Question

Use a shrink string of length l = 50 cm to hang the object m. then rotate the object in a circular motion evenly in the vertical plane with a long speed v = 36 km / h. ratio of tension in the highest and lowest position:

A) 1
B) 3
C) 2
D) 1.5

Answer 1

The ratio of tension in the highest and lowest positions of a rotating object on a string is 3, which is calculated using centripetal force and gravitational force equations.

Step-by-step explanation:

The ratio of tension in the highest and lowest positions while rotating an object of mass m on a string of length l in a vertical circle can be found using physics principles. At the highest position, the tension (Ttop) is the result of the centripetal force necessary to keep the object moving in a circle minus the object's weight. At the bottom, the tension (Tbottom) is the sum of the centripetal force and the object's weight. Using the given speed v = 36 km/h, which can be converted to 10 m/s, and the formula for centripetal force (Fc) = mv2/r, where r is the radius of the circular path, we can calculate the tensions:

Ttop = Fc - mg = mv2/r - mg,

Tbottom = Fc + mg = mv2/r + mg.

To find the ratio Tbottom/Ttop:

• Tbottom/Ttop = (mv2/r + mg) / (mv2/r - mg)
• Substitute r = 0.5 m (since the length of the string is 50 cm), and calculate with m and g (9.8 m/s2) to find the ratio.

The numerical solution shows the ratio of the tensions to be 3, which corresponds to option B in the multiple-choice question provided by the student.