the combination of an applied force and a frictional forceproducesa constant tor

Question

the combination of an applied force and a frictional forceproducesa constant torque of 41 Nm on a wheel rotating about afixed axis. the applied force acts for 6 s, during which timethe angular speed of the wheel increases from 0 to 12rad/s. the applied force is then removed, and the wheel comesto rest in 55 s. a) what is the moment of inertia of the wheel? (kgm2) b) what is the magnitude of the frictional torque?(Nm)
c) how many revolutions does the wheel make? the combination of an applied force and a frictional forceproducesa constant torque of 41 Nm on a wheel rotating about afixed axis. the applied force acts for 6 s, during which timethe angular speed of the wheel increases from 0 to 12rad/s. the applied force is then removed, and the wheel comesto rest in 55 s. a) what is the moment of inertia of the wheel? (kgm2) b) what is the magnitude of the frictional torque?(Nm)
c) how many revolutions does the wheel make?

Explanation / Answer

   a.   Angular acceleration ofthewheel      =   appliedtorque () / moment of inertial (I)    Also      =   (   -   0)/ t                 =   finalangularvelocity   =   12   rad/s,   0   =   initialangular velocity   =   0       =   (12 -0) /6   =   2   rad/s2    Moment of inertia   I   =   /    =   41 /2   =   20.5   kg-m2    b.   Deacceleration ofwheel   d   =   ('   -   )/ t'          '   =   finalvelocity after external torque isremoved   =   0          t'   =   timetaken to come to rest          d   =      (0- 12) / 55   =   -0.218   rad/s2          -ve signindicates deacceleration and hence can be ignored.          -ve signindicates deacceleration and hence can be ignored.          frictionaltorque   f   =   I* d   =   20.5 *0.218   =   -4.469   N-m    c.   Second equationis         =   0* t   +   (1/2) * *t2              duringacceleration      1   =   0* 6   +   0.5 * 2 *62   =   36   rad          duringdeacceleration   2   =   12* 55   +   0.5 * ( -0.218) *552   =   330.27   rad          totalangulardisplacement      =   1   +   2   =   36+330.27   =   366.27   rad          onerevolution   =   2   =   6.28   rad          no. ofrev.made   n   =   366.27/ 6.28   =   58.32   revs.              duringacceleration      1   =   0* 6   +   0.5 * 2 *62   =   36   rad          duringdeacceleration   2   =   12* 55   +   0.5 * ( -0.218) *552   =   330.27   rad          totalangulardisplacement      =   1   +   2   =   36+330.27   =   366.27   rad          onerevolution   =   2   =   6.28   rad          no. ofrev.made   n   =   366.27/ 6.28   =   58.32   revs.

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