Conservation of Energy in Rolling Motion: Videos & Practice Problems

In rolling motion, an object rotates while translating without slipping, requiring static friction to convert linear kinetic energy into rotational kinetic energy. The work done by static friction is zero, allowing conservation of energy to apply. For a solid cylinder rolling down an incline, the final linear speed can be derived using the equation: $\sqrt{\frac{4}{3}ghsin\left(\theta \right)}$. This highlights the relationship between linear and rotational dynamics, emphasizing the lower coefficients in rotational motion compared to linear motion.

Conservation of Energy in Rolling Motion

In rolling motion, an object not only rotates around its axis but also translates along a surface. This type of motion can be exemplified by a cylinder rolling down an inclined plane. The key to understanding rolling motion lies in the relationship between linear velocity and angular velocity, expressed by the equation:

$$v_{cm} = r \omega$$

where \(v_{cm}\) is the linear velocity of the center of mass, \(r\) is the radius, and \(\omega\) is the angular velocity. For an object to roll without slipping, static friction is essential. This frictional force converts some of the linear kinetic energy into rotational kinetic energy without dissipating energy, meaning the work done by static friction is zero:

$$W_{f_s} = 0$$

In scenarios where an object accelerates, static friction is necessary to produce angular acceleration (\(\alpha\)). The phrase "without slipping" indicates that kinetic friction is absent, allowing the object to roll smoothly. When applying the conservation of energy principle, the equation can be structured as follows:

$$KE_{initial} + PE_{initial} + W_{non-conservative} = KE_{final} + PE_{final}$$

For a solid cylinder released from rest at the top of an incline, the initial kinetic energy is zero, and potential energy is given by \(mgh_{initial}\). At the bottom of the incline, the potential energy is zero, and the kinetic energy consists of both translational and rotational components:

$$KE_{final} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$$

Using the moment of inertia for a solid cylinder, \(I = \frac{1}{2} mr^2\), and substituting \(\omega\) with \(\frac{v}{r}\), the equation simplifies to:

$$KE_{final} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mr^2\right \left(\frac{v}{r}\right)^2$$

After simplification, the equation leads to:

$$mgh_{initial} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2$$

Combining terms results in:

$$mgh_{initial} = \frac{3}{4} mv^2$$

From this, we can derive the final linear velocity:

$$v_{final} = \sqrt{\frac{4}{3}gh_{initial}}$$

To express height in terms of the incline length \(l\) and angle \(\theta\), we use:

$$h_{initial} = l \sin(\theta)$$

Thus, the final velocity becomes:

$$v_{final} = \sqrt{\frac{4gl \sin(\theta)}{3}}$$

This result highlights that the final velocity in rolling motion is similar in form to that of linear motion, but with a different coefficient. The coefficient in rolling motion is lower than that in linear motion, reflecting the dual nature of kinetic energy in rolling objects. This understanding aids in verifying calculations, ensuring that the derived coefficients remain consistent with the principles of energy conservation.

A solid sphere of mass M = 10 kg and radius R = 2 is rolling without slipping with speed V = 5 m/s on a flat surface when it reaches the bottom of an inclined plane that makes an angle of Θ = 37° with the horizontal. The plane has just enough friction to cause the sphere to roll without slipping while going up. What maximum height will the sphere attain? (Use g = 10 m/s².)

• A. 0.175 m
• B. 0.350 m
• C. 1.75 m
• D. 3.50 m

Answer: (C)

Sphere on rough and smooth hills

In the study of rolling motion and conservation of energy, consider a solid sphere rolling down a hill and then up another. The first hill has static friction, which allows for angular acceleration, causing the sphere to roll faster as it descends. Conversely, the second hill is smooth, meaning there is no static friction and thus no angular acceleration, allowing the sphere to maintain its rotational speed while it ascends.

