A solid cylinder with a mass m and radius r is mounted so that it can be rotated about an axis that passes through the center of both ends. At what angular speed, `omega` must the cylinder rotate to have the same total kinetic energy that it would have if it were moving horizontally with a speed v without rotation?

Options

A. `omega=v/r`
B. `omega=v/(2r)`
C. `omega=v^(2)/r^(2)`
D. `omega=v/rsqrt2`

Answer

To find the angular speed \( \omega \) at which a solid cylinder must rotate to have the same total kinetic energy as when it is moving horizontally with speed \( v \), we can follow these steps:

### Step 1: Calculate the translational kinetic energy of the cylinder
When the cylinder is moving horizontally with speed \( v \), its translational kinetic energy \( KE_{trans} \) is given by the formula:
\[ KE_{trans} = \frac{1}{2} mv^2 \]

### Step 2: Calculate the rotational kinetic energy of the cylinder
When the cylinder is rotating about its axis, its rotational kinetic energy \( KE_{rot} \) is given by the formula:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the cylinder. For a solid cylinder rotating about its central axis, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} mr^2 \]

### Step 3: Set the translational and rotational kinetic energies equal
To find the angular speed \( \omega \) that makes the rotational kinetic energy equal to the translational kinetic energy, we set:
\[ \frac{1}{2} mv^2 = \frac{1}{2} I \omega^2 \]
Substituting the expression for \( I \):
\[ \frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{1}{2} mr^2\right) \omega^2 \]

### Step 4: Simplify the equation
We can cancel \( \frac{1}{2} \) from both sides:
\[ mv^2 = \frac{1}{2} mr^2 \omega^2 \]
Next, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ v^2 = \frac{1}{2} r^2 \omega^2 \]

### Step 5: Solve for \( \omega \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{2v^2}{r^2} \]
Taking the square root of both sides gives:
\[ \omega = \sqrt{\frac{2v^2}{r^2}} = \frac{\sqrt{2}v}{r} \]

### Final Result
Thus, the angular speed \( \omega \) at which the cylinder must rotate is:
\[ \omega = \frac{\sqrt{2}v}{r} \]

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