A coin of mass 10 g rolls along a horizontal table with a velocity of 6 cm/s. Its total kinetic energy is

Options

A. `9muJ`
B. `18muJ`
C. `27muJ`
D. `36muJ`

Explanation

To find the total kinetic energy of a coin rolling along a horizontal table, we need to consider both its translational and rotational kinetic energy. Here’s a step-by-step solution:

### Step 1: Identify the mass and velocity of the coin
- The mass \( m \) of the coin is given as 10 g, which we need to convert to kilograms for standard SI units:
\[ m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \]
- The velocity \( v \) of the coin is given as 6 cm/s, which we also convert to meters per second:
\[ v = 6 \, \text{cm/s} = 6 \times 10^{-2} \, \text{m/s} = 0.06 \, \text{m/s} \]

### Step 2: Calculate the translational kinetic energy
The translational kinetic energy \( KE_{trans} \) is given by the formula:
\[ KE_{trans} = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE_{trans} = \frac{1}{2} \times 0.01 \, \text{kg} \times (0.06 \, \text{m/s})^2 \]
Calculating \( (0.06)^2 \):
\[ (0.06)^2 = 0.0036 \]
Now substituting this back into the equation:
\[ KE_{trans} = \frac{1}{2} \times 0.01 \times 0.0036 = 0.000018 \, \text{J} = 18 \, \mu J \]

### Step 3: Calculate the rotational kinetic energy
For a rolling object, the rotational kinetic energy \( KE_{rot} \) can be expressed as:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
Where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid disk (like a coin), the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]
Since the coin is rolling, we can relate \( \omega \) and \( v \) using the equation \( v = \omega r \), which gives:
\[ \omega = \frac{v}{r} \]
Substituting \( \omega \) into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE_{rot} = \frac{1}{4} m v^2 \]

### Step 4: Combine both kinetic energies
Now we can find the total kinetic energy \( KE_{total} \):
\[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 \]
Factoring out \( m v^2 \):
\[ KE_{total} = m v^2 \left(\frac{1}{2} + \frac{1}{4}\right) = m v^2 \left(\frac{2}{4} + \frac{1}{4}\right) = m v^2 \left(\frac{3}{4}\right) \]
Substituting the values:
\[ KE_{total} = 0.01 \, \text{kg} \times (0.06 \, \text{m/s})^2 \times \frac{3}{4} \]
Calculating \( m v^2 \):
\[ m v^2 = 0.01 \times 0.0036 = 0.000036 \, \text{J} \]
Now substituting this into the equation:
\[ KE_{total} = 0.000036 \times \frac{3}{4} = 0.000027 \, \text{J} = 27 \, \mu J \]

### Final Answer
The total kinetic energy of the coin is:
\[ \boxed{27 \, \mu J} \]