Question

A round body of mass \(M\), radius \(R,\) and moment of inertia \(I\) about its center of mass is struck a sharp horizontal blow along a line at height \(h\) above its center (with \(0 \leq h \leq R,\) of course). The body rolls away without slipping immediately after being struck. Calculate the ratio \(I /\left(M R^{2}\right)\) for this body.

Short answer

The ratio of the moment of inertia to mass and radius squared is given by the expression: \(\frac{I}{M R^2} = \frac{2 K_{total} - (p_{after})^2}{(p_{after} / M R)^2} \times \frac{1}{M R^2}\) where K_total is the total kinetic energy of the round body, and p_after is the momentum of the body after the collision. This ratio does not depend on the height h where the body was struck.

Step-by-step solution

Identify the kinetic energy of the round body after the collision

The body starts rolling without slipping immediately after the collision, so we can consider the kinetic energy of the round body as the sum of its translational kinetic energy and its rotational kinetic energy. Let's denote the velocity of the center of mass after the collision as v_cm and the angular velocity as w. The translational kinetic energy is given by: \(K_t = \frac{1}{2} M v_{cm}^2\) The rotational kinetic energy is given by: \(K_r = \frac{1}{2} I w^2\) The total kinetic energy of the round body is: \(K_{total} = K_t + K_r\)

Relationship between linear velocity and angular velocity

As the body rolls without slipping, we can establish a relationship between the linear velocity of the center of mass and the angular velocity as follows: \(v_{cm} = R w\)

Momentum conservation

Since the collision was horizontal and along a line with height h, we can establish the linear momentum conservation in horizontal direction. Let's denote the momentum before the collision is \(p_{before}\) and after the collision is \(p_{after}\). \(p_{before} = p_{after}\) \(p_{before} = M v_{cm}\), therefore: \(p_{after} = M v_{cm} = M R w\)

Express the kinetic energy in terms of momentum

Now we can express the total kinetic energy in terms of momentum using the linear momentum conservation formula: \(K_{total} = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I w^2 = \frac{1}{2} (p_{after})^2 / M + \frac{1}{2} I (p_{after}/M R)^2\)

Calculate the ratio of moment of inertia to mass and radius squared

To find the ratio I/MR^2, we need to express I in terms of MR^2, h, and known physical quantities. From the linear momentum conservation equation, we have: \(w = \frac{p_{after}}{M R}\) Now we can further express the total kinetic energy: \(K_{total}= \frac{1}{2} (p_{after})^2 / M + \frac{I}{2} (p_{after} / M R)^2 \) From the above equation, we can express I in terms of MR^2, h, and other known physical quantities: \(I = \frac{2 K_{total} - (p_{after})^2}{(p_{after} / M R)^2}\) Now plug the value of I back into the ratio I/MR^2: \(\frac{I}{M R^2} = \frac{2 K_{total} - (p_{after})^2}{(p_{after} / M R)^2} \times \frac{1}{M R^2}\) This is the final expression for the ratio of the moment of inertia to mass and radius squared. Note that in our analysis, we haven't used the height h explicitly, so this ratio doesn't depend on the height h where the body was struck.

Key concepts

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by a rotating object due to its motion. It's comparable to translational kinetic energy, which is the energy due to motion in a straight line. However, while translational kinetic energy depends on the mass and the square of the velocity of an object's center of mass, rotational kinetic energy depends on an object's moment of inertia and the square of its angular velocity.

Using the formula for rotational kinetic energy, \(K_r = \frac{1}{2} I \omega^2\), where \(I\) represents the moment of inertia and \(\omega\) the angular velocity, we can calculate the energy related to the object's rotation. The moment of inertia is essentially the 'rotational mass', and the angular velocity is the rotational equivalent of linear velocity.

For example, when a round body like a wheel or cylinder rolls without slipping, its rotational kinetic energy contributes to the total kinetic energy along with its translational kinetic energy. The conservation of energy and the relationship between translational and rotational motion are critical when analyzing the motion of rolling objects.

Understanding rotational kinetic energy is crucial as it helps explain the behavior of rotating systems and is a key part of problems involving rolling motion, such as the one in our exercise where a round body rolls away after being struck. Here, each point on the body is moving with both a translational velocity and an angular velocity around the center of mass, leading to a combination of kinetic energies.

Linear Momentum Conservation

Linear momentum conservation is a fundamental concept in physics, stating that in a closed system with no external forces, the total momentum remains constant over time. It's a direct consequence of Newton's third law of motion.

For a body that is rolling without slipping, the linear momentum is governed by its mass and the velocity of its center of mass. Before an external force is applied (like the sharp blow in our exercise), the body has a certain momentum. After the force is applied and immediately after the collision, the body's momentum must remain the same if there is no other external force acting on it. In our exercise, we denote the momentum before and after the collision as \(p_{before}\) and \(p_{after}\), respectively, with the conservation equation \(p_{before} = p_{after}\).

Momentum conservation is used here to establish a crucial link between the translational motion and the rotational motion by the equation \(p_{after} = M R \omega\). This relationship allows us to express quantities related to rotation, such as the angular velocity, in terms of linear momentum. This interconnection between linear and rotational motion is essential for solving problems where objects exhibit both types of motion, as is often the case in rolling without slipping scenarios.

Rolling Without Slipping

Rolling without slipping is a common motion for wheels and balls where there is a static frictional force between the rolling object and the surface, preventing any relative motion (slipping) at the point of contact. For an object to roll without slipping, the linear velocity \(v_{cm}\) at the center of mass and the angular velocity \(\omega\) must be related through the radius \(R\) of the object by the formula \(v_{cm} = R \omega\).

This condition ensures that as the object rotates, each point in contact with the surface is momentarily at rest relative to the surface, mimicking the behavior of an object that 'sticks' to the surface as it rolls. The rolling object adheres to this critical relationship, which allows for the determination of one type of motion knowing the other.

In our exercise, the round body begins to roll away without slipping immediately after being struck. This means that we can relate its translational kinetic energy and rotational kinetic energy using the link between \(v_{cm}\) and \(\omega\), allowing for a comprehensive understanding of its motion post-collision. Additionally, when an object is rolling without slipping, there is less energy dissipated as heat through friction compared to an object that slips, which often matters in conservation of energy problems.