Question

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

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

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 = \(KE_{initial} + KE_{rotational}\)

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.

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.

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: \(\frac{1}{2} m(Rw)^{2} + \frac{1}{2} Iw^2 = mgh\)

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, \[\frac{1}{2} + \frac{I}{2mR^2} = \frac{3}{4}\]

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.