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Chapter 5: Problem 623

Moment of inertia of a sphere of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is \(\mathrm{I}\). keeping mass constant if graph is plotted between \(\mathrm{I}\) and \(\mathrm{R}\) then its form would be.

Short Answer

Expert verified
The graph between the moment of inertia (I) and radius (R) of a sphere while keeping the mass constant (M) would be a parabolic curve with the equation: \( I(R) = \dfrac{2}{5}M R^2 \)

Step by step solution

01

Moment of Inertia of a Solid Sphere

The moment of inertia of a given object depends on its mass distribution. For a solid sphere of mass M and radius R, the moment of inertia is given by the formula: \( I = \dfrac{2}{5}MR^2 \)
02

Study the relationship between I and R

To analyze the relationship between I (moment of Inertia) and R (radius) while keeping the mass M constant, we can rewrite the formula as: \( I(R) = \dfrac{2}{5}M R^2 \) Note that M is a constant value, so the graph of I with respect to R will be determined by the R^2 term.
03

Graphing I(R)

To plot I as a function of R, we see that I increases with R^2. This means that the graph will be a parabolic curve, with the vertex at the origin (0,0) and opening upwards (since R^2 is always positive). The graph will have the form of y = kx^2, where k is a constant, in this case, k = (2/5) * M. Therefore, the graph between the moment of inertia (I) and radius (R) of a sphere while keeping the mass constant M would be a parabolic curve with the equation: \( I(R) = \dfrac{2}{5}M R^2 \)

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Most popular questions from this chapter

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be.... \(\\{\mathrm{A}\\} 15: 14\) \(\\{\mathrm{B}\\} 14: 15\) \(\\{\mathrm{C}\\} 5: 3\) \(\\{\mathrm{D}\\} 3: 5\)

The moment of inertia of a hollow sphere of mass \(\mathrm{M}\) and inner and outer radii \(\mathrm{R}\) and \(2 \mathrm{R}\) about the axis passing through its centre and perpendicular to its plane is \(\\{\mathrm{A}\\}(3 / 2) \mathrm{MR}^{2}\) \\{B \(\\}(13 / 32) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(31 / 35) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(62 / 35) \mathrm{MR}^{2}\)

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 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 \(\\{\mathrm{D}\\} 1\)
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