KCET · Physics · Laws of Motion
One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed \(v\), the net force on the particle (directed towards the centre) is ( \(T\) is the tension in the string)
- A \(T\)
- B \(T-\frac{m v^{2}}{l}\)
- C \(T+\frac{m v^{2}}{l}\)
- D zero
Answer & Solution
Correct Answer
(A) \(T\)
Step-by-step Solution
Detailed explanation
Consider a string of length \(l\) connected to a particle as shown in figure
As the particle is moving with velocity \(v\) in uniform circular motion, so the net force must be equal to the centripetal force, which is provided by the tension in the string.
\(\therefore\) Net force \(=\) Centripetal force \(\left(F_{C}\right)\)
\(=\) Tension in string \((T)\)
\(\Rightarrow \quad \frac{m v^{2}}{l}=T\)
As the particle is moving with velocity \(v\) in uniform circular motion, so the net force must be equal to the centripetal force, which is provided by the tension in the string.
\(\therefore\) Net force \(=\) Centripetal force \(\left(F_{C}\right)\)
\(=\) Tension in string \((T)\)
\(\Rightarrow \quad \frac{m v^{2}}{l}=T\)
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