A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim

Question

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s

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Khoii Minh 3 years 2021-08-18T00:15:53+00:00 1 Answers 58 views 0

Answers ( )

  1. thachthao
    0
    2021-08-18T00:17:20+00:00
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    Answer:

    The force required is equal to 30\pi N.

    Explanation:

    We know that

         Torque = force × perpendicular distance

                      = F × R = I×\alpha

                   I = M×\frac{R^{2} }{2}

      From above equation \alpha = \frac{FR}{I}              ………….. 1

          We know that

                  w=w_{o} +\alpha t                      ………………………2

                   Given that t= 2 sec

                                      w_{o} = 0

                                      w=0.4 rev/s = 0.8\pi rad/s

    From equation 1 and 2 , we get

                      F =  \frac{wI}{RT}  =  \frac{w\times m \times r}{2t}

        Upon substituting the above values we will be getting

              F = 30\pi N

                       

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