Question

A thin uniform cylindrical turntable of radius 2.2 m and mass 35 kg rotates in a horizontal plane with an initial angular speed of 11 rad/s. The turntable bearing is frictionless. A clump of clay of mass 17 kg is dropped onto the turntable and sticks at a point 1.5 m from the point of rotation.
Find the angular speed of the clay and turntable.

Answer 1

Explanation:

The given data is as follows.

    M = 35 kg,    radius (r) = 2.2 m,

     m = 17 kg,     = 11 rad/s

We assume that will be the final angular speed.

Now, according to the conservation of angular momentum.

         $L_{1} = L_{2}$

or,    $I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

Putting the given values into the above formula as follows.

  $I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

   

or,  

      = $\frac{(0.5 \times 35 \times (2.2)^{2}) \times 11}{(0.5 \times 35 \times (2.2)^{2} + 17 \times (1.5)^{2})}$

      = 7.58 rad/s

Thus, we can conclude that the angular speed of the clay and turntable is 7.58 rad/s.

Answer 2

Answer:

Explanation:

Mass of turn table, M = 35 kg

Radius of turn table, R = 2.2 m

initial angular velocity, ω = 11 rad/s

mass of clay, m = 17 kg

distance of clay from centre of table, r = 1.5 m

Let the final angular velocity is ω'.

By the use of conservation of angular momentum

I x ω = I' x ω'

where, I is the moment of inertia of the turn table = 0.5 MR²

I = 0.5 x 35 x 2.2 x 2.2 = 84.7 kg m²

I' is the moment of inertia of the table and the clay lump.

I' = I + mr² = 84.7 + 17 x 1.5 x 1.5 = 122.95 kg m²

Now

84.7 x 11 = 122.95 x ω'

ω' = 7.78 rad/s