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STD 11 - 6. system of particles and rotational motion — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6. system of particles and rotational motion4 Marks
MCQ
$A$ time varying force $F = 2t$ is applied on a spool rolling as shown in figure. The angular momentum of the spool at time $t$ about bottommost point is:
  • A
    $\frac{{{r^2}{t^2}}}{R}$
  • B
    $\frac{{{{(R + r)}^2}}}{r}\, t^2$
  • ✓
    $(R + r)t^2$
  • D
    data is insufficient
✓

Answer

Correct option: C.
$(R + r)t^2$
c
Change in angular momentum(L) = Torque $\frac{d L}{d t}=\tau$

$\Rightarrow d L=\tau d t$

Taking torque about lowest point distance of force applied from lowest point $=(R+$$r)$

$\tau=2 t(R+r)$$...(1)$

$\Rightarrow d L=2 t(R+r) d t$

at after time t, the angular momentum becomes

$L=\int_{0}^{L} d L=\int_{0}^{t} 2 t(R+r)$

$\Rightarrow L=(R+r) t^{2}$

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