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 address this, we utilize the principle of centripetal force in circular motion. The force required to maintain the bob's circular trajectory is supplied by the tension in the string. The centripetal force, denoted as \( F_c \), is calculated using the formula:

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

Where:

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

At the initial state, the bob's speed is \( \omega \) (adjusted from rpm to appropriate units), and the tension \( T \) equals the centripetal force:

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

 

When the bob's speed is doubled to \( 2\omega \), the updated tension, \( T' \), is given by:

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

 

Simplifying the expression for \( T' \):

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

 

It is observed that the term \(\frac{m \cdot \omega^2}{r}\) is identical to the initial tension \( T \). Consequently:

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

 

Therefore, at a doubled speed of \( 2\omega \) and maintaining the same radius, the string tension increases to 4T.