Question

A sphere of radius $R$ and mass $M$ sits on a horizontal tabletop. A horizontally directed impulse with magnitude $J$ is delivered to a spot on the ball a vertical distance $h$ above the tabletop. a) Determine the angular and translational velocity of the sphere just after the impulse is delivered. b) Determine the distance $h_0$ at which the delivered impulse causes the ball to immediately roll without slipping.

Short answer

Question: Determine the angular and translational velocities of the sphere just after the impulse is delivered and the distance at which the impulse causes the ball to roll without slipping instantly. Answer: The angular and translational velocities of the sphere just after the impulse are given by $v_f=\frac{J}{M}$ and ${\omega }_f=\frac{J\cdot h}{\frac{2}{5}M{R}^2}$. The distance at which the delivered impulse causes the ball to immediately roll without slipping is $h_0=\frac{2}{5}R$.

Step-by-step solution

Identify the given values

The problem provides us with the following information: - Sphere radius: R - Sphere mass: M - Impulse magnitude: J - Vertical distance above the tabletop: h

Determine the linear impulse delivered

Impulse-momentum theorem states that the impulse applied equals the change in linear momentum. The linear momentum is given by the product of the mass and velocity. Let's denote the linear impulse as J_lin. Since the sphere is initially at rest, the change in linear momentum is equal to the final linear momentum, $$J_{lin}=M{v}_f$$ We need to solve for the final linear velocity, $v_f$, so we divide by the mass M, $$v_f=\frac{J_{lin}}{M}$$

Determine the angular impulse delivered

Angular momentum is the product of the moment of inertia (I) of the sphere and its angular velocity (ω). The moment of inertia of a sphere is given by $$I=\frac{2}{5}M{R}^2$$ For the impulse J applied at a distance h from the center, we can determine the angular impulse by considering the torque (τ) caused by the impulse. The torque is the product of the perpendicular distance and the force. Let's denote the angular impulse as J_ang. Since the sphere is initially at rest, the change in angular momentum is equal to the final angular momentum, $$J_{ang}=I{\omega }_f$$ We need to find the final angular velocity (${\omega }_f$) so we divide by moment of inertia, $${\omega }_f=\frac{J_{ang}}{I}=\frac{J_{ang}}{\frac{2}{5}M{R}^2}$$

Obtain the angular and translational velocities

Now, we can substitute the angular and translational impulses (J_ang and J_lin, respectively) with the given impulse J. For the linear impulse, since it is applied horizontally, it is equal to the given impulse J, so $$v_f=\frac{J}{M}$$ For the angular impulse, it is given by the product of the applied impulse (J) and the perpendicular distance (h) from the center, $${\omega }_f=\frac{J\cdot h}{I}=\frac{J\cdot h}{\frac{2}{5}M{R}^2}$$ #b) Distance h0 for rolling without slipping#

Define a rolling condition without slipping

For a sphere to roll without slipping, its linear velocity and angular velocity should be related by the equation, $$v_f=R{\omega }_f$$

Replace vf and ωf with the expressions found in part a)

Now, we substitute the expressions we found for the linear and angular velocity above, $$\frac{J}{M}=R\left(\frac{J\cdot h_0}{\frac{2}{5}M{R}^2}\right)$$

Solve for h0

In this step, we will solve the equation for $h_0$. First, we can simplify the equation by canceling out J and dividing by R: $$\frac{1}{M}=\frac{h_0}{\frac{2}{5}MR}$$ Now, we solve for $h_0$, $$h_0=\frac{2}{5}MR\cdot \frac{1}{M}=\frac{2}{5}R$$ Upon solving the problem, we found that the angular and translational velocities of the sphere just after the impulse are given by: $$v_f=\frac{J}{M}$$ and $${\omega }_f=\frac{J\cdot h}{\frac{2}{5}M{R}^2}$$ We also found that the distance $h_0$ at which the delivered impulse causes the ball to immediately roll without slipping is: $$h_0=\frac{2}{5}R$$

Key concepts

Impulse and Momentum

Understanding impulse and momentum is crucial to solving physics problems involving motion. Impulse is the product of force and the time duration over which the force acts. It can be considered the change in momentum of an object. In this problem, an impulse $J$ was applied to a stationary sphere, resulting in a change in its linear momentum. The relationship is defined by the impulse-momentum theorem, expressed as $J=\mathrm{\Delta }p$, where $p$ is the momentum.

Momentum itself is defined as the product of an object's mass and its velocity, given by $p=mv$. In this scenario, since the sphere starts from rest, the impulse directly alters its final velocity with $v_f=\frac{J}{M}$.

This relation means the impulse determines how quickly the sphere moves after the impulse is applied, reflecting the direct correlation between impulse and momentum when external forces are applied to a mass.

Angular Velocity

Angular velocity describes how fast an object rotates around an axis. In this exercise, after the impulse was delivered at a height above the sphere's center, not only was linear velocity affected, but angular motion was introduced. Angular momentum $L$ is analogous to linear momentum for rotational motion and is the product of a rotating object's moment of inertia $I$ and its angular velocity $\omega$, expressed as $L=I\omega$.

The impulse delivered at a distance $h$ from the center generated torque, leading to angular momentum. The relationship between angular impulse and change in angular momentum can be expressed as $J_{\text{ang}}=I{\omega }_f$.

Knowing the moment of inertia for a sphere $I=\frac{2}{5}M{R}^2$ allows us to solve for ${\omega }_f$, representing how the impulse set the sphere spinning at a specific rate, ${\omega }_f=\frac{J\cdot h}{\frac{2}{5}M{R}^2}$. This demonstrates how applying force offsets from an object's center produces rotational motion.

Rolling Without Slipping

Rolling without slipping is an ideal condition in physics where an object rolls smoothly without any skidding. This condition ensures a fixed relationship between linear and angular velocities, crucial for understanding this sphere's motion post-impulse.

In this context, for a sphere to begin rolling without slipping immediately after the impulse, the sphere's linear velocity must equal the tangential speed at its surface due to rotation: $v_f=R{\omega }_f$. This condition ensures that each point on the contact surface has zero relative velocity concerning the tabletop.

By substituting the expressions for $v_f$ and ${\omega }_f$, we can solve for the height $h_0$ at which the impulse needs to be applied to achieve this motion: $h_0=\frac{2}{5}R$. This equation implies that the correct application of impulse at this height will cause the sphere to roll without slipping, demonstrating an essential principle in mechanics.

Moment of Inertia

Moment of inertia is a fundamental concept in rotational dynamics, representing how difficult it is to change an object's rotational state. It is dependent on the mass distribution relative to the rotational axis. For a solid sphere, this distribution is represented by $I=\frac{2}{5}M{R}^2$. This shows how the mass and radius of the sphere contribute to its resistance to angular acceleration.

In our scenario, understanding the moment of inertia informs us about the sphere's angular response to the applied impulse. When the impulse is delivered, the change in rotational motion, captured by the angular velocity ${\omega }_f$, is calculated using this moment of inertia.

This illustrates how crucial the concept of moment of inertia is to determining how an object will rotate about its axis when subject to external forces, bridging the fields of linear and rotational dynamics.