A wheel mounted on an axis that is not frictionless is...

A wheel mounted on an axis that is not frictionless is initially at rest. A constant external torque of 50 N·m is applied to the wheel for 20 s., giving the wheel an angular velocity of 600 rev/min. The external torque is then removed, leaving the frictional torque to bring the wheel to rest 120 s. later. A) Use angular kinematics to find the constant angular accelerations while the wheel’s angular velocity is increasing and while the wheel’s angular velocity is decreasing. B) Use the angular form of Newton’s second law, ?net = I?, for both time intervals to find the constant torque due to friction, ?fric, and the wheel’s moment of inertia, I.

Transcript

00:01 Hello, here a wheel mounted on an axis is not frictionless and initially it's repressed.

00:07 So it means that the initial rotational speed is zero radiance per second.

00:12 A constant external torque of 50 newton meter is applied to the wheel for delta t of 20 seconds, given the wheel an angular velocity of 600 revolution, per minute so let's convert it to radiance per second so we have to take this number multiplied by 2 pi and divided by 60 to convert it to radiance per second so that is 62 .8 radiance per second then the external force is removed and leaving the frictional torque to bring wheel to the rest for the delta t to which is 120 seconds so now we have to answer question a to find the constant angular speed acceleration of the wheel and so this angular acceleration alpha can be found as following alpha equals to omega 1 minus omega initial divided by delta t1 which is 6 which is 62 .8 radiance per second minus zero radiance per second divided by 20 seconds that is 3 .14 radiance per second squared and now let's find though that is alpha alpha alpha 1 while accelerating so that is while and now let's find alpha 2 which is angular acceleration during this last step so that it equals to 0 minus omega 1 divided by delta t 2 which is negative 62 .8 radiance per second divided by 120 seconds that is 0.

03:35 Negative 0 .524 radiance per second squared.

03:43 So alpha 2 is lower than 0 because the wheel comes to the rest.

03:58 Okay now we have to answer question b and we have to find the torque of friction and will inertia momentum so let me first save right right down the accelerations and here you know yeah it's better to report the negative sign for alpha 2 but absolute value of alpha 2 is of course positive it is 0 .522 radiance per second square but we need to keep in mind that in fact it's directed to the opposite direction with respect to the speed so we need to keep keep these accelerations to answer question two and now we will do it so let's answer now question two or question b here we have to calculate inertia and torque of the friction force so let's do this first we have to write down to systems of a equation so the first equation is external torque which is 50 newton meter minus friction torque of the friction force equals to momentum of inertia times alpha one and on the other hand negative friction for torque of the friction force equals to i times alpha 2 where alpha 2 is with negative sign therefore external friction torque of external friction plus i alpha 2 equals to i alpha 1 so from here external torque equals to i alpha 1 minus alpha 2 so that is therefore this momentum equals to external torque divided by alpha 1 minus alpha 2 which is 15 newton meter divided by alpha 1 that is 3 .14 radiance per second squared minus negative 0 .524 radiance per second squared now let's complete this calculation...

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