CBSE Class 11-science Answered
MI = MR2/2 = 2.5 kg m2
Torque = (10 N)(0.25 m) = = 2.5 N m
Torque = MI x Angular acceleration
(10 N)(0.25 m) = ((80 kg)(0.25 m)2/2) x Angular Acceleration,
Angular Acceleration = 1 rad/s2
Final Angular Velocity = Initial Angular Velocity + (Angular Acceleration)(Time)
= 5 + (1)(20) = 25 rad/s
Work done by the torque = Change in KE = I(wf2-wi2)/2 = 2.5(252-52)/2 = 750 J
Rigid Body and Centre of Mass - Part 1This video explains the nature of a rigid body, the centre of mass, the mot...
Torque - Part 1This video explains torque, magnitude of torque and the right hand rule on ...
Moment of Inertia - Part 1This video explains the moment of inertia, its relation with rotational kin...
Kinematics of Rotational Motion about a Fixed Axis - Part 1This video explains the analogy between linear and angular variables of mot...
Dynamics of Rotational Motion - Part 1This video explains the analogy between the torque of rotational and linear...