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Step-by-Step Solution
Step 1: Identify the Forms of Energy Involved
When the rolling cylinder compresses the spring, its initial kinetic energy (both translational and rotational) is completely converted into the elastic potential energy of the spring at maximum compression.
Step 2: Write the Expression for Initial Kinetic Energy
The cylinder has:
Translational kinetic energy: $ \\frac{1}{2} m v^2 $
Rotational kinetic energy: $ \\frac{1}{2} I \\omega^2 $
For a solid cylinder of mass $ m $ and radius $ R $, the moment of inertia about its central axis is
$ I = \\frac{1}{2} m R^2 $.
Since the cylinder rolls without slipping, $ v = R\\omega $, hence $ \\omega = \\frac{v}{R} $. Therefore, the rotational kinetic energy becomes:
$ \\frac{1}{2} \\left(\\frac{1}{2} m R^2\\right) \\left(\\frac{v}{R}\\right)^2 = \\frac{1}{4} m v^2 $.
Thus, the total initial kinetic energy is:
$ \\text{KE}_\\text{total} = \\frac{1}{2} m v^2 + \\frac{1}{4} m v^2 = \\frac{3}{4} m v^2.$
Step 3: Write the Expression for Spring Potential Energy
At maximum compression $ x $, all of the cylinder’s initial kinetic energy is converted to the elastic potential energy of the spring:
$ \\text{PE}_\\text{spring} = \\frac{1}{2} k x^2, $
where $ k $ is the spring constant.
Step 4: Equate the Energies and Solve for $ x $
Set the total kinetic energy equal to the spring’s potential energy:
$ \\frac{3}{4} m v^2 = \\frac{1}{2} k x^2. $
Substitute the given values: $ m = 3 \\text{ kg}, \\quad v = 4 \\text{ m/s}, \\quad k = 200 \\text{ N/m}.$
$ \\frac{3}{4} \\times 3 \\times 4^2 = \\frac{1}{2} \\times 200 \\times x^2 \\
\\implies \\frac{3}{4} \\times 3 \\times 16 = 100 \\times x^2 \\
\\implies 36 = 100 \\times x^2 \\
\\implies x^2 = 0.36 \\
\\implies x = 0.6 \\text{ m}.
$
Step 5: State the Final Answer
The maximum compression produced in the spring is $ 0.6 \\text{ m} $.
