Test 4 Practice

A hollow sphere of mass M = 1.5 kg and radius R = 0.2 m is rolling without slipping on a flat horizontal surface with translational velocity
v
0
=
2.8
m/s when it encounters a hill. The moment of inertia of the hollow sphere is
I
=
2
3
M
R
2
.

Assuming the hollow sphere rolls without slipping, what is the maximum height
H
it will roll up the hill?

(a)
0.4 m

(b)
1.07 m

(c)
0.27 m

(d)
0.67 m

(e)
0.8 m

To solve for the maximum height \( H \) that the hollow sphere will roll up the hill, we can use the principle of energy conservation. The initial kinetic energy of the sphere (which includes both translational and rotational components) will be completely converted into gravitational potential energy at the maximum height.

### Step 1: Write Down the Known Values
- Mass, \( M = 1.5 \ \text{kg} \)
- Radius, \( R = 0.2 \ \text{m} \)
- Translational velocity, \( v_0 = 2.8 \ \text{m/s} \)
- Moment of inertia for a hollow sphere, \( I = \frac{2}{3} M R^2 \)

### Step 2: Initial Energy (Kinetic Energy)

The initial kinetic energy \( K \) consists of both translational and rotational components:
\[
K = K_{\text{translational}} + K_{\text{rotational}}
\]

1. **Translational kinetic energy**:
   \[
   K_{\text{translational}} = \frac{1}{2} M v_0^2
   \]

2. **Rotational kinetic energy** (using \( \omega = \frac{v_0}{R} \) for rolling without slipping):
   \[
   K_{\text{rotational}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{2}{3} M R^2 \right) \left( \frac{v_0}{R} \right)^2
   \]

### Step 3: Substitute and Simplify

Substitute \( I = \frac{2}{3} M R^2 \) and \( \omega = \frac{v_0}{R} \):
\[
K_{\text{rotational}} = \frac{1}{2} \cdot \frac{2}{3} M R^2 \cdot \frac{v_0^2}{R^2} = \frac{1}{3} M v_0^2
\]

Thus, the total initial kinetic energy \( K \) is:
\[
K = K_{\text{translational}} + K_{\text{rotational}} = \frac{1}{2} M v_0^2 + \frac{1}{3} M v_0^2 = \frac{5}{6} M v_0^2
\]

### Step 4: Set Up Energy Conservation

At the maximum height \( H \), all kinetic energy is converted into gravitational potential energy \( U = M g H \):
\[
\frac{5}{6} M v_0^2 = M g H
\]

Cancel \( M \) from both sides:
\[
\frac{5}{6} v_0^2 = g H
\]

Solve for \( H \):
\[
H = \frac{5 v_0^2}{6 g}
\]

### Step 5: Plug in the Values

Using \( v_0 = 2.8 \ \text{m/s} \) and \( g = 9.8 \ \text{m/s}^2 \):
\[
H = \frac{5 \cdot (2.8)^2}{6 \cdot 9.8} = \frac{5 \cdot 7.84}{58.8} \approx 0.67 \ \text{m}
\]

### Answer

The correct answer is:
**(d) 0.67 m**

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Would you like further details or have questions about this solution?

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**Related Questions:**

1. How would the answer change if the sphere were solid instead of hollow?
2. What would happen if the sphere did not roll without slipping?
3. How does the radius affect the maximum height if all other factors remain the same?
4. Can we use this method for objects with different shapes (e.g., a cylinder)?
5. Why does the mass \( M \) cancel out in the energy conservation equation?

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**Tip:** When dealing with rolling objects, always account for both translational and rotational kinetic energy.

A hollow sphere of mass M = 1.5 kg and radius R = 0.2 m is rolling without slipping on a flat horizontal surface with translational velocity v0 = 2.8 m/s when it encounters a hill. The moment of inertia of the hollow sphere is I = 2/3 M R^2. Assuming the hollow sphere rolls without slipping, what is the maximum height H it will roll up the hill? (a) 0.4 m (b) 1.07 m (c) 0.27 m (d) 0.67 m (e) 0.8 m

Solve for the maximum height a hollow sphere of mass 1.5 kg and radius 0.2 m rolls up a hill, given its initial velocity of 2.8 m/s. Explore energy conservation principles and rotational motion with a detailed step-by-step solution. Suitable for high school students studying energy and motion.

Math Problem Statement

A hollow sphere of mass M = 1.5 kg and radius R = 0.2 m is rolling without slipping on a flat horizontal surface with translational velocity v 0

2.8 m/s when it encounters a hill. The moment of inertia of the hollow sphere is I

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