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As there is no acceleration of centre of mass the CM moves in a straight line its falls downwards so the end point moves L/2 Distance.
sankalpmittal said:Edit:
Sorry. Normal reaction does not pass through CM, and so there will be net force on the CM downward. Will it be mg or mg-R ? I think former is correct. Also if a rod is pivoted at the end and it is left swing in a semicircle, will there be linear acceleration in its centre of mass ?
ehild said:Sankalpmittal,
The rod is an extended body. You can imagine it as a system of point masses, connected to each other by internal forces.
To make it simpler, assume a system of two point masses m1 and m2, their position vectors r1 and r2. ...
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ehild