"A grinding wheel with a moment of inertia of 2.0 kg-m2 has a 2.5 N-m torque applied to it. What is its final kinetic energy 10 seconds after starting from rest?"

Ek = 156.25 J

Explanation:

Given:-

- The moment of inertia of the grinding wheel I = 2.0 kg-m2

- The torque applied T = 2.5 N-m

Find:-

What is its final kinetic energy 10 seconds after starting from rest?"

Solution:-

- The relationship between applied torque and the angular acceleration α of the grinding wheel is can be expressed as:

                                       T = I*α

- Solve for α , using the given data:

                                       2.5 = 2.0*α

                                       α = 1.25 rad / s^2

- Assuming constant acceleration ( constant applied torque T ) we can determine the angular velocity of the grinding wheel after time t = 10 sec using rotational kinematic equation of motion.

                                      wf = wi + α*t

Where, wf = Final angular velocity (rad/s)

            wi = Initial angular velocity (rad/s)

- Assuming the grinding wheel was initially at rest wi = 0 rad /s. We have:

                                     wf = 0 + 1.25*10

                                     wf = 12.5 rad/s

- The kinetic energy (Ek) of the grinding wheel after 10 seconds can b determined in rotational kinematics terms as follows:

                                     Ek = 0.5*I*wf^2

                                     Ek = 0.5*2.0*12.5^2

                                    Ek = 156.25 J

- The kinetic energy of the grinding wheel after 10 seconds is Ek = 156.25 J

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