physics question

Question

In an effort to rev up her class, Ms. Player does a demonstration with a bucket of water that has a total mass of 1.8 kg. Ms. Player whirls the bucket in a vertical circle, of radius 1.3 m, such that it has a speed of 3.9 m/s at the top of the loop and 6.4 m/s at the bottom of the loop.

Determine the tension force in Ms. Player's arm for the two locations.

Answer 1

At the top of the circle the equation to use is mv_a²/r =mg +T_a where T_a is the tension in the teacher's arm at the high point of the trajectory. Using the information in the problem 1.8(3.9)²/1.3 N =1.8(9.8) N + T_a. So 21.06N =17.64N + T_a and T_a = 3.42N. When the bucket is at the lowest point of the trajectory then T_b -mg = mv_b²/r where T_b is the tension in the teacher's arm at the lowest point of the trajectory. Using the information in the problem T_b = 17.64N +1.8(6.4)²/1.3 N. So T_b = 17.64N + 56.714N= 74.35N.

Answer 2

The centripetal force is .

1. At the top of the loop, the relationship between the the tension T, the bucket weight mg and the centripetal force is:

, so the tension force is .

1. At the bottom of the loop, the relationship between the the tension T, the bucket weight mg and the centripetal force is:

, so the tension force is .

Apply the values at each position into the equations above, and you will get your answer.