Question

A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a turning cylindrical shaft. If the shaft’s diameter can be no larger than 16.0 cm due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator?

Answer 1

Answer:

α =18.75  rad/s²

Explanation:

Given that

Acceleration a = 0.15 g

We know that   g =10 m/s²

a= 0.15 x 10 = 1.5  m/s²

d= 16 cm

Radius r= 8 cm

Lets take angular acceleration =α rad/s²

As we know that

a= α r

Now by putting the values

1.5 = α  x 0.08

α =18.75  rad/s²