A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?

Question

Accepted Answer

Correct answer: B. (b): m=3 kg ,r=40 cm=40 10^-2m ,F=30N Moment of inertia of hollow cylinder about its axis =mr^2=3 kg (O .4 )^2m^2=0.48 kg m^2 The torque is given by, = Iα where I=moment of inertia, =angular acceleration In the given case, = rF, as the force is acting perpendicularly to the radial vector. = l= Frmr^2= F mr= 303 40 10^-2= 30 1003 40 =25 rad s^-2

$A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?$

Solution

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