Question

a rod of length L is resting on a horizontal table such that 3/4 of the length is not on the table but one end is tied to the string and the rod is made horizontal when the the string is cut the rod rotates .find out the angular acceleration of the rod .

Answer 1

Answer:

the length is not the tavble but one end is tied to the string andvthe rod is made horizontal when the string is cut the rod rotates.

Answer 2

Given :

The length  of rod :

=L

Part of the length of the rod not on the table :

= $\frac{3}{4} L$

One end is tied to the string and the rod is made horizontal .

After cutting the string the rod starts rotating .

To Find :

Angular acceleration of the rod = ?

Solution :

Let the mass of the rod be :

= m

Let at a distance x be a small mass element of mass dm given as :

$dm = \frac{M}{L}. dx$

We can write the torque produced by the dm mass as :

$d\tau = \frac{M}{L} \times dx \times g \times x$

$d\tau = \frac{Mg}{L} x . dx$

Now the total torque on the rod will be given by integrating the above equation for limits x = 0 to x = $\frac{3L}{4}$  :

$\tau = \int\limits^\frac{3l}{4} _0 { \frac{Mg}{L} } \,x. dx$

$\tau = \frac{Mg}{L} \frac{x^{2} }{2}$

So on solving the above equation for limits x = 0 to x = $\frac{3L}{4}$ :

$\tau = \frac{Mg}{L} \times \frac{9L^{2} }{16\times 2}$

$\tau = \frac{9MgL}{32}$      -(1)

Since we know that formula for torque is :

$\tau = I\alpha$      -(2)

Now the moment of inertia of the rod about point P is :

$I = \frac{ML^{2} }{12} +\frac{ML^{2} }{16}$

  $= \frac{4ML^{2}+3ML^{2} }{48}$

  $= \frac{7ML^{2} }{48}$

So now using equations 1 and 2 for equating torque we get :

$\frac{7ML^{2} }{48} \times \alpha = \frac{9MgL}{32}$

⇒$\alpha = \frac{27g}{14L}$

Hence $\alpha = \frac{135}{7L}$

So finally the angular momentum α of the rod is $\frac{135}{7L}$ .