A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is n

Question

A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions?

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Amity 4 years 2021-08-04T23:23:12+00:00 1 Answers 21 views 0

Answers ( )

  1. Dulcie
    0
    2021-08-04T23:24:37+00:00
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    Answer:

    The torque must be applied to the wheel is 15.7 N-m.  

    Explanation:

    Given that,

    Mass of the wheel of cylinder, m = 50 kg

    Diameter of the wheel, d = 1 m

    Radius, r = 0.5 m      

    Initial speed off the wheel is 0

    Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

    Angular displacement, \theta=20\ rev=125.6\ rad

    The torque is given by :

    \tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m

    So, the torque must be applied to the wheel is 15.7 N-m.

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