Question

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

Short answer

Answer: The cylinder should be released from a height of \(\frac{15}{14}h_0\), where \(h_0\) is the initial height of the solid sphere.

Step-by-step solution

Write the Conservation of Mechanical Energy for the Sphere

The total mechanical energy is the sum of gravitational potential energy and kinetic energy. For the sphere, the initial mechanical energy at height \(h_0\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}I_{sphere}\omega^2$$

Find the Moment of Inertia and Angular Velocity for the Sphere and the Cylinder

Moment of inertia for a solid sphere is \(I_{sphere} = \frac{2}{5}MR^2\), and for a cylinder it is \(I_{cylinder} = \frac{1}{2}MR^2\). Since both objects are rolling without slipping, we can relate their linear and angular velocities by: \(v = R\omega\). We can substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v_{s}}{R}\right)^2$$

Simplify the Energy Conservation Equation for the Sphere

Now, we can simplify the equation for the sphere by cancelling out terms and collecting like terms: $$gh_0 = \frac{1}{2}v_{s}^2 + \frac{1}{5}v_{s}^2$$ Combine the fractions: $$gh_0 = \frac{7}{10}v_{s}^2$$ Solving for the final velocity of the sphere: $$v_{s} = \sqrt{\frac{10}{7}gh_0}$$

Write the Conservation of Mechanical Energy for the Cylinder

Now we need to repeat the process for the cylinder. The initial mechanical energy at height \(h\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}I_{cylinder}\omega^2$$ Substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v_{c}}{R}\right)^2$$

Simplify the Energy Conservation Equation for the Cylinder

Now, we can simplify the equation for the cylinder by cancelling out terms and collecting like terms: $$gh = \frac{1}{2}v_{c}^2 + \frac{1}{4}v_{c}^2$$ Combine the fractions: $$gh = \frac{3}{4}v_{c}^2$$

Set the Sphere's and Cylinder's Final Velocities Equal

Since we want the cylinder to have the same speed as the sphere at the bottom, we can set their final velocities equal: $$v_{s} = v_{c}$$ $$\sqrt{\frac{10}{7}gh_0} = \sqrt{\frac{4}{3}gh}$$

Solve for the Cylinder's Release Height

Now, we can simplify the equation and solve for the cylinder's release height: $$\frac{10}{7}gh_0 = \frac{4}{3}gh$$ Dividing both sides by \(g\) and multiplying by \(\frac{3}{4}\), we find the value for \(h\): $$h = \frac{3}{4}\left(\frac{10}{7}h_0\right)$$ $$h = \frac{15}{14}h_0$$ So the cylinder should be released from a height of \(\frac{15}{14}h_0\) to have the same speed as the sphere at the bottom of the incline.

Key concepts

Rolling Motion

Rolling motion involves objects such as spheres and cylinders rolling down an incline without slipping. When something rolls without slipping, it means the point of contact on the object is momentarily at rest while the object rotates around it. This is crucial because it connects linear and rotational motion through the relationship:

• Linear Velocity (\(v\)) relates to Angular Velocity (\(\omega\)) by: \(v = R\omega\). Here \(R\) is the radius of the object.

This relationship allows us to break down the motion into translational (straight-line) and rotational (spinning) kinetic energies. For instance, as a sphere rolls down an incline, its increased speed comprises both the speed at which its center moves and how fast it's spinning:

• Translational Kinetic Energy: \(\frac{1}{2}Mv^2\)
• Rotational Kinetic Energy: \(\frac{1}{2}I\omega^2\)

These are both components of the total kinetic energy we consider when applying the conservation of mechanical energy.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it as rotational mass for simplicity. It's affected by the distribution of mass within an object and is calculated differently for different shapes:

• For a solid sphere, \(I_{sphere} = \frac{2}{5}MR^{2}\)
• For a solid cylinder, \(I_{cylinder} = \frac{1}{2}MR^{2}\)

These formulas reflect how mass distribution matters. A sphere has a smaller moment of inertia compared to a cylinder of the same mass and radius, meaning it's easier to start spinning and faster for a given incline height. Understanding moment of inertia is essential when calculating how fast an object will move or how it reacts to forces. It combines with angular velocity (\(\omega\)) to find rotational kinetic energy which changes how we apply the conservation of mechanical energy.

Inclined Plane

An inclined plane is a flat, sloped surface, and it's a classic setting in physics problems to study motion. As objects like spheres and cylinders roll down an incline, gravity does work on them, converting potential energy to kinetic energy. Here's how it works:

• The slope angle (\(\theta\)) affects the component of gravitational force acting along the incline. Steeper slopes result in faster accelerations.
• Potential Energy at height \(h\): \(Mg h\), where \(M\) is mass, and \(g\) is gravitational acceleration. This energy is what gets converted to kinetic energy at the bottom.
• Kinetic Energy forms (both translational and rotational) increase as potential energy decreases, conserving total mechanical energy:

When tackling these problems, note that the height from which an object is released dictates its speed at the bottom. Different objects with different mass distributions may need to start from different heights to end at the same speed due to their moments of inertia. This is why the cylinder must start from a higher height (\(\frac{15}{14}h_0\)) to match the sphere's speed at the incline's bottom.