Hopefully:
Find:
The first flywheel stopped after 1 minute, that is, it made $\theta_1=\omega_1 \cdot t=63 \cdot 60 = $3780 rads during this period of time. The second flywheel made 360 revolutions until it came to a complete stop, which corresponds to a rotation angle of $\theta_2=360\cdot 2\pi=720\pi$ rad. Let's calculate the time during which the second flywheel made a given number of revolutions:
$$\theta_2=\omega_2 \cdot t \Rightarrow t=\frac{\theta_2}{\omega_2}=\frac{720\pi}{63}\approx36.13\text{ с}$$
Now you can determine the braking torque of the two flywheels. To do this, we use the formula that describes the movement of the flywheel in the presence of friction:
$$I\frac{d\omega}{dt}=M-M_{\text{тр}},$$
where $I$ is the moment of inertia of the flywheel, $M$ is the moment of external forces acting on the flywheel, $M_{\text{tr}}$ is the moment of friction.
Since the flywheels are identical, their moments of inertia are equal, which means that the moments of the forces acting on them at the same angular velocity are also equal. Thus, we can conclude that the braking torque is proportional to the friction torque. It can also be noted that the frictional torque acting on the flywheel is proportional to the friction force, which in turn is proportional to the normal force acting on the flywheel. In this case, the normal force is determined by the weight of the flywheel and is equal to $F_{\text{n}}=mg$, where $m$ is the mass of the flywheel, and $g$ is the acceleration of gravity.
Thus, we can write:
$$M_{\text{тр}}=\mu F_{\text{н}},$$
where $\mu$ is the friction coefficient.
Since the flywheels are the same, the friction coefficient in both cases will be equal, which means that the friction moments can be compared directly. In this case, you can use the formula obtained above to calculate the friction torque for each flywheel.
For the first flywheel:
$$M_{\text{тр},1}=\mu F_{\text{н},1}=\mu mg$$
For the second flywheel:
$$M_{\text{тр},2}=\mu F_{\text{н},2}=\mu mg$$
Since the masses of the flywheels are the same, it is possible to compare the friction moments using only the friction coefficient $\mu$. Let's compare the values of the friction moments for both flywheels:
$$M_{\text{тр},1}=M_{\text{тр},2}$$
Thus, the braking torques for both flywheels were the same.
Answer: The braking torques for both flywheels were the same.
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He told two identical flywheels at rest...
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The product description relates to problem 10720 involving two identical flywheels at rest that are given the same angular velocity of 63 rad/s.
As a result of the action of friction forces on the first flywheel, it stopped after one minute, and the second flywheel made 360 revolutions until it stopped completely.
It is necessary to determine which flywheel had the greater braking torque and by how many times.
To solve the problem, it is necessary to use the law of conservation of energy of rotational motion, which states that the total mechanical energy of the rotational motion of a rigid body remains constant in the absence of external moments of forces.
You should also take into account the formula for the moment of frictional force, which is expressed through the coefficient of friction and normal force.
The calculation formula for determining the braking torque will look like this:
M = I * (ω_f - ω_i) / t
where M is the braking torque, I is the moment of inertia of the flywheel, ω_f and ω_i are the final and initial angular velocities, respectively, t is the time of movement.
To determine the moment of inertia of a flywheel, you can use its geometric parameters (mass, dimensions, location of the axes) and the formula for the moment of inertia of a solid body relative to the axis of rotation.
Using the obtained values of the braking torque for each flywheel, you can calculate how many times the braking torque was greater for one flywheel compared to the other.
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