A uniform rod of length 2l is placed with one end in contact with horizontal and is then inclined at an angle
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Hello dear,
● Answer -
w = √(3gsinα/2l) rad/s
● Explaination-
Here, we'll talk about rod and earth as a system.
By law of conservation of energy -
P.E. = K.E.
mglsinα = 1/2 Iw^2
w^2 = 2mglsinα / I
For a rod, moment of inertia perpendicular to its length is-
I = m(2l)^2 / 3 = 4ml^2 / 3
Putting this,
w^2 = 2mglsinα / 4ml^2/3
w = √(3gsinα/2l) rad/s
Angular velocity is √(3gsinα/2l) rad/s.
Hope this is useful...
● Answer -
w = √(3gsinα/2l) rad/s
● Explaination-
Here, we'll talk about rod and earth as a system.
By law of conservation of energy -
P.E. = K.E.
mglsinα = 1/2 Iw^2
w^2 = 2mglsinα / I
For a rod, moment of inertia perpendicular to its length is-
I = m(2l)^2 / 3 = 4ml^2 / 3
Putting this,
w^2 = 2mglsinα / 4ml^2/3
w = √(3gsinα/2l) rad/s
Angular velocity is √(3gsinα/2l) rad/s.
Hope this is useful...
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