Energy of Rolling Motion: Videos & Practice Problems

Rolling motion involves both linear and rotational velocities, characterized by the relationship between the center of mass velocity (v_cm) and angular velocity (ω), expressed as $v^{\mathrm{cm}}=R\omega$. For objects rolling without slipping on a surface, kinetic energy is the sum of linear kinetic energy ($\frac{1}{2}\mathrm{mv}{v}^2$) and rotational kinetic energy ($\frac{1}{2}I{\omega }^2$), leading to total kinetic energy calculations.

Energy of Rolling Motion (Surface vs Air)

Rolling motion occurs when an object, such as a cylinder or sphere, moves along a surface while simultaneously rotating around its own axis. This type of motion is characterized by two types of velocities: linear velocity, which pertains to the movement of the center of mass, and rotational velocity, which describes the spinning of the object. The relationship between these velocities is expressed by the equation:

v_{cm} = r \cdot \omega

Here, v_{cm} represents the linear velocity of the center of mass, r is the radius of the object, and \omega is the angular velocity. This equation holds true only when the object is rolling without slipping on a surface. If the object is in the air, the linear and rotational velocities are not directly related.

In the context of kinetic energy, rolling objects possess both linear kinetic energy and rotational kinetic energy. The formulas for these energies are:

KE_{linear} = \frac{1}{2} m v_{cm}^2

KE_{rotational} = \frac{1}{2} I \omega^2

Where I is the moment of inertia of the object. For a solid sphere, the moment of inertia is given by:

I = \frac{2}{5} m r^2

To illustrate these concepts, consider a solid sphere with a mass of 2 kg and a radius of 0.3 m rolling without slipping on a horizontal surface at a linear velocity of 10 m/s. The linear kinetic energy can be calculated as:

KE_{linear} = \frac{1}{2} \cdot 2 \cdot 10^2 = 100 \text{ joules}

Next, to find the rotational kinetic energy, we first need to determine the angular velocity:

\omega = \frac{v_{cm}}{r} = \frac{10}{0.3} \approx 33.33 \text{ rad/s}

Substituting this value into the rotational kinetic energy formula, we have:

KE_{rotational} = \frac{1}{2} \cdot \frac{2}{5} \cdot 2 \cdot (0.3^2) \cdot (33.33^2) = 40 \text{ joules}

Finally, the total kinetic energy of the rolling sphere is the sum of its linear and rotational kinetic energies:

KE_{total} = KE_{linear} + KE_{rotational} = 100 + 40 = 140 \text{ joules}

This example highlights the importance of understanding both types of kinetic energy in rolling motion and reinforces the relationship between linear and rotational velocities when rolling without slipping on a surface.

A 150-g baseball, 3.85 cm in radius, leaves the pitcher's hand with 30 m/s horizontal and 20 rad/s clockwise. Calculate the ball's linear, rotational, and total kinetic energy.

• A. K_L = 6.75 × 10⁴ J ;K_R = 1.78 × 10⁵ J ;K_tot = 2.45 × 10⁵ J
• B. K_L = 67.5 J ;K_R = 1.16 J ;K_tot = 68.66 J
• C. K_L = 67.5 J ;K_R = 0.46 J ;K_tot = 68.96 J
• D. K_L = 67.5 J ;K_R = 0.018 J ;K_tot = 67.52 J

Answer: (D)

Ratio of energies of cylinder on surface

In rotational dynamics, understanding the relationship between different forms of kinetic energy is crucial. For a solid cylinder rolling without slipping on a horizontal surface, we can analyze the ratio of its rotational kinetic energy to its total kinetic energy using specific equations. The moment of inertia \( I \) for a solid cylinder is given by the formula:

\[ I = \frac{1}{2} m r^2 \]

where \( m \) is the mass and \( r \) is the radius of the cylinder. The rolling motion allows us to relate the linear velocity \( v_{cm} \) of the center of mass to the angular velocity \( \omega \) through the equation:

\[ v_{cm} = r \omega \]

To find the desired ratio, we set up the expression for the ratio of rotational kinetic energy \( K_{R} \) to total kinetic energy \( K_{total} \). The rotational kinetic energy is defined as:

\[ K_{R} = \frac{1}{2} I \omega^2 \]

And the total kinetic energy is the sum of linear and rotational kinetic energies:

\[ K_{total} = K_{L} + K_{R} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

Substituting the expression for \( I \) into the equation for \( K_{R} \) gives:

\[ K_{R} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega^2 \]

Next, we replace \( \omega \) with \( \frac{v}{r} \) to eliminate the angular variable:

\[ K_{R} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{4} mv^2 \]

Now, substituting this back into the total kinetic energy expression, we have:

\[ K_{total} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

Now we can set up the ratio:

\[ \text{Ratio} = \frac{K_{R}}{K_{total}} = \frac{\frac{1}{4} mv^2}{\frac{3}{4} mv^2} \]

Here, the mass \( m \) and \( v^2 \) cancel out, simplifying the ratio to:

\[ \text{Ratio} = \frac{1}{3} \]

This result indicates that for a solid cylinder, the rotational kinetic energy constitutes one-third of the total kinetic energy. It's important to note that this ratio can vary for different shapes, as the moment of inertia \( I \) will differ, affecting the final ratio of kinetic energies.

A hollow sphere of mass M and radius R rolls without slipping on a horizontal surface with angular speed W. Calculate the ratio of its linear kinetic energy to its total kinetic energy.

• A. K_L/K_tot =0.15
• B. K_L/K_tot = 0.60
• C. K_L/K_tot = 0.71
• D. K_L/K_tot = 1.67

Answer: (B)

Here’s what students ask on this topic:

What is rolling motion in physics?

Rolling motion in physics refers to the movement of an object that rotates around its own axis while also moving linearly along a surface. This type of motion combines both linear velocity (v_cm) and angular velocity (ω). A common example is a wheel or a ball rolling on the ground. The relationship between the linear velocity of the center of mass and the angular velocity is given by the equation v_cm = Rω, where R is the radius of the object. This relationship holds true only if the object is rolling without slipping on a surface.

How do you calculate the total kinetic energy of a rolling object?

The total kinetic energy of a rolling object is the sum of its linear kinetic energy and rotational kinetic energy. The linear kinetic energy (K_linear) is given by the equation $\frac{1}{2}\mathrm{mv}{v}^2$, where m is the mass and v is the linear velocity. The rotational kinetic energy (K_rotational) is given by the equation $\frac{1}{2}I{\omega }^2$, where I is the moment of inertia and ω is the angular velocity. Therefore, the total kinetic energy (K_total) is $K$_total=K_linear+K_rotational.

What is the relationship between linear velocity and angular velocity in rolling motion?

In rolling motion, the linear velocity (v_cm) of the center of mass and the angular velocity (ω) are related by the equation v_cm = Rω, where R is the radius of the rolling object. This relationship holds true only if the object is rolling without slipping on a surface. If the object is rolling in the air or slipping, this equation does not apply.

What is the moment of inertia for a solid sphere?

The moment of inertia (I) for a solid sphere is given by the equation $\frac{2}{5}m{R}^2$, where m is the mass of the sphere and R is its radius. This value is crucial for calculating the rotational kinetic energy of the sphere when it is rolling.

How do you calculate the rotational kinetic energy of a rolling object?

The rotational kinetic energy (K_rotational) of a rolling object is calculated using the equation $\frac{1}{2}I{\omega }^2$, where I is the moment of inertia and ω is the angular velocity. The moment of inertia depends on the shape and mass distribution of the object. For example, for a solid sphere, I = $\frac{2}{5}m{R}^2$.