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 2 years agoRelated Solutions
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