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Chapter 5: Problem 611

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? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)

Short Answer

Expert verified
The correct answer is: \(\boxed{\mathrm{(A)}\, \mathrm{L}}\).

Step by step solution

01

Determine the Change in Angular Momentum

To find the change in angular momentum, we will subtract the initial angular momentum from the final angular momentum: ΔL = L_final - L_initial = 5L - 2L = 3L
02

Calculate Torque

Now that we have the change in angular momentum, we can use the formula for torque to find the magnitude of the torque acting on the wheel: Torque = ΔL / Δt = 3L / 3s = L The magnitude of the torque acting on the wheel is L. The correct answer is: \(\boxed{\mathrm{(A)}\, \mathrm{L}}\).

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Most popular questions from this chapter

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

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight \(\mathrm{mg}\) is attached to the end of the cord and falls from the rest. After falling through the distance \(\mathrm{h}\). the angular velocity of the wheel will be.... \(\\{B\\}\left[2 m g h /\left(I+m r^{2}\right)\right]\) $\\{\mathrm{C}\\}\left[2 \mathrm{mgh} /\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}$ \(\\{\mathrm{D}\\} \sqrt{(2 \mathrm{gh})}\)

Initial angular velocity of a circular disc of mass \(\mathrm{M}\) is \(\omega_{1}\) Then two spheres of mass \(\mathrm{m}\) are attached gently two diametrically opposite points on the edge of the disc what is the final angular velocity of the disc? \(\\{\mathrm{A}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{B}\\}[(\mathrm{M}+4 \mathrm{~m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{C}\\}[\mathrm{M} /(\mathrm{M}+4 \mathrm{~m})] \omega_{1}\) \\{D\\} \([\mathrm{M} /(\mathrm{M}+2 \mathrm{~m})] \omega_{1}\)

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)
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