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Grade 12General Physics

A rod of length l lie on the smooth ground and has masss m. It makes 45 degree with vertical. Then the accleration of the end in contact with ground will be?g=10m/s^2

nehal

8 Years agoGrade 12

1 Answer

Eshan

8 Years ago

Cosider the point in contact with the ground, the accleration of which has been shown on the right.

Since the point is constraint to not move in the vertical direction(since it can only slide along the surface), $A_{CM}=\dfrac{l}{2}\alpha sin45^{\circ}$

From the free body diagram of the rod,

$mg-N=mA_{CM}$
Torque acting about the center of mass is $N\dfrac{l}{2}cos45^{\circ}=I\alpha=\dfrac{ml^2}{12}\alpha$
$\implies N=\dfrac{\sqrt{2}ml}{6}\alpha$

From the three equations,

$l\alpha=\dfrac{6\sqrt{2}g}{5}$

Hence the acceleration of the point= $\dfrac{l}{2}\alpha sin45^{\circ}=\dfrac{3}{5}g$

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