A time-varying force F = 2t is applied on a s | vedclass

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(C) The rate of change of angular momentum $(L)$ is equal to the net torque $(	au)$ acting on the system about the point of rotation.$\frac{dL}{dt} =

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$(R + r)t^2$

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(C) The rate of change of angular momentum $(L)$ is equal to the net torque $(	au)$ acting on the system about the point of rotation.$\frac{dL}{dt} = 	au$$\Rightarrow dL = 	au dt$Taking torque about the bottommost point (point of contact with the ground),the perpendicular distance of the line of action of the force $F$ from the bottommost point is $(R + r)$.$	au = F 	imes (R + r) = 2t(R + r)$Substituting this into the angular momentum equation:$dL = 2t(R + r) dt$Integrating from $t = 0$ to $t$ (assuming initial angular momentum is zero):$L = \int_{0}^{t} 2t(R + r) dt$$L = 2(R + r) \int_{0}^{t} t dt$$L = 2(R + r) \left[ \frac{t^2}{2} ight]_{0}^{t}$$L = (R + r)t^2$

$A$ time-varying force $F = 2t$ is applied on a spool rolling as shown in the figure. The angular momentum of the spool at time $t$ about the bottommost point is:

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