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Chapter 5: Problem 621

A small object of uniform density rolls up a curved surface with initial velocity 'u'. It reaches up to maximum height of $3 \mathrm{v}^{2} / 4 \mathrm{~g}$ with respect to initial position then the object is \(\\{\mathrm{A}\\}\) ring \(\\{B\\}\) solid sphere \(\\{\mathrm{C}\\}\) disc \\{D\\} hollow sphere

Short Answer

Expert verified
The short answer is: The object is a solid sphere (option B). This is determined by analyzing the given maximum height and relating this information to the moment of inertia, which corresponds to a solid sphere (\(I = \frac{1}{2} mR^2\)).

Step by step solution

01

Write down the initial Kinetic Energy (KE) and Potential Energy (PE) equations

Assuming the object starts from the ground level, its initial KE is \(KE_{initial} = \frac{1}{2} mv^{2}\), where 'm' is the mass, and 'v' is the initial velocity. Also, since the object is rolling, it also has rotational kinetic energy given by \(KE_{rotational} = \frac{1}{2} Iw^2\), where 'I' is the moment of inertia and 'w' is the angular velocity. The object's initial PE is \(PE_{initial} = 0\) as the object is at ground level. So, Total Initial KE = \(KE_{initial} + KE_{rotational}\)
02

Write the relation between linear velocity and angular velocity

The linear velocity 'v' and angular velocity 'w' for a rolling object are related by the equation \(v = R*w\), where 'R' is the radius of the object.
03

Write down the Potential Energy at the maximum height

At maximum height 'h', all kinetic energy would be converted to potential energy (neglecting air friction and energy loss due to rolling). Thus, the potential energy at the maximum height would be: \(PE_{max} = mgh\), where 'g' is the acceleration due to gravity.
04

Equate the Total Initial KE to PE at maximum height

At maximum height, \(Total\: Initial\: KE = PE_{max}\) Thus, \(\frac{1}{2} mv^{2} + \frac{1}{2} I*w^2 = mgh\) Now, substituting \(v = Rw\), we get: \(\frac{1}{2} m(Rw)^{2} + \frac{1}{2} Iw^2 = mgh\)
05

Rearrange the equation to find the moment of inertia (I)

Rearranging the equation, we get: \(\frac{1}{2} mR^2w^2 + \frac{1}{2} Iw^2 = mgh\) Divide both sides of the equation by \(\frac{1}{2} mR^2w^2\): \(\frac{1}{2} + \frac{I}{2mR^2} = \frac{gh}{v^2}\) According to the given information, the object reaches a maximum height of \(3v^2/4g\), so substituting \(h = 3v^2/4g\): \(\frac{1}{2} + \frac{I}{2mR^2} = \frac{g}{v^2} * \frac{3v^2}{4g}\) Now simplifying, \[\frac{1}{2} + \frac{I}{2mR^2} = \frac{3}{4}\]
06

Calculate the moment of inertia (I) and determine the shape of the object

Solving for I in the last equation: \(\frac{I}{2mR^2} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}\) \(I = \frac{1}{2} mR^2\) The moment of inertia \(I = \frac{1}{2} mR^2\) corresponds to a solid sphere. Therefore, the correct choice is 'B' solid sphere.

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Most popular questions from this chapter

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

Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)

Two spheres each of mass \(\mathrm{M}\) and radius \(\mathrm{R} / 2\) are connected with a mass less rod of length \(2 \mathrm{R}\) as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? \(\\{\mathrm{A}\\}(21 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(2 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(5 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(5 / 21) \mathrm{MR}^{2}\)

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

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 \(\\}(13 / 32) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(31 / 35) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(62 / 35) \mathrm{MR}^{2}\)
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