Question

problem 7: a torque of 2.5 nm is applied to a ring of mass 200 g and radius 12 cm. first find the moment of inertia of the ring and then find the angular acceleration of the ring.

Explanation:

Step1: Calculate moment of inertia

Moment of inertia formula for a ring: \(I = m r^{2}\).
Given \(m=200\ g = 0.2\ kg\), \(r = 12\ cm=0.12\ m\).
\(I=0.2\times(0.12)^{2}\)
\(I = 0.2\times0.0144=0.00288\ kg\cdot m^{2}\)

Step2: Calculate angular acceleration

Torque formula \(\tau=I\alpha\), so \(\alpha=\frac{\tau}{I}\).
Given \(\tau = 2.5\ N\cdot m\), \(I = 0.00288\ kg\cdot m^{2}\).
\(\alpha=\frac{2.5}{0.00288}\approx868\ rad/s^{2}\)

Answer:

Moment of inertia \(I = 0.00288\ kg\cdot m^{2}\), angular acceleration \(\alpha\approx868\ rad/s^{2}\)