SATHEE: Motion in a Plane - Result Question 50

Motion in a Plane - Result Question 50

53. A particle moves along a circle of radius $(\frac{20}{π}) m$ with constant tangential acceleration. If the velocity of the particle is $80 m / s$ at the end of the second revolution after motion has begun, the tangential acceleration is [2003]

(a) $40 π m / s^{2}$

(b) $40 m / s^{2}$

(d) $160 π m / s^{2}$

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Answer:

Correct Answer: 53. (b)

Solution:

(b) Given, $r = \frac{20}{π} m$

$v_{j} = 80 m / s e c \Rightarrow w_{f} = \frac{8 θ π}{2 θ} \Rightarrow 4 π$

$θ = 2$ re $N = 4 π$ radian

From equation,

$w_{f}^{2} = w_{0}^{2} + 2 α θ [∵ w_{0} = 0]$

$(4 π)^{2} = 0 + 2 . α .4 π$

$α = 2 π$

tangential acceleration

$\begin{aligned} a_{t} = α \cdot r \\ a t = 2 π \cdot \frac{20}{π} = 40 m / s e c^{2} \end{aligned}$

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Motion in a Plane - Result Question 50

53. A particle moves along a circle of radius (20π)m with constant tangential acceleration. If the velocity of the particle is 80m/s at the end of the second revolution after motion has begun, the tangential acceleration is [2003]

Answer:

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53. A particle moves along a circle of radius $(\frac{20}{π}) m$ with constant tangential acceleration. If the velocity of the particle is $80 m / s$ at the end of the second revolution after motion has begun, the tangential acceleration is [2003]