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

Answer 1

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 = $30\pi N$