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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

The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

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A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)
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