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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

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 \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)

If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be .... \(\\{\mathrm{A}\\}[365 \times 3]\) \(\\{\mathrm{B}\\}[365 \times 27]\) \(\\{\mathrm{C}\\}[365 \times(3 \sqrt{3})]\) \(\\{\mathrm{D}\\}[365 /(3 \sqrt{3})]\)

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

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 thin circular ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity \(\mathrm{w}\). Two objects each of mass \(\mathrm{m}\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity.... $\\{\mathrm{A}\\}[\\{\omega(\mathrm{M}-2 \mathrm{~m})\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]$ \(\\{\mathrm{B}\\}[\\{\omega \mathrm{M}\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]\) \(\\{C\\}[\\{\omega M)\\} /\\{M+m\\}]\) \(\\{\mathrm{D}\\}[\\{\omega(\mathrm{M}+2 \mathrm{~m})\\} / \mathrm{M}]\)
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