quest

Independent variable is something that is changed in a experiment.

Another question on Physics

Physics, 22.06.2019 03:00

Which of the following harmful chemicals are found in tobacco smoke? a. carbon monoxide b. carbon dioxide c. nicotine b. carbon dioxide d. both a and c

Answers: 2

Answer

Physics, 22.06.2019 20:20

Area around a charged object where the objects can exert a force on another charged objects is an electric

Answers: 2

Answer

Physics, 23.06.2019 00:00

Boiling water is an example of physical change. true or false

Answers: 2

Answer

Physics, 23.06.2019 00:30

What is the relationship between wavelength of light and the quantity of energy per photon?