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.

Key concepts

Rotational Kinetic Energy

When an object is rolling, it doesn't just move forward. It also spins around an axis. This spinning gives the object rotational kinetic energy. Calculating this energy is important for understanding how objects move.
An object's rotational kinetic energy is given by the formula \(KE_{rotational} = \frac{1}{2} Iw^2\). Here, \(I\) stands for the moment of inertia. This tells us how the mass is distributed in the object. The more spread out the mass, the larger the moment of inertia. \(w\) is the angular velocity, which is how fast the object spins.
For example, if you compare a ring and a solid sphere, the ring has a higher moment of inertia because its mass is distributed further from the center. That's why it spins differently even if both have the same mass and angular velocity. Understanding rotational kinetic energy helps us explain why different shapes roll differently.

Potential Energy

Potential energy is the energy stored due to an object's position relative to another position. For objects near the Earth’s surface, this is often due to height.
The potential energy at a height \(h\) is calculated by \(PE = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height. When a rolling object reaches its maximum height, all its initial kinetic energy has transformed into potential energy, provided there's no energy loss.
To put it simply, potential energy is what you get when a rolling object climbs up a hill. The higher it goes, the more potential energy it has. At the peak, this potential energy can then convert back into kinetic energy as the object rolls back down. This concept helps us understand energy conservation and transformation in rolling objects.

Linear and Angular Velocity Relation

Understanding the relationship between linear velocity and angular velocity is key to analyzing rolling objects. Linear velocity (\(v\)) describes how fast the object moves along a straight path, while angular velocity (\(w\)) tells us how fast it spins.
For a perfectly rolling object without slipping or skidding, the relation between these two velocities is given by \(v = Rw\). Here, \(R\) is the radius of the object. This equation shows that the linear velocity depends on both how fast the object rotates (angular velocity) and its size (radius).
Imagine a skateboard wheel rolling down a ramp. The radius and its rotational speed determine how fast it rolls on the ground. The bigger the wheel and the faster it spins, the quicker it moves. Grasping this concept lets us solve dynamics problems where rolling motion is involved.