204k views
0 votes
A bucket of water is swung in a circle in a vertical plane. The radius of the circle is 0.85 m. What is the minimum speed that will just prevent the water from falling out of the bucket when it is upside down (i.e., at the top of the circle)

User Roni
by
7.9k points

1 Answer

4 votes

Answer:2.88 m/s

Explanation:

Given

radius of circle r=0.85 m

At highest Point weight will Provide the centripetal Force

thus


weight=mg


Centripetal\ Force=(mv^2)/(r)

where m is the mass of water in bucket


mg=(mv^2)/(r)


v=√(gr)


v=√(9.8* 0.85)


v=2.88 m/s

User Iskuskov Alexander
by
9.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories