Respuesta :
(a) [tex]20.8 kg m^2[/tex]
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:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
where
[tex]\omega_f = 10.1 rad/s[/tex] is the final angular velocity
[tex]\omega_i=0[/tex] is the initial angular velocity
t = 5.90 s is the time taken
Substituting,
[tex]\alpha=\frac{10.1 rad/s-0}{5.90 s}=1.71 rad/s^2[/tex]
Now we can find the moment of inertia of the wheel by using the equivalent of Newton's second law for rotational motions:
[tex]\tau = I\alpha[/tex] (1)
where
[tex]\tau=35.6 Nm[/tex] is the torque applied
I is the moment of inertia
[tex]\alpha=1.71 rad/s^2[/tex] is the angular acceleration
Solving the equation for I,
[tex]I=\frac{\tau}{\alpha}=\frac{35.6 Nm}{1.71 rad/s^2}=20.8 kg m^2[/tex]
(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
[tex]\alpha=\frac{\omega_f-\omega_i}{t}=\frac{0-(10.1 rad/s)}{59.0 s}=-0.17 rad/s^2[/tex]
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:
[tex]\tau=I \alpha=(20.8 kg m^2)(-0.17 rad/s^2)=-3.5 Nm[/tex]
(c) 52.5 revolutions
The total angular displacement covered by the wheel in the first part of the motion is given by:
[tex]\omega_f^2 - \omega_i^2 = 2 \alpha_1 \theta_1[/tex]
where
[tex]\omega_f = 10.1 rad/s\\\omega_i = 0\\\alpha_1 = 1.71 rad/s^2[/tex]
Solving for [tex]\theta_1[/tex],
[tex]\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[/tex]
The total angular displacement covered by the wheel in the second part of the motion is given by:
[tex]\omega_f^2 - \omega_i^2 = 2 \alpha_2 \theta_2[/tex]
where
[tex]\omega_f = 0 rad/s\\\omega_i = 10.1 rad/s\\\alpha_2 = -0.17 rad/s^2[/tex]
Solving for [tex]\theta_2[/tex],
[tex]\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[/tex]
So the total angular displacement in radians is
[tex]\theta=\theta_1+\theta_2=29.8 rad+300.0 rad=329.8 rad[/tex]
And since [tex]1 rev = 2\pi rad[/tex]
the angle convered in revolutions is
[tex]\theta=\frac{329.8 rad}{2\pi rad/rev}=52.5 rev[/tex]
(a) The moment of inertia of wheel is [tex]20.81 \;\rm kg-m^{2}[/tex].
(b) The magnitude of torque due to friction is [tex]-3.558 \;\rm N-m^{2}[/tex].
(c) The number of revolutions made by wheel during its entire journey is 52.
Given data:
Magnitude of torque on wheel is, [tex]T = 35.6 \;\rm N-m[/tex].
Initial angular speed during first half is, [tex]\omega_{1}= 0\;\rm rad/s[/tex].
Final angular speed during the first half is, [tex]\omega_{2}= 10.1\;\rm rad/s[/tex].
Time interval during the first half is, [tex]t=5.90 \;\rm s[/tex].
Time interval during the second half is, [tex]t'=59.0 \;\rm s[/tex].
(a)
Use the expression of torque to calculate the moment of inertia (I) as,
[tex]T = I \times \alpha[/tex]
Here, [tex]\alpha[/tex] is the angular acceleration of wheel. And its value is,
[tex]\alpha =\dfrac{\omega_{2}-\omega_{1}}{t}\\\alpha =\dfrac{10.1-0}{5.90}\\\alpha =1.71 \;\rm rad/s^{2}[/tex]
Then, moment of inertia is,
[tex]35.6 = I \times 1.71\\I = 20.81 \;\rm kg-m^{2}[/tex]
Thus, the moment of inertia of wheel is [tex]20.81 \;\rm kg-m^{2}[/tex].
(b)
The angular acceleration due to friction is,
[tex]\alpha'=\dfrac{\omega'_{2}-\omega_{2}}{t'}\\\alpha'=\dfrac{0-10.1}{59.0}\\\alpha'=-0.171 \;\rm rad/s^{2}[/tex]
Then torque due to friction is,
[tex]T'= I \times \alpha'\\T'= 20.81 \times (-0.171)\\T'=-3.558 \;\rm N-m^{2}[/tex]
Thus, the magnitude of torque due to friction is [tex]-3.558 \;\rm N-m^{2}[/tex].
(c)
For first half, apply the third rotational equation of motion to obtain angular distance as,
[tex]\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[/tex]
For second half, again apply the third rotational equation of motion to obtain angular distance as,
[tex]\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[/tex]
Total angular distance is,
[tex]\theta_{t}=\theta + \theta'\\\theta_{t}=29.82 + 298.27\\\theta_{t}=328.09 \;\rm rad[/tex]
And number of revolution during the entire journey is,
[tex]N=\dfrac{\theta_{t}}{2 \pi} \\N=\dfrac{328.09}{2 \pi} \\N \approx 52[/tex]
Thus, the number of revolutions made by wheel during its entire journey is 52.
Learn more about rotational motion here:
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