Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)


a.) Determine the linear regression

b.) At this given angular velocity, what is the rotational kinetic energy?

Sagot :

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.