A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string.If the angular speed at the bottom is 8.00 rad/s,what is the tension in the string when the object is at the bottom of the circle?
A) 5.60 N
B) 10.5 N
C) 16.7 N
D) 19.6 N

Question

Quizplus · Accepted Answer

At the bottom of the circle, the tension in the string would be equal to the weight of the object plus the centripetal force.

The weight of the object can be calculated by multiplying its mass by the acceleration due to gravity:
Weight = 0.400 kg x 9.81 m/s^2 = 3.924 N

The centripetal force can be calculated using the formula:
Centripetal force = mass x (angular speed)^2 x radius

Radius = length of the string = 0.500 m

Centripetal force = 0.400 kg x (8.00 rad/s)^2 x 0.500 m = 12.8 N

Therefore, the tension in the string at the bottom of the circle would be:
Tension = Weight + Centripetal force = 3.924 N + 12.8 N = 16.7 N

Hence, the answer is option C (16.7 N).

A 0400-Kg Object Is Swung in a Circular Path and in in a Vertical