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        A ladder of length l is slipping with its ends against a vertical wall and a horizontal floor .At a certain moment, speed of the end in contact with the horizontal floor is v and the ladder makes an angle 30 with the horizontal . Then the speed of ladders centre must be
2 years ago

Niloy Mazumder
24 Points
							Let’s take VG  is the speed of the center of the ladder .Instantaneous angular velocity$\omega =\frac{v}{\frac{l}{2}}=\frac{v_{G}}{lsin30}$so v=vGv=Gv⇒∘30sinlGv=2lv=ω" id="MathJax-Element-38-Frame" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">$v=v_{G}$ω$v=v_{G}$=vl2=vGlsin30∘⇒vG=v
`
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions