A uniform cylinder rests on a cart as shown. | vedclass

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(B) Let the mass of the cylinder be $m$,diameter $d = 4\, cm$ (radius $r = 2\, cm$),and height $h = 10\, cm$.$1$. Condition for sliding:The maximum ac

Vedclass · Accepted Answer

$4\, m/s^2$

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(B) Let the mass of the cylinder be $m$,diameter $d = 4\, cm$ (radius $r = 2\, cm$),and height $h = 10\, cm$.$1$. Condition for sliding:The maximum acceleration $a_{max}$ before the cylinder slides is given by $f_{max} = \mu N = m a_{max}$. Since $N = mg$,we have $m a_{max} = \mu mg$,so $a_{max} = \mu g = 0.5 	imes 10 = 5\, m/s^2$.$2$. Condition for tipping:To tip over,the torque due to the pseudo-force $ma$ about the edge $A$ must be greater than or equal to the torque due to the weight $mg$ about the same edge $A$.The pseudo-force $ma$ acts at the center of mass,which is at a height $h/2 = 5\, cm$ from the base. The weight $mg$ acts at a horizontal distance $r = 2\, cm$ from the edge $A$.Taking torque about edge $A$:$(ma) 	imes (h/2) \geq (mg) 	imes r$$a 	imes (5\, cm) \geq g 	imes (2\, cm)$$a \geq g 	imes (2/5) = 10 	imes 0.4 = 4\, m/s^2$.Since the acceleration required to tip $(4\, m/s^2)$ is less than the acceleration required to slide $(5\, m/s^2)$,the cylinder will tip over at $a = 4\, m/s^2$.

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