Question #d8ba2

1 Answer
Apr 1, 2017

Answer:

I got #6.4revs#

Explanation:

I would use the general relationship for rotational motion (analogous to the linear one):
#omega^2=omega_0^2+2alphatheta#
where:
#omega=# angular velocity (initial and final, with #omega=0# the final one);
#alpha# is angular acceleration;
#theta# is the angle in radians.
We get:
#0=20^2-2*5*theta# as it is slowing down #alpha<0#;
rearranging:
#theta=400/10=40rad#
Giving:
#40/(2pi)=6.36~~6.4revs#