A solid sphere of mass 1 kg and radius 1 m ro | vedclass

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(D) Given: Mass $M = 1 kg$,Radius $R = 1 m$,$	heta = 30^{\circ}$,$g = 10 m s^{-2}$.Component of weight down the plane: $Mg \sin 	heta = 1 	imes 10

Vedclass · Accepted Answer

$2.86$

Vedclass · Answer

(D) Given: Mass $M = 1 kg$,Radius $R = 1 m$,$	heta = 30^{\circ}$,$g = 10 m s^{-2}$.Component of weight down the plane: $Mg \sin 	heta = 1 	imes 10 	imes \sin 30^{\circ} = 10 	imes 0.5 = 5 N$.Let $f$ be the friction force acting up the plane. The two external forces of $1 N$ each act in opposite directions relative to the center,creating a net torque.Equation of motion for translation: $Mg \sin 	heta - f - F_{ext} = Ma$. Here,the two $1 N$ forces act such that one aids motion and one opposes it,but they create a net torque. Looking at the diagram,the net force along the incline is $Mg \sin 	heta = 5 N$. The two $1 N$ forces cancel each other out in terms of net translational force.So,$5 - f = 1 	imes a \Rightarrow f = 5 - a$.Equation of motion for rotation about the center of mass $(COM)$: $	au_{net} = I \alpha$.The torque due to friction $f$ is $fR$ (counter-clockwise). The torque due to the two $1 N$ forces is $2 	imes (1 N 	imes 0.5 m) = 1 N m$ (clockwise).Thus,$fR - 1 = I \alpha$,where $I = \frac{2}{5} MR^2 = 0.4 kg m^2$ and $\alpha = \frac{a}{R} = a$.$f(1) - 1 = 0.4 a \Rightarrow f = 1 + 0.4 a$.Equating the two expressions for $f$: $5 - a = 1 + 0.4 a$.$4 = 1.4 a \Rightarrow a = \frac{4}{1.4} = \frac{40}{14} = \frac{20}{7} \approx 2.86 m s^{-2}$.

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