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.

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

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, \[J_{lin} = Mv_{f}\] We need to solve for the final linear velocity, \(v_{f}\), so we divide by the mass M, \[v_{f} = \frac{J_{lin}}{M}\]

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 \[I = \frac{2}{5}MR^2\] 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, \[\omega_{f} = \frac{J_{ang}}{I} = \frac{J_{ang}} {\frac{2}{5}MR^2}\]

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, \[\omega_{f} =\frac{J \cdot h}{I}=\frac{J\cdot h}{\frac{2}{5}MR^2}\] #b) Distance h0 for rolling without slipping#

For a sphere to roll without slipping, its linear velocity and angular velocity should be related by the equation, \[v_{f} = R\omega_{f}\]

Now, we substitute the expressions we found for the linear and angular velocity above, \[\frac{J}{M} = R\left(\frac{J\cdot h_{0}}{\frac{2}{5}MR^2}\right)\]

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 \[\omega_{f} =\frac{J\cdot h}{\frac{2}{5}MR^2}\] We also found that the distance \(h_{0}\) at which the delivered impulse causes the ball to immediately roll without slipping is: \[h_{0} =\frac{2}{5}R\]