Question

A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes:

• A. T
• B. 4T
• C. \(\frac{T}{4}\)
• D. √2T

Answer 1

The Correct Option is B

When an object is whirled in a horizontal plane, the tension in the string provides the centripetal force required to keep the object in circular motion. The centripetal force is given by the formula:

                                                                              F = m ⋅ ω² ⋅ r

where m is the mass of the object, ω is the angular velocity, and r is the radius of the circular path.

Initially, the tension T in the string is equal to the centripetal force:

                                                                              T = m ⋅ ω² ⋅ r

When the speed becomes 2ω, the new tension T' in the string is:

                                                                               T' = m ⋅ (2ω)² ⋅ r

Simplifying, we get:

                                                                               T' = m ⋅ 4ω² ⋅ r

Therefore, the new tension T' is four times the initial tension T:

                                                                                     T' = 4T

Step-by-Step Solution:

Step 1: Identify the initial tension in the string

                                                                               T = m ⋅ ω² ⋅ r

Step 2: Determine the new angular velocity:

                                                                                       2ω

Step 3: Calculate the new tension in the string using the new angular velocity:

                                                                               T' = m ⋅ (2ω)² ⋅ r

Step 4: Simplify the expression to find the new tension:

                                                                                      T' = 4T

Final Answer:

The tension in the string becomes 4T.