Question

A particle moves along a circle of radius {20/(22/7)} m with constant tangential acceleration. If the speed of the particle is 80m/sec at the end of the second revolution after motion has begun, the tangential acceleration is
(1)160(22/7)
(2)40(22/7)
(3)40
(4)640(22/7)
All are in m/sec^2

Answer 1

Answer on Question#50070 – Physics – Mechanics | Kinematics | Dynamics

(3) 40 ms⁻²

Solution

Equations:

$$C = 2 \pi R$$ $$v = v _ {0} + a t$$ $$S = v _ {0} t + \frac {a t ^ {2}}{2}$$

, where $v$ – tangential velocity, $v_{0}$ – initial tangential velocity, $R$ – radius of circle, $a$ – tangential acceleration, $S$ – length of path.

Assume initial tangential velocity $v_{0} = 0$.

Two rotations $\equiv$ ($S_{2} = 2C = 4\pi R$)

$$v = a t$$ $$S = \frac {a t ^ {2}}{2}$$

We know $v_{2}$ and $R$ from task, $\pi \approx \frac{22}{7}$.

$$t _ {2} = \frac {v _ {2}}{a}$$ $$S _ {2} = \frac {a \left(\frac {v _ {2}}{a}\right) ^ {2}}{2} = \frac {v _ {2} ^ {2}}{2 a} = 4 \pi R$$ $$a = \frac {v _ {2} ^ {2}}{8 \pi R} \approx \frac {v _ {2} ^ {2}}{8 \left(\frac {2 2}{7}\right) R}$$ $$a = \frac {8 0 ^ {2}}{8 \left(\frac {2 2}{7}\right) \left(\frac {2 0}{\frac {2 2}{7}}\right)} = \frac {6 4 0 0}{1 6 0} = 4 0 \left(\frac {m}{s ^ {2}}\right)$$

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