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

To solve this problem, we apply the concept of centripetal force in circular motion. The tension in the string provides the necessary centripetal force to keep the bob in its circular path. The formula for the centripetal force \( F_c \) is given by:

\[ F_c = \frac{mv^2}{r} \]

 

Where:

• \( m \) is the mass of the bob.
• \( v \) is the tangential speed of the bob.
• \( r \) is the radius of the circular path.

Initially, the speed of the bob is \( \omega \) (converted to relevant units from rpm), and the tension \( T \) is the centripetal force:

\[ T = \frac{m(\omega)^2}{r} \]

 

When the speed of the bob increases to \( 2\omega \), the new tension \( T' \) can be expressed as:

\[ T' = \frac{m(2\omega)^2}{r} \]

 

Simplifying \( T' \):

\[ T' = \frac{m \cdot 4\omega^2}{r} \]

 

Notice that \(\frac{m \cdot \omega^2}{r}\) is equivalent to the initial tension \( T \). Thus:

\[ T' = 4\left(\frac{m \cdot \omega^2}{r}\right) = 4T \]

 

Therefore, when the speed becomes \( 2\omega \), maintaining the same radius, the tension in the string becomes 4T.

Answer 2

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.