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Chapter 5: Problem 666

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first two second it rotate through an angle \(\theta_{1}\), in the next \(2 \mathrm{sec}\). it rotates through an angle \(\theta_{2}\), find the ratio \(\left(\theta_{2} / \theta_{1}\right)=\) \(\\{\mathrm{A}\\} 1\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} 3\) \(\\{\mathrm{D}\\} 4\)

Short Answer

Expert verified
The ratio \(\left(\theta_{2} / \theta_{1}\right) = 3\), which corresponds to option C.

Step by step solution

01

Find the angle θ₁

To find the angle \(\theta_1\) rotated within the first 2 seconds, we use the equation of motion mentioned above. Since the initial angular velocity \(\omega_0\) is zero, we can simplify the equation to: $$\theta_1 = \frac{1}{2} \alpha t^2$$ Where \(\alpha\) is the angular acceleration, and \(t = 2\,\text{sec}\).
02

Find the angle θ₂

To find the angle \(\theta_2\) rotated within the next 2 seconds, first calculate the angular velocity \(\omega\) at the end of the first 2 seconds using the formula: $$\omega = \omega_0 + \alpha t$$ Since \(\omega_0 = 0\), the formula becomes \(\omega = \alpha t\). Now, we have to calculate the angle rotated in the next 2 seconds with the given angular velocity \(\omega\) and angular acceleration \(\alpha\). Again, we apply the equation of motion mentioned above: $$\theta_2 = \omega t + \frac{1}{2}\alpha t^2$$ Replacing \(\omega\) in the above equation, we get: $$\theta_2 = \alpha t \cdot t + \frac{1}{2}\alpha t^2$$
03

Find the ratio of angles θ₂ and θ₁

To find the ratio \(\left(\frac{\theta_2}{\theta_1}\right)\), simply divide the equations of Step 2 by Step 1: $$\frac{\theta_2}{\theta_1} = \frac{\alpha t \cdot t + \frac{1}{2}\alpha t^2}{\frac{1}{2} \alpha t^2}$$ By simplification, we get: $$\frac{\theta_2}{\theta_1} = \frac{2 \alpha t \cdot t + \alpha t^2}{\alpha t^2}$$ The term \(\alpha\) can be canceled out from both the numerator and denominator. Further simplification provides: $$\frac{\theta_2}{\theta_1} = \frac{2t \cdot t + t^2}{t^2} = \frac{3t^2}{t^2} = 3$$ Thus, the ratio \(\left(\theta_{2} / \theta_{1}\right) = 3\), which corresponds to option C.

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Most popular questions from this chapter

Statement \(-1\) - Two cylinder one hollow and other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow will reach the bottom of inclined plane first. Statement \(-2-\mathrm{By}\) the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \\{B \\} Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1\). \\{C\\} Statement - 1 is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true

A cylinder of mass \(\mathrm{M}\) has length \(\mathrm{L}\) that is 3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? \(\\{\mathrm{A}\\} 1\) \(\\{\mathrm{B}\\}(1 / \sqrt{3})\) \(\\{\mathrm{C}\\} \sqrt{3}\) \(\\{\mathrm{D}\\}(\sqrt{3} / 2)\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

Two discs of the same material and thickness have radii \(0.2 \mathrm{~m}\) and \(0.6 \mathrm{~m}\) their moment of inertia about their axes will be in the ratio \(\\{\mathrm{A}\\} 1: 81\) \(\\{\mathrm{B}\\} 1: 27\) \(\\{C\\} 1: 9\) \(\\{\mathrm{D}\\} 1: 3\)

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. If the disc is replaced by a ring of the same mass \(\mathrm{M}\) and the same radius \(R\), the ratio of the frictional force on the ring to that on the disc will be \(\\{\mathrm{A}\\} 3 / 2\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} \sqrt{2}\) \(\\{\mathrm{D}\\} 1\)
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