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:
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
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.
[tex]L_{1} = L_{2}[/tex]
or, [tex]I_{1} \times \omega_{1} = I_{2} \times \omega_{2}[/tex]
Putting the given values into the above formula as follows.
[tex]I_{1} \times \omega_{1} = I_{2} \times \omega_{2}[/tex]
or,
= [tex]\frac{(0.5 \times 35 \times (2.2)^{2}) \times 11}{(0.5 \times 35 \times (2.2)^{2} + 17 \times (1.5)^{2})}[/tex]
= 7.58 rad/s
Thus, we can conclude that the angular speed of the clay and turntable is 7.58 rad/s.