The combination of an applied force and a constant frictional forceproduces a co
ID: 1753810 • Letter: T
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.Related Questions
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