Question 286: Centre of Mass and Rotational Motion

Question

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) :

Accepted Answer

40 + f = m(R$$\alpha $$)         . . . .(i)
40 $$ 	imes $$ R $$-$$ f $$ 	imes $$ R = mR
2
$$\alpha $$
40 $$-$$ f = mR$$\alpha $$         . . . .(ii)
From (i) and (ii)
$$\alpha $$ = $${{40} \over {mR}}$$ = 16

Answer

16 rad/s
2

Answer

20 rad/s
2

Answer

10 rad/s
2

Answer

12 rad/s
2

Centre of Mass and Rotational Motion

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