MCQ
Disc $A$ and disc $B$ with a hole have equal mass and radius and a string is wrapped around them as shown. Equal force $F$ is applied on the string of each body. Friction is sufficient for rolling. After time $t$ velocity of $A$ is $v_A$ and that of $B$ is $v_B$ and kinetic energy $k_A$ & $k_B$ , then
  • A
    $v_A = v_B$
  • B
    $v_A < v_B$
  • $k_A > k_B$
  • D
    $k_A < k_B$

Answer

Correct option: C.
$k_A > k_B$
c
$\mathrm{F}+\mathrm{f}=\mathrm{ma}$

$(\mathrm{F}-\mathrm{f}) \mathrm{R}=\mathrm{I} \alpha$

$\Rightarrow \mathrm{F}-\mathrm{f}=\frac{\mathrm{I} \alpha}{\mathrm{R}^{2}} \quad(\mathrm{a}=\mathrm{R} \alpha)$

adding

$F=\left(m+\frac{I}{R^{2}}\right) \frac{a}{2}$

$v=0+a t$

$=\frac{2 \mathrm{F}}{\left(\mathrm{m}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right)} \mathrm{t}$

$\mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

$=\frac{1}{2} \mathrm{V}^{2}\left(\mathrm{m}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right), \omega=\frac{\mathrm{V}}{\mathrm{R}}$

$=\frac{1}{2}\left(\frac{2 \mathrm{Ft}}{\mathrm{m}+\frac{\mathrm{I}}{\mathrm{R}^{2}}}\right)^{2}\left(\mathrm{m}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right)=\frac{2 \mathrm{F}^{2} \mathrm{t}^{2}}{\left(\mathrm{m}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right)}$

$\mathrm{Also}, \mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{A}}$                    

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