Question

A body having moment of inertia about rotating body is the same equal to 3 rad/s. Kinetic energy of this rotating body is the same at a speed of

• A. 1.0 m/s
• B. 0.5 m/s
• C. 1.5 m/s
• D. 2.0 m/s

Answer 1

Answer: (A)

a) 1.0 m/s

Answer 2

Note: The question phrasing is incorrect. Based on the options and typical physics problems, it is assumed that the question intended to ask: "A body having moment of inertia 3 kg-m² is rotating with angular velocity 3 rad/s. Kinetic energy of this rotating body is the same as that of a body of mass 27 kg moving with a speed of". This interpretation aligns with the provided similar question. We will solve the problem based on this interpretation.

Solution: The rotational kinetic energy of the rotating body is given by $KE_{rot} = \frac{1}{2} I \omega^2$. Given moment of inertia $I = 3$ kg-m² and angular velocity $\omega = 3$ rad/s. $KE_{rot} = \frac{1}{2} \times 3 \times (3)^2 = \frac{1}{2} \times 3 \times 9 = \frac{27}{2}$ J.

The translational kinetic energy of a body of mass $m$ moving with speed $v$ is given by $KE_{trans} = \frac{1}{2} m v^2$. Assuming the mass of the translating body is $m = 27$ kg (based on the similar question), we have: $KE_{trans} = \frac{1}{2} \times 27 \times v^2$.

According to the question, the kinetic energies are the same: $KE_{rot} = KE_{trans}$ $\frac{27}{2} = \frac{1}{2} \times 27 \times v^2$

Multiply both sides by 2: $27 = 27 v^2$

Divide both sides by 27: $v^2 = \frac{27}{27} = 1$

Taking the square root: $v = \sqrt{1} = 1$ m/s (since speed is a non-negative quantity).

Comparing this result with the given options, we find that it matches option a).

Explanation: The rotational kinetic energy is calculated using the given moment of inertia and angular velocity. The translational kinetic energy is calculated using the assumed mass and the unknown speed. By equating the two kinetic energies, an equation is formed that can be solved for the speed $v$.