Rotational Dynamics and Energy – Physics Exam Study Guidance

Q1. A uniform cylinder has mass M = 4.0 kg, radius R = 0.060 m, and moment of inertia for rotation about an axis through its center given by $I = \frac{1}{2}MR^2$. The cylinder rolls without slipping up an incline. The translational speed of the cylinder at the bottom of the incline is 4.00 m/s. What maximum height above its initial position does the cylinder reach before starting to slide back down?

Background

Topic: Conservation of Energy in Rotational Motion

This question tests your understanding of how to apply conservation of mechanical energy to a rolling object, accounting for both translational and rotational kinetic energy as it moves up an incline.

Key Terms and Formulas

• Translational Kinetic Energy: $K_{trans} = \frac{1}{2}Mv^2$

• Rotational Kinetic Energy: $K_{rot} = \frac{1}{2}I\omega^2$

• Moment of Inertia for Cylinder: $I = \frac{1}{2}MR^2$

• Relationship for Rolling Without Slipping: $v = \omega R$

• Potential Energy at Height h: $U = Mgh$

Step-by-Step Guidance

1. Write the conservation of energy equation, equating the total kinetic energy at the bottom to the potential energy at the maximum height:

   $K_{trans} + K_{rot} = U$

   $\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = Mgh$

2. Substitute the moment of inertia for a cylinder and the rolling condition $v = \omega R$ into the equation:

   $I = \frac{1}{2}MR^2$ and $\omega = \frac{v}{R}$

   So, $K_{rot} = \frac{1}{2} \left(\frac{1}{2}MR^2\right) \left(\frac{v}{R}\right)^2$

3. Simplify the rotational kinetic energy term and combine it with the translational kinetic energy:

   $K_{rot} = \frac{1}{4}Mv^2$

   Total kinetic energy: $\frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$

4. Set the total kinetic energy equal to the gravitational potential energy at height $h$ and solve for $h$:

   $\frac{3}{4}Mv^2 = Mgh$

   Divide both sides by $Mg$ to isolate $h$.

Try solving on your own before revealing the answer!

Final Answer: 0.61 m

$h = \frac{3v^2}{4g}$

Plugging in $v = 4.00$ m/s and $g = 9.8$ m/s$^2$ gives $h \approx 0.61$ m. The energy is shared between translational and rotational forms, so the cylinder does not reach as high as if it were sliding without rotation.

Q2. A uniform sphere has mass = 2.0 kg, radius = 0.060 m, and moment of inertia for rotation about an axis through its center given by $I = \frac{2}{5}MR^2$. If it is rolling without slipping on a horizontal surface, and the translational speed of the sphere is 2.00 m/s, what is its total kinetic energy?

Background

Topic: Rotational and Translational Kinetic Energy

This question tests your ability to calculate the total kinetic energy of a rolling object, which includes both translational and rotational components.

Key Terms and Formulas

• Translational Kinetic Energy: $K_{trans} = \frac{1}{2}Mv^2$

• Rotational Kinetic Energy: $K_{rot} = \frac{1}{2}I\omega^2$

• Moment of Inertia for Sphere: $I = \frac{2}{5}MR^2$

• Rolling Without Slipping: $v = \omega R$

Step-by-Step Guidance

1. Calculate the translational kinetic energy using the given mass and speed:

   $K_{trans} = \frac{1}{2}Mv^2$

2. Express the rotational kinetic energy in terms of $v$ using $I$ and $\omega = \frac{v}{R}$:

   $K_{rot} = \frac{1}{2} \left(\frac{2}{5}MR^2\right) \left(\frac{v}{R}\right)^2$

3. Simplify the rotational kinetic energy expression:

   $K_{rot} = \frac{1}{5}Mv^2$

4. Add the translational and rotational kinetic energies to get the total kinetic energy:

   $K_{total} = K_{trans} + K_{rot}$

Try solving on your own before revealing the answer!

Final Answer: 4.0 J

$K_{total} = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$

Plugging in $M = 2.0$ kg and $v = 2.00$ m/s gives $K_{total} = 4.0$ J. Both translational and rotational energies contribute to the total.