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Chapter 5: Problem 581

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

Expert verified
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

01

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) = \(25 rad/s^{2}\)
02

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: \(1500 = \frac{1}{2}(1.2)ω^{2}\)
03

Solve for final angular velocity

: Rearrange the equation and solve for ω: \(ω^{2} = \frac{3000}{1.2}\) Calculate ω: \(ω = \sqrt{\frac{3000}{1.2}} \approx 50 rad/s\)
04

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: \(50 = 0 + (25)t\)
05

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) \(2s\).

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Most popular questions from this chapter

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

A cylinder of mass \(5 \mathrm{~kg}\) and radius \(30 \mathrm{~cm}\), and free to rotate about its axis, receives an angular impulse of $3 \mathrm{~kg} \mathrm{M}^{2} \mathrm{~S}^{-1}$ initially followed by a similar impulse after every \(4 \mathrm{sec}\). what is the angular speed of the cylinder 30 sec after initial impulse? The cylinder is at rest initially. \(\\{\mathrm{A}\\} 106.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{B\\} \(206.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{C\\} \(107.6 \mathrm{rad} \mathrm{S}^{-1}\) \\{D \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight \(\mathrm{mg}\) is attached to the end of the cord and falls from the rest. After falling through the distance \(\mathrm{h}\). the angular velocity of the wheel will be.... \(\\{B\\}\left[2 m g h /\left(I+m r^{2}\right)\right]\) $\\{\mathrm{C}\\}\left[2 \mathrm{mgh} /\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}$ \(\\{\mathrm{D}\\} \sqrt{(2 \mathrm{gh})}\)
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