A solid sphere of Mass M and Radius R rolling (pure) encounters a protrusion on the surface of height R/4. Its in the form of a step. What can be the maximum velocity V of sphere so that it doesn't leave contact from surface and final velocity?

To determine the maximum velocity \( V \) of a solid sphere of mass \( M \) and radius \( R \) that can roll over a step of height \( \frac{R}{4} \) without losing contact with the surface, we need to analyze the forces and energy involved in this scenario. The key here is to ensure that the sphere maintains contact with the surface as it rolls over the step.

Understanding the Dynamics of the Sphere

When the sphere rolls over the step, it will pivot around the edge of the step. At the point of contact with the step, the sphere experiences a change in its center of mass height, which affects the gravitational potential energy and the kinetic energy of the sphere.

Energy Considerations

Initially, the sphere has kinetic energy due to its velocity \( V \) and gravitational potential energy due to its height. As it approaches the step, we can express the total mechanical energy as:

• Kinetic Energy: \( KE = \frac{7}{10} MV^2 \) (for a solid sphere, considering both translational and rotational motion)
• Potential Energy: \( PE = MgH \), where \( H \) is the height of the center of mass above the reference point.

When the sphere reaches the top of the step, its center of mass will rise by \( \frac{R}{4} \). Therefore, the change in potential energy as it climbs the step is:

Change in Potential Energy: \( \Delta PE = Mg \cdot \frac{R}{4} \)

Applying Conservation of Energy

Using the conservation of energy principle, the initial kinetic energy must equal the sum of the potential energy at the top of the step and the kinetic energy at that point:

\( \frac{7}{10} MV^2 = Mg \cdot \frac{R}{4} + KE_{top} \)

At the top of the step, the sphere will have some velocity \( v_{top} \). The kinetic energy at that point can be expressed as:

\( KE_{top} = \frac{7}{10} Mv_{top}^2 \)

Finding the Maximum Velocity

To ensure the sphere does not lose contact with the step, we need to analyze the forces acting on it at the top of the step. The gravitational force must provide the necessary centripetal force to keep the sphere in contact:

At the top of the step, the gravitational force acting downwards is \( Mg \), and the required centripetal force is \( \frac{Mv_{top}^2}{R} \). For the sphere to maintain contact, we need:

\( Mg \geq \frac{Mv_{top}^2}{R} \)

From this, we can derive that:

\( g \geq \frac{v_{top}^2}{R} \)

Thus, \( v_{top}^2 \leq gR \) or \( v_{top} \leq \sqrt{gR} \).

Combining the Equations

Now, substituting \( v_{top} \) back into the energy equation:

\( \frac{7}{10} MV^2 = Mg \cdot \frac{R}{4} + \frac{7}{10} M(\sqrt{gR})^2 \)

After simplifying, we get:

\( \frac{7}{10} V^2 = g \cdot \frac{R}{4} + \frac{7}{10} gR \)

Combining the terms gives:

\( \frac{7}{10} V^2 = gR \left( \frac{1}{4} + \frac{7}{10} \right) \)

Calculating the right side:

\( \frac{1}{4} + \frac{7}{10} = \frac{5}{20} + \frac{14}{20} = \frac{19}{20} \)

Thus, we have:

\( \frac{7}{10} V^2 = \frac{19}{20} gR \)

Solving for \( V^2 \):

\( V^2 = \frac{19}{20} \cdot \frac{10}{7} gR = \frac{19gR}{14} \)

Finally, taking the square root gives us the maximum velocity:

\( V = \sqrt{\frac{19gR}{14}} \)

Final Velocity After the Step

After the sphere rolls over the step, it will continue to roll with a new velocity that can be calculated using the conservation of energy principle again, considering the height change and the kinetic energy at the top of the step. However, the maximum velocity before encountering the step is crucial for ensuring it maintains contact.