Initially, at point A, the sphere is at rest at a height \( h_1 \). As it rolls down to point B, the potential energy at the top converts into kinetic energy at the bottom. The potential energy at point A can be expressed as \( PE = mgh_1 \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. At point B, the sphere has both translational and rotational kinetic energy, given by:

\[ KE = \frac{1}{2} mv_B^2 + \frac{1}{2} I \omega_B^2 \]

For a solid sphere, the moment of inertia \( I \) is \( \frac{2}{5} mr^2 \). The relationship between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r\omega \), allowing us to express \( \omega \) in terms of \( v \). Substituting these into the kinetic energy equation and simplifying leads to the expression for the velocity at point B:

\[ v_B = \sqrt{\frac{10gh_1}{7}} \]

Next, as the sphere moves from point C (the bottom of the hill) to point D (the top of the second hill), it retains its rotational kinetic energy while gaining potential energy. The energy conservation equation from C to D can be expressed as:

\[ KE_C + PE_C = KE_D + PE_D \]

At point C, the kinetic energy includes both translational and rotational components, while at point D, the sphere has potential energy \( PE_D = mgh_2 \) and retains its rotational kinetic energy. Notably, the rotational kinetic energy at point D is equal to that at point B due to the absence of friction. This leads to the equation:

\[ \frac{1}{2} mv_B^2 + \frac{1}{2} I \omega_B^2 = mgh_2 \]

By substituting the expressions for \( v_B \) and \( I \), and simplifying, we find:

\[ h_2 = \frac{10}{14} h_1 = 0.71 h_1 \]

This result indicates that the maximum height attained on the second hill is less than the initial height due to the energy conversion into rotational motion. The energy initially stored as potential energy is partially converted into rotational kinetic energy, which explains why the sphere does not reach the same height on the second hill. In essence, some of the energy that could have contributed to height is instead used for rotation, resulting in a lower maximum height.

You may remember that the lowest speed that an object may have at the top of a loop-the-loop of radius R, so that it completes the loop without falling, is √gR . Determine the lowest speed that a solid sphere must have at the bottom of a loop-the-loop, so that it reaches the top with enough speed to complete the loop. Assume the sphere rolls without slipping.

• A. $\sqrt{5gR}$
• B. $\sqrt{\frac{13gR}{5}}$
• C. $\sqrt{\frac{27gR}{7}}$
• D. $\sqrt{10gR}$

Answer: (C)

Here’s what students ask on this topic:

What is rolling motion and how does it differ from pure rotation?

Rolling motion is a type of motion where an object rotates around its own axis while also translating along a surface without slipping. This is different from pure rotation, where the object only spins around a fixed axis without any translational movement. In rolling motion, the velocity at the point of contact with the surface is zero, and the velocity of the center of mass (v_cm) is related to the angular velocity (ω) by the equation v_cm = rω, where r is the radius of the object.

How does static friction play a role in rolling motion?

Static friction is crucial in rolling motion as it prevents slipping and allows the object to roll. It converts some of the linear kinetic energy into rotational kinetic energy without dissipating energy. This means the work done by static friction is zero. Static friction ensures that the point of contact between the rolling object and the surface is momentarily at rest, enabling the object to roll smoothly.

How is the conservation of energy applied in rolling motion problems?

In rolling motion problems, conservation of energy is applied by considering both translational and rotational kinetic energies. The total mechanical energy is conserved, meaning the sum of initial kinetic and potential energies equals the sum of final kinetic and potential energies. For a rolling object, the kinetic energy includes both translational (1/2 mv²) and rotational (1/2 Iω²) components. Static friction does not do any work, so it does not affect the energy conservation equation.

What is the final linear speed of a solid cylinder rolling down an incline?

The final linear speed (v_f) of a solid cylinder rolling down an incline can be derived using conservation of energy. The equation is given by:

$\sqrt{\frac{4gl\sin \left(\theta \right)}{3}}$

where g is the acceleration due to gravity, l is the length of the incline, and θ is the angle of the incline with the horizontal. This equation shows the relationship between the linear and rotational dynamics of the rolling cylinder.

Why is the coefficient in rotational motion lower than in linear motion?

The coefficient in rotational motion is lower than in linear motion because a rolling object has both translational and rotational kinetic energies. This means that part of the energy is used for rotation, resulting in a lower translational speed compared to an object that only translates. For example, the final velocity of a rolling object is derived as $\sqrt{\frac{4gh}{3}}$, which is less than the final velocity of a purely translating object, $\sqrt{2gh}$.