JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 1)
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :
_11th_January_Evening_Slot_en_1_1.png)
16 rad/s2
20 rad/s2
10 rad/s2
12 rad/s2
Explanation
_11th_January_Evening_Slot_en_1_2.png)
40 + f = m(R$$\alpha $$) . . . .(i)
40 $$ \times $$ R $$-$$ f $$ \times $$ R = mR2$$\alpha $$
40 $$-$$ f = mR$$\alpha $$ . . . .(ii)
From (i) and (ii)
$$\alpha $$ = $${{40} \over {mR}}$$ = 16
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