A particle is projected with a speed v_0 = sq | vedclass

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(C) At the bottom of the bowl,the normal force $N$ and gravity $mg$ act on the particle. The net force provides the centripetal acceleration:$N - mg =

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$\sqrt{2} g 
warrow$

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(C) At the bottom of the bowl,the normal force $N$ and gravity $mg$ act on the particle. The net force provides the centripetal acceleration:$N - mg = \frac{mv_0^2}{R}$Given $v_0 = \sqrt{gR}$,we have:$N - mg = \frac{m(gR)}{R} = mg$$N = 2mg$The kinetic friction force is $f_k = \mu N = 0.5 	imes 2mg = mg$.The tangential acceleration $a_T$ is caused by friction:$ma_T = f_k = mg \implies a_T = g$ (directed horizontally,opposite to velocity).The normal acceleration $a_N$ is the centripetal acceleration:$a_N = \frac{v_0^2}{R} = \frac{gR}{R} = g$ (directed vertically upwards).The total acceleration $a$ is the vector sum of $a_T$ and $a_N$:$a = \sqrt{a_T^2 + a_N^2} = \sqrt{g^2 + g^2} = \sqrt{2}g$.The direction is at an angle of $45^\circ$ to the horizontal,pointing upwards and backwards,represented as $
warrow$.

$A$ particle is projected with a speed $v_0 = \sqrt{gR}$ at the bottom of a hemispherical bowl. The coefficient of friction between the particle and the hemispherical surface is $\mu = 0.5$. Then,the initial acceleration of the particle is:

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