A rotating uniform cylindrical platfrm of mass 220kg andradius 5.5 m slows down

Question

A rotating uniform cylindrical platfrm of mass 220kg andradius 5.5 m slows down from 3.8 rev/sec to rest in 16 seconds whenthe driving mtor is disconnected. Estimate the power output of themotor (hp) required to maintain a steady speed of 3.8rev/sec. Here is what I did: P=F*v I=1/2*110*5.5=605kgm^2 3.8 rev/sec=23.9 rad/sec =f   ------v=R---->v=5.5*23.9=131.5=v so, =-1.49rad/sec =I*---->605*-1.49 =-903 F*d=903--->F*5.5=903 F=-164.1 P=F*v P=164.1*131.5 P=21571watts ----> 21571/746=28.9hp Is anyone getting the same answer?? Thank you!!! A rotating uniform cylindrical platfrm of mass 220kg andradius 5.5 m slows down from 3.8 rev/sec to rest in 16 seconds whenthe driving mtor is disconnected. Estimate the power output of themotor (hp) required to maintain a steady speed of 3.8rev/sec. Here is what I did: P=F*v I=1/2*110*5.5=605kgm^2 3.8 rev/sec=23.9 rad/sec =f   ------v=R---->v=5.5*23.9=131.5=v so, =-1.49rad/sec =I*---->605*-1.49 =-903 F*d=903--->F*5.5=903 F=-164.1 P=F*v P=164.1*131.5 P=21571watts ----> 21571/746=28.9hp Is anyone getting the same answer?? Thank you!!! so, =-1.49rad/sec =I*---->605*-1.49 =-903 F*d=903--->F*5.5=903 F=-164.1 P=F*v P=164.1*131.5 P=21571watts ----> 21571/746=28.9hp Is anyone getting the same answer?? Thank you!!!

Explanation / Answer

Hmmm... your approach is a little misguided. You're usingconcepts involved with linear motion (force, velocity, distance)when this is a rotational motion problem. Also, it seems likeyou're just trying to randomly place information in the equations(like using the radius of 5.5 meters for your "d") . I always tell my students, when starting a physicsproblem: . 1. Determine what type of problem it is and what concepts areinvolved. . 2. Draw a picture (or several) to show what is happeningand label all relevant information. . 3. Choose equations that are relevant to the giveninformation and the type of problem. . 4. Use the equations to solve for what you are asked tofind. .     In this case, you are asked forpower... which is  energy per time. So thisis an ENERGY problem. Specifically, it deals with the changingkinetic energy of the drum. What changes the drum's KE? In thefirst case (motor off), it is just friction that does work on thedrum. In the second case (motor on), we know the KE remainsconstant so the motor must do enough work to offset the effect ofthe friction. So... . 1. calculate the work done by friction to slowthe drum when the motor is off. .    2. recognize that the work done by themotor must equal the work done by friction to keep the drum turningat constant speed. . So for #1...    work equals change inkinetic energy. .    Initial KE of drum = (1/2) I2   = (1/2) (1/2) m r22 = .                       = (1/4) * 220 * 5.52 * (3.8 *2)2 =    948456 Joules . Final KE of drum = 0 .     Work done by friction = 0 -948456 =   -948456 Joules. . Power exerted by friction = work / time = 948456 J / 16 s =    59278 Watts . So...   the motor must equal the slowing power ofthe friction. So the motor must exert power of 59278 W which is. .          59278 W * 1 hp / 746 W =     79.5 hp   for themotor

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