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A rod of length 50cm is pivoted at one end. It is raised such that it makes an angle of 30 degrees from the horizontal and is then released from rest. Its angular speed when it passes through the horizontal (in rad/sec) will be (g=10 ms -2 ) √15 √30 √10 √30 /2 Plz explain....
Work done by gravity from initial to finalposition is, W=mg 2ℓ sin30 o = 4mgℓ According to work-energy theorem W= 21 Iω 2 ⇒ 21 3mℓ 2 ω 2 = 4mgℓ ω= 2ℓ3g = 2×0.53×10 ω= 30 rad/sec
Dear student A rod of length 50cm is provided at one end. It is raised such that if makes an angle of 30^{\circ} from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal ( in rad s-1) will be (g = 10ms-2)Explanation:By work energy theorum,mgl/4=1/2×Iw²mgl/4 =ml²w²/3×2w²=√3g/2l=√30
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