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

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answered: 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:

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answered: 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:

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