A disc of circumference s is at rest at a poi | vedclass

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(D) Let $F$ be the applied force,$m$ be the mass,and $R$ be the radius of the disc. Circumference $s = 2\pi R$.Between $A$ and $B$,the disc rolls with

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All of the above.

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(D) Let $F$ be the applied force,$m$ be the mass,and $R$ be the radius of the disc. Circumference $s = 2\pi R$.Between $A$ and $B$,the disc rolls without slipping. The equations of motion are:$F - f = ma_1$ and $fR = I\alpha = (\frac{1}{2}mR^2)\alpha$. Since $a_1 = R\alpha$,we get $f = \frac{1}{2}ma_1$. Substituting this,$F - \frac{1}{2}ma_1 = ma_1 \implies F = \frac{3}{2}ma_1 \implies a_1 = \frac{2F}{3m}$.The distance $AB = s = 2\pi R = \frac{1}{2}a_1 T^2$. Thus,$T^2 = \frac{4\pi R}{a_1} = \frac{4\pi R}{2F/3m} = \frac{6\pi mR}{F}$.At point $B$,the linear velocity $v_B = a_1 T$ and angular velocity $\omega_B = \alpha T = \frac{a_1}{R}T$.To the right of $B$,the surface is smooth,so friction $f = 0$. The new linear acceleration $a_2 = F/m$. Since $F = \frac{3}{2}ma_1$,$a_2 = \frac{3}{2}a_1 > a_1$. Thus,linear acceleration increases.In time $T$ after $B$,the distance covered is $d = v_B T + \frac{1}{2}a_2 T^2 = (a_1 T)T + \frac{1}{2}(\frac{3}{2}a_1)T^2 = a_1 T^2 + \frac{3}{4}a_1 T^2 = \frac{7}{4}a_1 T^2$. Since $s = \frac{1}{2}a_1 T^2$,$d = \frac{7}{2}s > s$.Since there is no friction to the right of $B$,the angular velocity remains constant at $\omega_B = \frac{a_1 T}{R}$. The time taken for one rotation is $t = \frac{2\pi}{\omega_B} = \frac{2\pi R}{a_1 T} = \frac{s}{a_1 T} = \frac{\frac{1}{2}a_1 T^2}{a_1 T} = \frac{T}{2}$.All statements are correct.

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