Question

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

Short answer

The angular acceleration must be applied for a duration of \(2s\) to produce a rotational kinetic energy of \(1500J\). The correct answer is (B) \(2s\).

Step-by-step solution

Identify the given values

: Moment of inertia (I) = \(1.2 kgm^{2}\) Initial angular velocity (ω_initial) = \(0 rad/s\) (at rest) Target rotational kinetic energy (K.E.) = \(1500 J\) Angular acceleration (α) = \(25 rad/s^{2}\)

Use the rotational kinetic energy formula

: The formula for rotational kinetic energy is: K.E. = \(\frac{1}{2}Iω^{2}\) Where: K.E. is the rotational kinetic energy I is the moment of inertia ω is the final angular velocity We have been provided the target rotational kinetic energy (1500 J) and the moment of inertia (1.2 kgm²). Our goal is to find the final angular velocity (ω). Plug the known values into the equation: \(1500 = \frac{1}{2}(1.2)ω^{2}\)

Solve for final angular velocity

: Rearrange the equation and solve for ω: \(ω^{2} = \frac{3000}{1.2}\) Calculate ω: \(ω = \sqrt{\frac{3000}{1.2}} \approx 50 rad/s\)

Use the equation of motion for rotational motion

: The equation of motion for rotational motion is: \(ω = ω_{initial} + αt\) Where: ω is the final angular velocity ω_initial is the initial angular velocity α is the angular acceleration t is the time for which the angular acceleration is applied We know the initial angular velocity (0 rad/s), the final angular velocity calculated in step 3, and the angular acceleration (25 rad/s²). Now, we can solve for t: \(50 = 0 + (25)t\)

Solve for the duration

: Rearrange the equation and solve for t: \(t = \frac{50}{25}\) Calculate t: \(t = 2s\) So, the angular acceleration must be applied for a duration of \(2s\) to produce a rotational kinetic energy of \(1500J\). The correct answer is (B) \(2s\).