A 160-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 31° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?
____rad/s
____rad/s
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The mass of 160N sphere in a standard (constant) gravitation field is 160/9.81=16.3 kg
Assuming no friction, at any time , the center of rotation for the sphere is the point at which it is in contact with the inclined surface.
The magnitude of the force in the direction of motion is 160sin31=82.4N
Now, using F=ma
82.4=16.3*a
This gives a=5.1m/s^2
using the constant acceleration formula v^2=u^2+2ax , where u=0m/s (starts from rest) a=5.1, x=6m
this gives v=7.8 m/s (note that this is the velocity of the center of the sphere)
now using v=w*r (w is the angular speed omega) where v=7.8m/s and r =0.2m
this gives w=39.1 rad/s
Assuming no friction, at any time , the center of rotation for the sphere is the point at which it is in contact with the inclined surface.
The magnitude of the force in the direction of motion is 160sin31=82.4N
Now, using F=ma
82.4=16.3*a
This gives a=5.1m/s^2
using the constant acceleration formula v^2=u^2+2ax , where u=0m/s (starts from rest) a=5.1, x=6m
this gives v=7.8 m/s (note that this is the velocity of the center of the sphere)
now using v=w*r (w is the angular speed omega) where v=7.8m/s and r =0.2m
this gives w=39.1 rad/s