Physics 111 Exam 3 Study Guidance: Impulse, Collisions, Rotational Motion, Center of Mass, and Moment of Inertia

Q7. A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0° slope. Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping.

Background

Topic: Rotational Dynamics & Rolling Motion

This question tests your understanding of rolling motion, the forces involved, and how to calculate acceleration and friction for a rolling object.

Key Terms and Formulas:

• Moment of inertia for a hollow sphere: $I = \frac{2}{3}MR^2$

• Newton's second law for translation: $F = ma$

• Newton's second law for rotation: $\tau = I\alpha$

• Relationship between linear and angular acceleration for rolling without slipping: $a = \alpha R$

• Static friction force: $f_s$

• Minimum coefficient of static friction: $\mu_s = \frac{f_s}{N}$

Step-by-Step Guidance

1. Draw a free-body diagram for the sphere, showing gravity, normal force, and friction.

2. Write Newton's second law for the forces parallel to the slope: $Mg\sin\theta - f_s = Ma$

3. Write the rotational equation: $f_s R = I \alpha$ and use $a = \alpha R$ to relate linear and angular acceleration.

4. Substitute $I$ for a hollow sphere and solve for $a$ in terms of $g$ and $\theta$.

5. Set up the equations to solve for $f_s$ and $\mu_s$ using the values found above.

Try solving on your own before revealing the answer!

Final Answer:

Acceleration: $a = 3.62\ \text{m/s}^2$ Friction force: $f_s = 4.83\ \text{N}$ Minimum coefficient of static friction: $\mu_s = 0.313$

These results are found by applying Newton's laws and the rolling condition, and substituting the values for mass, angle, and moment of inertia.

Q14. A system consists of three particles with masses m1 = m2 = 1.0 kg and m3 = 2.0 kg. located as shown in Figure. Find the coordinates of the center of mass of the system.

Background

Topic: Center of Mass

This question tests your ability to calculate the center of mass for a system of particles using their masses and positions.

Key Terms and Formulas:

• Center of mass coordinates: $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$, $y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$

• Each mass's position is given in the diagram.

Step-by-Step Guidance

1. Identify the positions $(x_i, y_i)$ for each mass from the diagram.

2. Write the formula for $x_{cm}$ and $y_{cm}$ using the masses and their coordinates.

3. Plug in the values for $m_1$, $m_2$, $m_3$ and their respective positions.

4. Calculate the numerator and denominator for each coordinate.

Try solving on your own before revealing the answer!

Final Answer:

$x_{cm} = 0.75\ \text{m}$, $y_{cm} = 1.0\ \text{m}$

The center of mass is calculated by weighting each position by its mass and dividing by the total mass.

Q16. In order to find the speed of a fast bullet, it is fired into a 4.0 kg wooden block on a horizontal surface. The bullet gets stuck in the block which starts moving with the speed of Vf=2.5 m/s. Find the original speed of the bullet Vi if its mass is 7.54 gram.

Background

Topic: Conservation of Momentum (Inelastic Collision)

This question tests your understanding of momentum conservation in a perfectly inelastic collision, where the bullet embeds in the block.

Key Terms and Formulas:

• Conservation of momentum: $m_b v_i + m_w v_{w,i} = (m_b + m_w) v_f$

• Initial velocity of block $v_{w,i} = 0$

• Final velocity $v_f$ is given

Step-by-Step Guidance

1. Write the conservation of momentum equation for the collision.

2. Plug in the values for the masses and final velocity.

3. Rearrange the equation to solve for the initial velocity of the bullet $v_i$.

4. Convert the bullet mass to kilograms before substituting.

Try solving on your own before revealing the answer!

Final Answer:

$v_i = 1330\ \text{m/s}$

By applying conservation of momentum and substituting the values, you find the bullet's original speed.

Q20. A part of a mechanical linkage has a mass of 3.6 kg. Its moment of inertia IP about an axis 0.15 m from its center of mass is IP = 0.132 kg m2. What is the moment of inertia Icm about a parallel axis through the center of mass?

Background

Topic: Rotational Inertia & Parallel Axis Theorem

This question tests your ability to use the parallel axis theorem to relate the moment of inertia about two parallel axes.

Key Terms and Formulas:

• Parallel axis theorem: $I_P = I_{cm} + M d^2$

• $I_P$ = moment of inertia about axis P

• $I_{cm}$ = moment of inertia about center of mass

• $M$ = mass, $d$ = distance between axes

Step-by-Step Guidance

1. Write the parallel axis theorem equation.

2. Plug in the values for $I_P$, $M$, and $d$.

3. Rearrange the equation to solve for $I_{cm}$.

4. Calculate $M d^2$ and subtract from $I_P$ to find $I_{cm}$.

Try solving on your own before revealing the answer!

Final Answer:

$I_{cm} = 0.051\ \text{kg}\cdot\text{m}^2$

Using the parallel axis theorem, you subtract $M d^2$ from $I_P$ to find the moment of inertia about the center of mass.