JEE MainPhysicsModerateCircular motion

A particle of mass $m = \frac{5}{42}kg$ is tied to an end of an
inextensible string and is whirled in a vertical circle as shown below.
Then, the difference in tension of string at A and B i.e.
$T_{A} - T_{B}$ in Newton is: (particle completes the circle)

Show answer and solution

\Delta T = \frac{3mg\Delta h}{I}

h_{B} - h_{A} = I(\frac{3}{5} + \frac{4}{5}) = \frac{7I}{5}

T_{A} - B_{B} = \frac{3mg}{I}(h_{B} - h_{A})

= \frac{3mg}{I} \times \frac{7I}{5}

= \frac{21mg}{5} = \frac{21}{5} \times \frac{5}{42} \times 10

= 5N

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.