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Chapter 5: Problem 612

A uniform disc of mass \(500 \mathrm{~kg}\) and radius \(2 \mathrm{~m}\) is rotating at the rate of \(600 \mathrm{r}\).p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of \(100 \mathrm{sec}\) ? \(\\{\mathrm{A}\\} 600 \pi \mathrm{Nm}\) \\{B \(\\} 500 \pi \mathrm{Nm}\) \\{C \(\\} .400 \pi \mathrm{Nm}\) \(\\{\mathrm{D}\\} 300 \pi \mathrm{Nm}\)

Short Answer

Expert verified
The torque required to rotate the disc in the opposite direction with the same angular speed in a time of 100 sec is approximately \(400\pi Nm\).

Step by step solution

01

Calculate the moment of inertia of the disc

The formula for the moment of inertia of a uniform disc is given by: \[I = \frac{1}{2}MR^2\] where M is the mass of the disc, R is its radius, and I is the moment of inertia. Given the mass M = 500 kg and radius R = 2 m, we can plug these values into the formula: \[I = \frac{1}{2} \times 500 \times (2)^2 = 1000\,kg.m^2\]
02

Convert angular speed from RPM to rad/s

Currently, the disc is rotating at 600 r.p.m., which should be converted to rad/s in order to find the angular acceleration later on. 1 revolution equals \(2\pi\) radians, and 1 minute equals 60 seconds. Therefore, \[\omega_1 = 600 \frac{rev}{min} \times \frac{2\pi\, rad}{1\, rev} \times \frac{1\, min}{60\, s} = 62.83\, rad/s\] where \(\omega_1\) is the initial angular speed.
03

Calculate the final angular speed in the opposite direction

As the final angular speed has the same magnitude but is in the opposite direction, we can represent it as the negative of the initial angular speed. \[\omega_2 = -\omega_1 = -62.83\, rad/s\]
04

Calculate the angular acceleration

We have the initial and final angular speeds as well as the time taken for the change. Using the equation \(\alpha = \frac{\omega_2 - \omega_1}{t}\), we can find the angular acceleration. Given the time t = 100s, we can calculate the angular acceleration as: \[\alpha = \frac{-62.83 - 62.83}{100} = -1.2566\, rad/s^2\]
05

Calculate the torque required

To calculate the torque required, we can use the formula: \[\tau = I\alpha\] Plugging in the moment of inertia I = 1000 kg.m² and angular acceleration \(\alpha = -1.2566\, rad/s^2\), we get: \[\tau = 1000 \times (-1.2566) = -1256.6\, Nm\] Since torque is a vector quantity, we need to consider its magnitude only. So, the required torque magnitude is: \[\tau = 1256.6\, Nm\] Now, let us see which option matches our result: \(600\pi Nm \approx 1884.96 Nm\) \\ \(500\pi Nm \approx 1570.80 Nm\) \\ \(400\pi Nm \approx 1256.64 Nm\) \\ \(300\pi Nm \approx 942.48 Nm\) As our result is very close to 1256.64 Nm, we choose the closest option C: \[ \tau = 400\pi Nm\]

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Most popular questions from this chapter

A constant torque of \(1500 \mathrm{Nm}\) turns a wheel of moment of inertia \(300 \mathrm{~kg} \mathrm{~m}^{2}\) about an axis passing through its centre the angular velocity of the wheel after 3 sec will be.......... $\mathrm{rad} / \mathrm{sec}$ \(\\{\mathrm{A}\\} 5\) \\{B \\} 10 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)

A Pulley of radius \(2 \mathrm{~m}\) is rotated about its axis by a force \(F=\left(20 t-5 t^{2}\right) N\) where \(t\) is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is $10 \mathrm{KgM}^{2}$, the number of rotations made by the pulley before its direction of motion is reversed is : \(\\{\mathrm{A}\\}\) more than 3 but less then 6 \(\\{\mathrm{B}\\}\) more than 6 but less then 9 \(\\{\mathrm{C}\\}\) more than 9 \\{D \\} Less then 3

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.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm} .\) is removed from tve \(\mathrm{x}\) edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate. $\\{\mathrm{A}\\}-7 \mathrm{~cm} \quad\\{\mathrm{~B}\\}+9 \mathrm{~cm} \quad\\{\mathrm{C}\\}-9 \mathrm{~cm} \quad\\{\mathrm{D}\\}+7 \mathrm{~cm}$

The angular momentum of a wheel changes from \(2 \mathrm{~L}\) to $5 \mathrm{~L}$ in 3 seconds what is the magnitudes of torque acting on it? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)
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