In summary, the maximum velocity \( V \) of the sphere to ensure it does not lose contact while rolling over the step of height \( \frac{R}{4} \) is given by:

\( V = \sqrt{\frac{19gR}{14}} \)

To determine the maximum velocity \( V \) of a solid sphere of mass \( M \) and radius \( R \) rolling without slipping as it encounters a step of height \( R/4 \), we need to analyze the forces and energy involved in this scenario. The key is to ensure that the sphere maintains contact with the surface as it rolls over the step. Let's break this down step by step.

Understanding the Problem

When the sphere rolls towards the step, it will experience a change in its motion as it climbs over the height of the step. The critical point occurs when the sphere is just about to leave the surface, which happens when the normal force becomes zero. At this point, the sphere is in free fall, and we need to find the conditions under which this occurs.

Applying Energy Conservation

We can use the principle of conservation of mechanical energy to analyze the motion of the sphere. Initially, the sphere has kinetic energy due to its velocity and potential energy due to its height. As it climbs the step, some of this kinetic energy is converted into potential energy.

• The initial kinetic energy \( KE_i \) of the sphere is given by:

  KE_i = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2

  where \( I \) is the moment of inertia of the sphere and \( \omega \) is its angular velocity. For a solid sphere, \( I = \frac{2}{5} MR^2 \) and \( \omega = \frac{V}{R} \).
• The potential energy \( PE \) gained when the sphere climbs the height \( h = \frac{R}{4} \) is:

  PE = Mgh = Mg \cdot \frac{R}{4}

Setting Up the Energy Equation

At the point just before the sphere leaves the surface, the total mechanical energy can be expressed as:

KE_i = PE + KE_f

Where \( KE_f \) is the kinetic energy just after climbing the step. At this point, the sphere's velocity will be less than \( V \) due to the energy conversion. We can express this as:

\frac{1}{2} M V^2 = Mg \cdot \frac{R}{4} + \frac{1}{2} M v_f^2 + \frac{1}{2} I \omega_f^2

Substituting \( I \) and \( \omega_f \) gives:

\frac{1}{2} M V^2 = Mg \cdot \frac{R}{4} + \frac{1}{2} M v_f^2 + \frac{1}{2} \cdot \frac{2}{5} MR^2 \cdot \left(\frac{v_f}{R}\right)^2

Finding the Final Velocity

Rearranging the equation to isolate \( v_f \) will help us find the final velocity of the sphere after it has climbed the step. The equation simplifies to:

V^2 = g \cdot \frac{R}{2} + \frac{5}{7} v_f^2

To ensure the sphere does not lose contact, we need to consider the centripetal force required to keep it in contact with the surface as it climbs. The condition for maintaining contact is that the gravitational force must provide enough centripetal force at the top of the step.

Condition for Maintaining Contact

At the top of the step, the centripetal force \( F_c \) required is:

F_c = \frac{M v_f^2}{R}

And the gravitational force acting on the sphere is:

F_g = Mg

For the sphere to remain in contact, we need:

Mg \geq \frac{M v_f^2}{R}

This leads to:

gR \geq v_f^2

Substituting \( v_f^2 \) from our earlier energy equation gives us a relationship between \( V \) and the height of the step.

Final Calculation

After substituting and simplifying, we find that the maximum initial velocity \( V \) can be expressed as:

V = \sqrt{gR \cdot \frac{3}{2}}

This means that if the sphere's initial velocity exceeds this value, it will lose contact with the surface as it climbs the step. Therefore, the maximum velocity \( V \) for the sphere to maintain contact while rolling over the step of height \( R/4 \) is:

V = \sqrt{gR \cdot \frac{3}{2}}

In summary, by applying the principles of energy conservation and analyzing the forces acting on the sphere, we can determine the conditions under which it maintains contact with the surface while rolling over the step. This approach not only illustrates the physics involved but also emphasizes the importance of understanding the interplay between kinetic and potential energy in motion.