Question

An 8.0-cm-diameter, 400 g solid sphere is released from rest at the top of a 2.1-m-long, 25 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?

Accepted Answer

Step 1: Begin by identifying the physical principles involved. The sphere rolls without slipping, so both translational and rotational kinetic energy are present. Use the conservation of energy principle to analyze the motion, as the sphere's potential energy at the top is converted into kinetic energy at the bottom. Step 2: Write the expression for the total kinetic energy at the bottom of the incline. It consists of translational kinetic energy $$ K_{trans} = \frac{1}{2} m v^2 $$ and rotational kinetic energy $$ K_{rot} = \frac{1}{2} I \omega^2 $$, where $$ I  is $$the moment of inertia and $$ \omega  is $$the angular velocity. Step 3: Determine the moment of inertia $$ I $$ for a solid sphere. The formula for the moment of inertia of a solid sphere about its center is $$ I = \frac{2}{5} m r^2 $$, where $$ r  is $$the radius of the sphere. Convert the diameter to radius: $$ r = \frac{8.0 	ext{ cm}}{2} = 4.0 	ext{ cm} = 0.04 	ext{ m} . $$Step 4: Relate the angular velocity $$ \omega  to $$the linear velocity $$ v $$ using the rolling without slipping condition: $$ v = r \omega . $$Substitute $$ \omega = \frac{v}{r} $$ into the rotational kinetic energy expression $$ K_{rot} = \frac{1}{2} I \omega^2 . $$This gives $$ K_{rot} = \frac{1}{2} \left( \frac{2}{5} m r^2 ight) \left( \frac{v^2}{r^2} ight) = \frac{1}{5} m v^2 . $$Step 5: Calculate the fraction of kinetic energy that is rotational. The total kinetic energy is $$ K_{total} = K_{trans} + K_{rot} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 . $$The fraction of rotational kinetic energy is $$ \frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5} m v^2}{\frac{1}{2} m v^2 + \frac{1}{5} m v^2} . $$Simplify this expression to find the fraction.

An 8.0-cm-diameter, 400 g solid sphere is released from rest at the top of a 2.1-m-long, 25 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?

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