A disc is given an angular velocity omega_0 a | vedclass

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(D) The forces acting on the disc along the incline are the component of gravity $mg \sin 	heta$ acting downwards and the kinetic friction $f_k$ acti

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$\alpha R = 2 g \sin 	heta$

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(D) The forces acting on the disc along the incline are the component of gravity $mg \sin 	heta$ acting downwards and the kinetic friction $f_k$ acting upwards.The equation of motion for the centre of mass is: $mg \sin 	heta - f_k = ma$.The torque equation about the centre of mass is: $f_k R = I \alpha = \left(\frac{mR^2}{2}ight) \alpha$.From the torque equation,we get $f_k = \frac{mR \alpha}{2}$.Substituting $f_k = \mu N = \mu mg \cos 	heta$ into the torque equation:$\frac{mR \alpha}{2} = \mu mg \cos 	heta$.Since $\mu = 	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have:$\frac{mR \alpha}{2} = \left(\frac{\sin 	heta}{\cos 	heta}ight) mg \cos 	heta = mg \sin 	heta$.Solving for $R \alpha$:$R \alpha = 2 g \sin 	heta$.

$A$ disc is given an angular velocity $\omega_0$ as shown in the figure and is kept on an inclined plane of inclination $\theta$ having a coefficient of friction $\mu = \tan \theta$. If the initial acceleration of the centre of mass is $a$ and the angular acceleration is $\alpha$,then:

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