A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant?
A) 9.5 N
B) 3.0 N
C) 8.1 N
D) 5.6 N
E) 4.7 N

Question

Quizplus · Accepted Answer

First, let's draw a free body diagram at the instant when the mass is at the 40 degree below the horizontal position. We have the tension T in the string, the gravitational force mg acting downwards, and the centripetal force Fc acting towards the center of the circle.

Since the mass is moving in a vertical circle, we know that the centripetal force is provided by the tension in the string:

Fc = T

To find the tension, we need to first find the centripetal force. We can use the circular motion equations:

Fc = mv^2/r

where m is the mass, v is the speed, and r is the radius of the circle.

Substituting in the values given, we have:

Fc = (0.40 kg)(5.0 m/s)^2 / 1.8m = 6.94 N

Therefore, the tension in the string is also 6.94 N. However, this is not the answer yet, because the question asks for the tension at the instant when the string is at an angle of 40 degrees below the horizontal.

We can use trigonometry to find the vertical component of the tension (since the tension is not purely horizontal or vertical).

T cos(40°) = Fc

T cos(40°) = 6.94 N

T = 6.94 N / cos(40°) = 8.11 N

So the magnitude of the tension in the string at this instant is 8.11 N, which is answer choice C.

A 040-Kg Mass Attached to the End of a String Swings