A uniform thin rod of weight w is supported horizontal by two vertical props at its ends . At t=0, one of these supports is kicked out . The force on the other support immediately there after is
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Let the center of mass of the rod have downward acceleration a. Then
W-F=ma ---------(1)
The angular momentum equation implies
WL/2 = Iω, ----------(2)
where I=moment of inertia about an end = 1/3mL2.
One has ω=a/(L/2),
a relation that is true for small ω, i.e short times.
From (1) and (2),
F=W/4
W-F=ma ---------(1)
The angular momentum equation implies
WL/2 = Iω, ----------(2)
where I=moment of inertia about an end = 1/3mL2.
One has ω=a/(L/2),
a relation that is true for small ω, i.e short times.
From (1) and (2),
F=W/4
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