1.

A constant couple of 500 Nm turns a wheel of moment of inertia 100 kg m^(2) about an axis through its centre, the angular velocity gained in two second is :

Answer»

5 rad `s^(-1)`
`100ms^(-1)`
`200ms^(-1)`
10 rad `s^(-1)`

Solution :`alpha=(tau)/(I)=(500)/(100)=5" rad/"s^(2)`
`omega=omega_(0)+alphat.=0+5xx2=10" rad/s"`


Discussion

No Comment Found

Related InterviewSolutions