An athlete throws a discus from rest to a fin | vedclass

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(A) Given: Initial angular velocity $\omega_{0} = 0 \, rad \, s^{-1}$,final angular velocity $\omega = 15 \, rad \, s^{-1}$,time $t = 0.270 \, s$,and

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$45$

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(A) Given: Initial angular velocity $\omega_{0} = 0 \, rad \, s^{-1}$,final angular velocity $\omega = 15 \, rad \, s^{-1}$,time $t = 0.270 \, s$,and radius $r = 0.810 \, m$.First,we calculate the angular acceleration $\alpha$ using the formula $\omega = \omega_{0} + \alpha t$.Since $\omega_{0} = 0$,we have $\alpha = \frac{\omega}{t} = \frac{15}{0.270} \, rad \, s^{-2}$.The tangential acceleration $a_{t}$ is given by $a_{t} = r \alpha$.Substituting the values: $a_{t} = 0.810 	imes \frac{15}{0.270} = 0.810 	imes 55.55 = 45 \, m \, s^{-2}$.Thus,the acceleration of the discus is $45 \, m \, s^{-2}$.

An athlete throws a discus from rest to a final angular velocity of $15 \, rad \, s^{-1}$ in $0.270 \, s$ before releasing it. During acceleration,the discus moves along a circular arc of radius $0.810 \, m$. The acceleration of the discus before it is released is ....... $m \, s^{-2}$.

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