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Grade 12th passMechanics

Question is in the photo please help ??

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Profile image of Damini Parihar
11 Years agoGrade 12th pass
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Profile image of Sandeep Pathak
ApprovedApproved Tutor Answer11 Years ago
There are three forces on the rod
  • Normal force, N between rod and surface – vertically upwards
  • Gravitational force, mg – vertically downwards
  • frictional force, f – horizontally towards right
Also, it is important to note that since there is no slip at the point of contact of rod with the surface, the rod is doing pure rotation about A which act as a hinge point. So direction of acceleration (Let’s call it a) is perpendicular to rod downwards.

Now it is straightforward to calculate a and limiting value of friction. Let’s apply equation of motion for translation in horizontal and vertical directions:
f=ma\sin\theta\\ mg-N=ma\cos\theta
Usingf=\mu N, we get
a = \frac{\mu g}{\mu\cos\theta+\sin\theta}

Now, considering moment of mg about A, we get
mg\cos\theta\times \frac{L}{2}=I\alpha=\frac{mL^2}{3}\times\frac{a}{L/2}\Rightarrow a = \frac{3g}{4}\cos\theta
Using two expressions for a, we get
\mu = \frac{3\sin\theta\cos\theta}{4-3\cos^2\theta}

Instantaneous acceleration of COM is a, while that of B is 2a usinga=r\alpha.