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Chapter 5: Problem 592

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

Short Answer

Expert verified
The rotational kinetic energy of the rod is \(K.E = \frac{1}{6} A L^3 D \omega^2\).

Step by step solution

01

Find the moment of inertia

To find the moment of inertia \(I\) of the rod, we can use the following formula: \[I = \int_0^L dm \, x^2\] Since the mass is uniformly distributed along the length, we can write \(dm = \frac{M}{L} dx\) , where \(M = A*L*D\) is the total mass of the rod. Now, the moment of inertia is given by: \[I = \int_0^L \frac{M}{L} x^2 dx\]
02

Calculate the integral

The integral will be: \[I = \frac{M}{L} \int_0^L x^2 dx\] \[I = \frac{M}{L} \left[\frac{x^3}{3}\right]_0^L\] \[I = \frac{M}{L} \left(\frac{L^3}{3}\right)\] \[I = \frac{1}{3} M L^2\] Now, substituting the value of M, we get: \[I = \frac{1}{3} (A * L * D) L^2\] \[I = \frac{1}{3} A L^3 D\]
03

Find the rotational kinetic energy

The formula for rotational kinetic energy is: \(K.E = \frac{1}{2} I \omega^2\) Now, substituting the value of I, we get: \(K.E = \frac{1}{2} (\frac{1}{3} A L^3 D) \omega^2\) \(K.E = \frac{1}{6} A L^3 D \omega^2\) So, the correct answer is \(B) (1 / 6) AL^{3} D \omega^{2}\).

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Most popular questions from this chapter

The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is \(\mathrm{I}_{1}\). The same rod is bent in to a ring and its moment of inertia about a diameter is \(\mathrm{I}_{2}\), Then \(\left[\mathrm{I}_{1} / \mathrm{I}_{2}\right]\) is. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

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 \(\\{\mathrm{A}\\} 3 / 2\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} \sqrt{2}\) \(\\{\mathrm{D}\\} 1\)

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)

A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)
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