A thin uniform metal rod of mass 'm' and length ‘l’ standing verti

1.	A thin uniform metal rod of mass 'm' and length ‘l’ standing vertically is now allowed to fall due to its own weight. The velocity with which its mid-point will strike ground will be : (1) √5/2 gl(2) √5 gl(3) √3 gl(4) √3/4 gl
Answer» Explain: Loss of gravitational = P.E mg l/2 Gain in rotational = KE = 1/2 lW² Now l = 1/3 ml² So, mg 1/2 = 1/2 x 1/2 ml² x W² ⇒ w = √3g/l So, the velocity of mid point ⇒v = w l/2 =√3g/l. l/2 = √3gl/4.

A thin uniform metal rod of mass 'm' and length ‘l’ standing vertically is now allowed to fall due to its own weight. The velocity with which its mid-point will strike ground will be : (1) √5/2 gl(2) √5 gl(3) √3 gl(4) √3/4 gl

Answer»

Explain: Loss of gravitational = P.E mg l/2

Gain in rotational = KE = 1/2 lW²

Now l = 1/3 ml²

So, mg 1/2 = 1/2 x 1/2 ml² x W² ⇒ w = √3g/l

So, the velocity of mid point ⇒v = w l/2

=√3g/l. l/2

= √3gl/4.

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A thin uniform metal rod of mass 'm' and length ‘l’ standing vertically is now allowed to fall due to its own weight. The velocity with which its mid-point will strike ground will be : (1) √5/2 gl(2) √5 gl(3) √3 gl(4) √3/4 gl

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