SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 12

12. A particle starting from rest, moves in a circle of radius ’ $r$ ‘. It attains a velocity of $V_{0} m / s$ in the $n^{th}$ round. Its angular acceleration will be,

[NEET Odisha 2019]

(a) $\frac{V_{0}^{2}}{4 π n r} r a d / s^{2}$

(b) $\frac{V_{0}}{n} r a d / s^{2}$

(d) $\frac{V_{0}^{2}}{4 π n r^{2}} r a d / s^{2}$

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Answer:

Correct Answer: 12. (d)

Solution:

(d) $V^{2} = u^{2} + 2 a s$

$V_{0}^{2} = 0 + 2 r α (2 π r) n \Rightarrow α = \frac{V_{0}^{2}}{4 π n r^{2}}$

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System of Particles and Rotational Motion - Result Question 12

12. A particle starting from rest, moves in a circle of radius ’ r ‘. It attains a velocity of V0m/s in the nth round. Its angular acceleration will be,

Answer:

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12. A particle starting from rest, moves in a circle of radius ’ $r$ ‘. It attains a velocity of $V_{0} m / s$ in the $n^{th}$ round. Its angular acceleration will be,