A wheel, with rotational inertia I, mounted on a vertical shaft with negligible

Question

A wheel, with rotational inertia I, mounted on a vertical shaft with negligible relational inertia, is rotating with angular speed omega o. A nonrotation wheel with rotational inertia 2l is suddenly dropped onto the same shaft as shown.. The resultant combination of the two wheels and shaft will rotate at: A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs 720 N and the shoe weighs 4.0 N, the man moves to the left at approximately:

Explanation / Answer

A man is marooned at rest on level frictionless ice. In desperation he hurls his shoe to the right 2. Given that The moment of inertia of rotational wheel = I The moment of inertia of non rotational wheel = 2I The initial angular velocity = 0 The final angular velocity = The final angular velocity =       since there is no externalforce on the system then the angular momentumis conserved                            Initialangular momentum = final angular momentum                                                       I1*1 = I2*2                                                        I * o = ( I +2I )                                                               = I * o / ( I +2I )                                                                   = o / 3 So the answer is option ( C ) 7) Given that   Weight of man   =  M * g   =   720N    So the mass of man,   M   =   720/ 9.8   kg = 73.46 kg    Mass of shoe, m   =   4.0 /9.8   kg = 0.408 kg The velocity of the shoe, v = 15 m/s Let the velocity of the man = V    According to law of conservation of energy we have    momentum of shoe   =   momentum of man                        m * v   =   M* V      From the above The velocity of the man, V = mv / M = 0.083 m/s = 8.3 * 10^-2 m/s So the answer is option ( C ) 7) Given that   Weight of man   =  M * g   =   720N    So the mass of man,   M   =   720/ 9.8   kg = 73.46 kg    Mass of shoe, m   =   4.0 /9.8   kg = 0.408 kg The velocity of the shoe, v = 15 m/s Let the velocity of the man = V    According to law of conservation of energy we have    momentum of shoe   =   momentum of man                        m * v   =   M* V      From the above The velocity of the man, V = mv / M = 0.083 m/s = 8.3 * 10^-2 m/s So the answer is option ( C )   Weight of man   =  M * g   =   720N    So the mass of man,   M   =   720/ 9.8   kg = 73.46 kg    Mass of shoe, m   =   4.0 /9.8   kg = 0.408 kg The velocity of the shoe, v = 15 m/s Let the velocity of the man = V    According to law of conservation of energy we have    momentum of shoe   =   momentum of man                        m * v   =   M* V      From the above The velocity of the man, V = mv / M = 0.083 m/s = 8.3 * 10^-2 m/s So the answer is option ( C )

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