A particle moves along a circle of radius lef | vedclass

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Question

(A) Given: Radius $r = \frac{20}{\pi} \, m$,initial angular velocity $\omega_i = 0$,and total angular displacement $	heta = 2 	imes (2 \pi) = 4 \pi

Vedclass · Accepted Answer

$40 \, m/s^2$

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(A) Given: Radius $r = \frac{20}{\pi} \, m$,initial angular velocity $\omega_i = 0$,and total angular displacement $	heta = 2 	imes (2 \pi) = 4 \pi \, rad$ (since it completes two revolutions).Final linear velocity $v = 80 \, m/s$.The final angular velocity is $\omega_f = \frac{v}{r} = \frac{80}{20/\pi} = 4 \pi \, rad/s$.Using the rotational kinematic equation $\omega_f^2 = \omega_i^2 + 2 \alpha 	heta$,we get:$(4 \pi)^2 = 0^2 + 2 \alpha (4 \pi)$$16 \pi^2 = 8 \pi \alpha$$\alpha = 2 \pi \, rad/s^2$.The tangential acceleration is $a_t = r \alpha = \left( \frac{20}{\pi} ight) 	imes (2 \pi) = 40 \, m/s^2$.

$A$ particle moves along a circle of radius $\left( \frac{20}{\pi} \right) \, m$ 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

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