A pendulum consists of a bob of mass m=0.1 k | vedclass

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(D) The angular momentum $J$ about the point of suspension is given by $J = r_{\perp} 	imes p$,where $r_{\perp}$ is the perpendicular distance from t

Vedclass · Accepted Answer

$0.18, 0.16$

Vedclass · Answer

(D) The angular momentum $J$ about the point of suspension is given by $J = r_{\perp} 	imes p$,where $r_{\perp}$ is the perpendicular distance from the pivot to the line of action of the impulse. Here,$r_{\perp} = H = 0.9 \ m$ and $p = P = 0.2 \ kg \cdot m/s$. Thus,$J = 0.9 	imes 0.2 = 0.18 \ kg \cdot m^2/s$.When the string becomes taut,the velocity component along the string is lost due to the impulsive tension. The bob is at an angle $	heta$ such that $\cos 	heta = H/L = 0.9/1.0 = 0.9$. The velocity of the bob just before the string becomes taut is $v = P/m = 0.2/0.1 = 2 \ m/s$. The component of velocity perpendicular to the string is $v_{\perp} = v \cos 	heta = 2 	imes 0.9 = 1.8 \ m/s$. The kinetic energy just after the lift-off is $K = \frac{1}{2} m v_{\perp}^2 = \frac{1}{2} 	imes 0.1 	imes (1.8)^2 = 0.05 	imes 3.24 = 0.162 \ J$. Rounding to two decimal places,$K \approx 0.16 \ J$.

Column-$I$	Column-$II$
$(1)$ $SI$ unit of torque	$(a)$ $m$
$(2)$ $SI$ unit of radius of gyration	$(b)$ $N\,m$
	$(c)$ $Js^{-2}$

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ $SI$ unit of torque $(a)$ $m$

$(2)$ $SI$ unit of radius of gyration $(b)$ $N\,m$

$(c)$ $Js^{-2}$

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