
Problem 8:  A ball with an initial velocity of 7.2 m/s rolls up a hill without slipping.

Randomized Variables
v = 7.2 m/s

Part (a) Treating the ball as a spherical shell, calculate the vertical height it reaches, in meters.
Numeric   : A numeric value is expected and not an expression.
h1 = __________________________________________


Part (b) Repeat the calculation for the same ball if it slides up the hill without rolling.
Numeric   : A numeric value is expected and not an expression.
h2 = __________________________________________



To solve this problem, we’ll need to consider two cases separately:

1. **Rolling without slipping** (spherical shell) - energy conservation takes into account both translational and rotational kinetic energy.
2. **Sliding without rolling** - energy conservation only involves translational kinetic energy.

### Given Data
- Initial velocity, \( v = 7.2 \, \text{m/s} \)

### Part (a): Rolling without Slipping
For a ball rolling without slipping, the total mechanical energy is conserved. The initial kinetic energy includes both translational and rotational kinetic energy, which is converted into gravitational potential energy at the maximum height.

1. **Translational Kinetic Energy (TKE):**
   \[
   \text{TKE} = \frac{1}{2} m v^2
   \]

2. **Rotational Kinetic Energy (RKE):**
   Since the ball is a spherical shell, the moment of inertia \( I \) is given by:
   \[
   I = \frac{2}{3} m R^2
   \]
   And rotational kinetic energy is:
   \[
   \text{RKE} = \frac{1}{2} I \omega^2 = \frac{1}{2} \cdot \frac{2}{3} m R^2 \cdot \left( \frac{v}{R} \right)^2 = \frac{1}{3} m v^2
   \]

3. **Total Initial Kinetic Energy (TKE + RKE):**
   \[
   \text{Total Initial KE} = \frac{1}{2} m v^2 + \frac{1}{3} m v^2 = \frac{5}{6} m v^2
   \]

4. **Gravitational Potential Energy (GPE) at Maximum Height:**
   At the maximum height \( h \), all kinetic energy is converted into potential energy:
   \[
   mgh = \frac{5}{6} m v^2
   \]
   Simplifying and solving for \( h \):
   \[
   h_1 = \frac{5 v^2}{6g}
   \]

Substitute \( v = 7.2 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \):
   \[
   h_1 = \frac{5 \cdot (7.2)^2}{6 \cdot 9.8} \approx 4.42 \, \text{m}
   \]

### Part (b): Sliding without Rolling
In this case, there is only translational kinetic energy (no rotational kinetic energy). Therefore, all the initial translational kinetic energy is converted into gravitational potential energy.

1. **Translational Kinetic Energy (only):**
   \[
   \frac{1}{2} m v^2
   \]

2. **Gravitational Potential Energy at Maximum Height:**
   \[
   mgh = \frac{1}{2} m v^2
   \]
   Solving for \( h \):
   \[
   h_2 = \frac{v^2}{2g}
   \]

Substitute \( v = 7.2 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \):
   \[
   h_2 = \frac{(7.2)^2}{2 \cdot 9.8} \approx 2.65 \, \text{m}
   \]

### Final Answers
- **(a)** \( h_1 = 4.42 \, \text{m} \)
- **(b)** \( h_2 = 2.65 \, \text{m} \)

Would you like further details on any specific steps, or have additional questions?

#### Further Questions:
1. What is the effect of different moments of inertia on height?
2. How would the answer change if the object were a solid sphere?
3. What if the ball rolled on a surface with friction but still had rotational energy?
4. How does the incline angle affect the motion if known?
5. What are the differences in energy conservation for different shapes?

#### Tip:
When solving energy problems, always identify all forms of kinetic and potential energy relevant to the object's motion.

Problem 8: A ball with an initial velocity of 7.2 m/s rolls up a hill without slipping.

Part (a) Treating the ball as a spherical shell, calculate the vertical height it reaches, in meters.

Part (b) Repeat the calculation for the same ball if it slides up the hill without rolling.

Calculate the vertical height a ball reaches as it rolls up a hill with an initial velocity of 7.2 m/s. Explore energy conservation concepts, including translational and rotational kinetic energy, and understand the differences between rolling and sliding motion. Perfect for high school physics students in Grades 11-12.

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