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04 Rolling Down a Wide Gutter

Aim

To show the effect of rolling radius on the amount of rotational kinetic energy.

Subjects

Diagram

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Figure 1:.

Equipment

Presentation

Both gutters are inclined about 77^{\circ}. The balls start rolling down at the same time. The ball that rolls down the wider gutter takes much more time to reach the horizontal plane as the one rolling down the narrow gutter.

Explanation

One ball (‘A’) rolls with a larger radius than the other (‘B’). So ‘A’ makes less rotations along the ramp than ‘B’. That’s why ball ‘A’ needs a smaller part of the available potential energy for its rotation to reach the end of the track and so more energy is available for its translation.

Epot =Etransl +Erot E_{\text {pot }}=E_{\text {transl }}+E_{\text {rot }}

Etransl =1/2mv2;Erot =1/2Iω2E_{\text {transl }}=1 / 2 m v^{2} ; E_{\text {rot }}=1 / 2 I \omega^{2}, with I=2/3mR2I=2 / 3 m R^{2} (thin-walled hollow sphere).

When a ball rolls down the gutter, then v=ωrv=\omega r ( rr being the radius of rotation) and we find: Etransl /Erot =(3/2)(I2/R2)E_{\text {transl }} / E_{\text {rot }}=(3 / 2)\left(I^{2} / R^{2}\right).

When the gutter is very narrow, r=Rr=R and Etransl /Erot =3/2E_{\text {transl }} / E_{\text {rot }}=3 / 2. So 60%60 \% of the pot. energy is transformed in translation of the ball and 40%40 \% in rotation.

Our wider gutter has dimensions such that r=1/2Rr=1 / 2 R, and so Etransl /Erot =3/8E_{\text {transl }} / E_{\text {rot }}=3 / 8. Now 27%27 \% of the pot. energy is transformed into translation of the ball and 73%73 \% in rotation.

Comparing these two rolling balls, ball ‘AA’ obtains 2.2 times as much energy for its translation as ball ‘B’ does. This means that ball ‘AA’ has at the end of the gutter a transl. speed almost 50%50 \% higher than ball ’ BB(2.21/2=1.48)\left(2.2^{1 / 2}=1.48\right). Then the time ball ’ BB ’ needs to travel along the gutter will also be 50%50 \% higher.

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