QuestionMay 16, 2025

A metal ring (mass 0.37 kg and radius 5.8 cm) rolls without slipping down a 3 meter ramp that has an angle of 40^circ What will its velocity be at the bottom of the ramp?

A metal ring (mass 0.37 kg and radius 5.8 cm) rolls without slipping down a 3 meter ramp that has an angle of 40^circ What will its velocity be at the bottom of the ramp?
A metal ring (mass 0.37 kg and radius 5.8 cm) rolls without slipping down a 3 meter ramp that has an angle of 40^circ What will its velocity be at the bottom of the ramp?

Solution
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Answer

v \approx 5.42 \, \text{m/s} Explanation 1. Calculate gravitational potential energy at the top PE = mgh, where m = 0.37 \, \text{kg}, g = 9.8 \, \text{m/s}^2, h = 3 \sin(40^\circ). 2. Calculate kinetic energy at the bottom Total energy is conserved: PE_{\text{top}} = KE_{\text{bottom}}. KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2. 3. Relate linear and angular velocity For rolling without slipping, v = r\omega. Use I = mr^2 for a ring. 4. Solve for velocity Substitute I and \omega into KE equation and solve for v.

Explanation

1. Calculate gravitational potential energy at the top<br /> $PE = mgh$, where $m = 0.37 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $h = 3 \sin(40^\circ)$.<br />2. Calculate kinetic energy at the bottom<br /> Total energy is conserved: $PE_{\text{top}} = KE_{\text{bottom}}$. $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.<br />3. Relate linear and angular velocity<br /> For rolling without slipping, $v = r\omega$. Use $I = mr^2$ for a ring.<br />4. Solve for velocity<br /> Substitute $I$ and $\omega$ into $KE$ equation and solve for $v$.
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