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`        A uniform rod of length 2l is placed with one end in contact with the horizontal table and is then inclined at an angle α to horizontal and allowed to fall.When it becomes horizontal ,it's angular velocity will be `
one year ago

Vikas TU
11137 Points
```							From energy conservation, when it is inclined,U.E = K.Emg((lsinα)/2) = 0.5*I*w^2mglsinα = Iw^2where I is the moment of inertia about the centre of the rod.I = m(2l)^2/12 => ml^2/3mglsinα = ((ml^2)*w)/3gsinα = lw/3w = (3gsinα)/l rad/s
```
one year ago
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• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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