Disc A and disc B (with a hole) have equal ma | vedclass

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(C) For a body of mass $m$,radius $R$,and moment of inertia $I$ undergoing rolling motion under force $F$ applied at the top:$F + f = ma$$(F - f)R = I

Vedclass · Accepted Answer

$k_A > k_B$

Vedclass · Answer

(C) For a body of mass $m$,radius $R$,and moment of inertia $I$ undergoing rolling motion under force $F$ applied at the top:$F + f = ma$$(F - f)R = I\alpha$Since $a = R\alpha$,we have $f = \frac{I\alpha}{R} = \frac{Ia}{R^2}$.Substituting $f$ in the first equation: $F + \frac{Ia}{R^2} = ma \Rightarrow F = a(m + \frac{I}{R^2})$.Thus,the acceleration is $a = \frac{F}{m + \frac{I}{R^2}}$.The velocity after time $t$ is $v = at = \frac{Ft}{m + \frac{I}{R^2}}$.The kinetic energy is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}v^2(m + \frac{I}{R^2})$.Substituting $v$: $K = \frac{1}{2} (\frac{Ft}{m + \frac{I}{R^2}})^2 (m + \frac{I}{R^2}) = \frac{F^2t^2}{2(m + \frac{I}{R^2})}$.Since disc $B$ has a hole,its mass is distributed further from the center,so $I_B > I_A$.As $I_B > I_A$,the denominator $(m + \frac{I}{R^2})$ is larger for $B$ than for $A$.Therefore,$v_A > v_B$ and $k_A > k_B$.

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