A hoop of radius `2 m` weight `100 kg`.It rolls along a horizontal floor so that its centre of mass has a speed of `20 cm s^-1`. How much work has to be done to stop it ?

Options

A. 2 J
B. 4 J
C. 6 J
D. 8 J

Explanation

To solve the problem of how much work has to be done to stop a hoop rolling along a horizontal floor, we can follow these steps:

### Step 1: Identify the given data
- Radius of the hoop, \( R = 2 \, \text{m} \)
- Weight of the hoop (mass), \( m = 100 \, \text{kg} \)
- Speed of the center of mass, \( V = 20 \, \text{cm/s} = 0.2 \, \text{m/s} \)

### Step 2: Calculate the angular speed (\( \omega \))
For a rolling object, the relationship between linear speed \( V \) and angular speed \( \omega \) is given by:
\[ \omega = \frac{V}{R} \]
Substituting the values:
\[ \omega = \frac{0.2 \, \text{m/s}}{2 \, \text{m}} = 0.1 \, \text{rad/s} \]

### Step 3: Calculate the kinetic energy
The total kinetic energy (\( KE \)) of the hoop consists of translational kinetic energy and rotational kinetic energy. The formulas for these are:
- Translational kinetic energy:
\[ KE_{\text{trans}} = \frac{1}{2} m V^2 \]
- Rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
Where \( I \) is the moment of inertia of the hoop. For a hoop, the moment of inertia is:
\[ I = mR^2 \]
Substituting this into the rotational kinetic energy formula gives:
\[ KE_{\text{rot}} = \frac{1}{2} (mR^2) \omega^2 \]

### Step 4: Substitute the values into the kinetic energy formulas
First, calculate the translational kinetic energy:
\[ KE_{\text{trans}} = \frac{1}{2} \times 100 \, \text{kg} \times (0.2 \, \text{m/s})^2 = \frac{1}{2} \times 100 \times 0.04 = 2 \, \text{J} \]

Next, calculate the rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2} \times (100 \, \text{kg} \times (2 \, \text{m})^2) \times (0.1 \, \text{rad/s})^2 \]
Calculating \( I \):
\[ I = 100 \times 4 = 400 \, \text{kg m}^2 \]
Now substituting into the rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2} \times 400 \times (0.1)^2 = \frac{1}{2} \times 400 \times 0.01 = 2 \, \text{J} \]

### Step 5: Calculate the total kinetic energy
Now, we can find the total kinetic energy:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 2 \, \text{J} + 2 \, \text{J} = 4 \, \text{J} \]

### Step 6: Determine the work done to stop the hoop
According to the work-energy theorem, the work done to stop the hoop is equal to the total kinetic energy:
\[ \text{Work done} = KE_{\text{total}} = 4 \, \text{J} \]

### Final Answer
The work that has to be done to stop the hoop is \( 4 \, \text{J} \).

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