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A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.30 kg is attached to one end and a second mass m2 = 3.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
A) What is the moment of inertia of the system?B) If the rod rotates with an angular speed of 2.00 rad/s, how much kinetic energy does the system have?C) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?D) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.00 rad/s?

Respuesta :

Answer:

Explanation:

Moment of inertia of the rod = 1/12 m L²

m is mass of the rod and L is its length

= 1/2 x 2.3 x 2 x 2

= 4.6 kg m²

Moment of inertia of masses attached with the rod

= m₁ d² + m₂ d²

m₁ and m₂ are masses attached , and d is their distance from the axis of rotation

= 5.3 x 1² + 3.5 x 1²

= 8.8 kg m²

Total moment of inertia = 13.4 kg m²

B )

Rotational kinetic energy = 1/2 I ω²

I is total moment of inertia and ω is angular velocity

= .5 x 13.4 x 2²

= 26.8 J .

C )

when mass of rod is negligible , moment of inertia will be due to masses only

Total moment of inertia of masses

= 8.8 kg m²

D )

kinetic energy of the system

= .5 x 8.8 x 2²

= 17.6 J .

(A) Total moment of inertia is 13.4 kgm²

(B) Total kinetic energy is 26.8J

(C) Moment of inertia is  8.8 kgm²
(D) Kinetic energy is 17.6J

Rotational motion:

(A) The moment of inertia of the rod is given by:

I = 1/12 mL²

where m is the mass of the rod

and L is the length

So,

I = (1/12) × 2.3 × 2²

I = 4.6 kgm²

Now, the moment of inertia of masses attached to the rod is given by:

I' = m₁ d² + m₂d²

where m₁ and m₂ are masses

and d is their distance from the axis of rotation

I' = 5.3 × 1² + 3.5 × 1²

I' = 8.8 kgm²

The total moment of inertia of the system is given by:

I(tot) = I + I'

I(tot) = 13.4 kgm²

(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:

KE = 1/2 I(tot)ω²

KE = 0.5 × 13.4 × 2²

KE = 26.8J

(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be

I(tot) = I' = 8.8 kgm²

(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:

KE = 1/2 I'ω²

KE = 0.5 × 8.8 × 2²

KE = 17.6J

Learn more about rotational motion:

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