Question

In: Physics

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod...

A person holds a ladder horizontally at its center.

Treating the ladder as a uniform rod of length 3.10 m and mass 8.56 kg , find the torque the person must exert on the ladder to give it an angular acceleration of 0.238 rad/s2 .

Solutions

Expert Solution

If a person holds a ladder horizontally at its center.

      Given mass of the rod M=8.56kg

         Length of the rod      l= 3.10m

Then the moment of inertia of the rod I=Ml2/12

                                                         =8.56(3.10)2/12

                                                          =6.855kgm2

      The torque the person must exert on the ladder =I

                  Given =0.238rad/sec2

                                                                         = (6.855)(0.238)

                                                                            =1.6315N-m

=        

0

genius_generous answered 1 year ago
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