3, 4, 5 ### Rotational Dynamics Problem SetA solid disk with a mass of 7.0 kg and a radius of 30 cm gradually increases its rotational speed from 45 revolutions per minute (rpm) to 120 rpm over a period of 5.0 seconds. Analyze the following questions based on this scenario:3. **Find the angular acceleration in rad/s².**4. **Calculate the number of revolutions the disk completes in these 5.0 seconds.**5. **Determine the change in rotational kinetic energy.**6. **Calculate the moment of inertia of the disk in kg·m².**7. **Find the magnitude of torque needed to achieve the change in rpm.**8. **Calculate the tangential acceleration of a point \( P \) on the outer rim of the disk.**#### Diagram ExplanationThe accompanying diagram illustrates a solid disk. The disk has a radius \( R \) and rotates about its center. It shows the angular velocity \( \omega \) and angular acceleration \( \alpha \). Point \( P \) is marked on the outer rim, indicating the location where tangential acceleration needs to be calculated.

Name: 3, 4, 5 ### Rotational Dynamics Problem SetA solid disk with a mass of
Uploaded: 2024-03-20T06:58:04.000Z
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Description: 3, 4, 5 ### Rotational Dynamics Problem SetA solid disk with a mass of 7.0 kg and a radius of 30 cm gradually increases its rotational speed from 45 revolutions per minute (rpm) to 120 rpm over a period of 5.0 seconds. Analyze the following questions

Given: m = 7 kg r = 30 cm = 0.30 m w0 = 45 rpm =45* 2pi/60 rad/s w = 120 rpm = 120*2pi/60 rad/s t = 5 s (3) alpha= (w-w0)/t = ((120- 45)*2pi/60)/5 = 1.57 rad/s^2 (4) th = w0t + 1/2*alpha*t^2 = (45*2pi/60)*5 + (1/2*1.57*5^2) = 43.2 rad = 43.2/2pi rev = 6.87 rev (5) I = 1/2*mr^2 =1/2*7*0.30^2 = 0.315 kg-m^2 dKE = 1/2*I*(w^2 - w0^2) = 1/2*0.315*((120*2pi/60)^2 - (45*2pi/60)^2) = 21.4 J