Question

A system consists of two point masses, each of 2 kg, connected by a weightless rod
of length 50 cm. Determine the radius of gyration of the system about an axis
perpendicular to the rod and passing through a point at 10 cm from one end.​

Answer 1

Answer:

a body consists of two point masses ,each of 2kg connected by a weightless rod of length 50cm determine its moment of inertia and radius of gyration ...

Answer 2

Solution:

Two-point masses each of mass 2 kg connected by a weightless rod.

Length of the rod = 50 cm = 0.5 m

As per the question, an imaginary axis is placed perpendicular to the rod and passing through a point of 10 cm(0.1 m) from one end,

Distance of the second mass from the other end = 0.5 - 0.1 = 0.4m.

Total moment of inertia of the system,

$\implies\sf{I = I_1 + I_2 }$

$\implies\sf{I = MR_1^2+ MR_2^2 }$

$\implies\sf{I = M(R_1^2+ R_2^2) }$

Now let's substitute the given values,

$\implies\sf{I = 2 (0.01+ 0.16) }$

$\implies\sf{I = 2 (0.17) }$

$\implies\sf{I = 0.34 \: kg m^2}$

Now,

$\implies\sf{I = (M+M) K^2 }$

$\implies\sf{I = 2M \times K^2 }$

$\implies\sf{ K^2 = \dfrac{I}{2M} }$

Now let's substitute the given values,

$\implies\sf{ K^2 = \dfrac{0.34}{2\times 2 } }$

$\implies\sf{ K^2 = \dfrac{0.34}{4 } }$

$\implies\sf{ K^2 = 0.085 }$

$\implies\boxed{\sf{ K = 0.3 m}}$

The radius of gyration of the system is 0.3 m.