The combination of an applied force and a constant frictional forceproduces a co

Question

The combination of an applied force and a constant frictional forceproduces a constant total torque of 36.4 N·m on a wheel rotating about afixed axis. The applied force acts for 5.93 s. During this time the angular speed ofthe wheel increases from 0 to 10.3 rad/s. The applied force is thenremoved, and the wheel comes to rest in 59.8 s. (a) Findthe moment of inertia of the wheel.

kg·m2

(b) Find the magnitude of the frictional torque.

N·m

(c) Find the total number of revolutions of the wheel.
rev
(a) Findthe moment of inertia of the wheel.

kg·m2

(b) Find the magnitude of the frictional torque.

N·m

(c) Find the total number of revolutions of the wheel.
rev

Explanation / Answer

   a.   First equation forangular motion is             =   0+ * t    angularacceleration         =   (10.3   -   0) /5.93   =   1.74   rad/s2    also   torque      =   I*    Moment of inertia ofwheel   I   =   /    =   36.4 /1.74   =   20.92   kg-m2    b.   Frictional torque causesdeacceleration of wheel.    '   =      +   '* t'    ' refers to quatities duringdeacceleration.    0   =   10.3   +   '* 59.8    deacceleration   '   =   -10.3 / 59.8   =   -0.172   rad/s2    - ve sign could be dropped as it onlyindicates deacceleration.       frictionaltorque   =   ' *I   =   0.172 *20.92   =   3.60   kg-m2    c.   second equation is          =   0* t   +  (1/2) * *t2    duringacceleration         0   =   0,   t   =   5.93s,      =   1.74   rad/s2       1   =   0* 5.93   +   0.5 * 1.74 *5.932   =   30.59   rad.       duringdeacceleration   0   =      =   10.3   rad/s,   t   =   59.8s,      =   -0.172   rad/s2       2   =   10.3* 59.8   +   0.5 * ( - 0.172) *59.82   =   308.40   rad       total angulardisplacement      =   1   +   2   =   30.59+308.40   =   338.99   rad    also   1   rev   =   2   =   6.28   rad       no. of rev.competed   n   =   / 2   =   338.99 /6.28   =   53.98   rev.       

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