Calculating Revolutions of a Ball Dropped from a Ramp

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smilingsteph
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If I have a ball/sphere is released from rest, rolls (without slipping) down a ramp, then drops from the table the ramp is on, how many revolutions does the ball make during the fall?

The diameter of the ball, height of the ramp, and height of the table are known. And I calculated the horizontal distance the ball travels before landing correctly. I think I calculated the w, v, and centripetal acceleration correctly. But I am not sure how to solve for the number of revolutions the ball makes before falling on the ground. This problem is probably relatively simple, but I might be having issues with it from looking at it for so long and confusing myself. Any help would be appreciated, thanks!
 
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smilingsteph said:
If I have a ball/sphere is released from rest, rolls (without slipping) down a ramp, then drops from the table the ramp is on, how many revolutions does the ball make during the fall?

The diameter of the ball, height of the ramp, and height of the table are known. And I calculated the horizontal distance the ball travels before landing correctly. I think I calculated the w, v, and centripetal acceleration correctly. But I am not sure how to solve for the number of revolutions the ball makes before falling on the ground. This problem is probably relatively simple, but I might be having issues with it from looking at it for so long and confusing myself. Any help would be appreciated, thanks!
Once the ball has left the ramp, are there any external torques acting on the ball?
 
to my knowledge no.

cylinder:
m= 1.8kg
r= 0.12m
l= 0.50m

ramp:
h= 0.6m
l= 5.0m

"A cylinder is released from rest at the top of a ramp and allowed to roll without slipping. What is the rotational kinetic energy?"

I already correctly solved that the total kinetic energy at the bottom of the ramp is 10.59J.
 
smilingsteph said:
to my knowledge no.

cylinder:
m= 1.8kg
r= 0.12m
l= 0.50m

ramp:
h= 0.6m
l= 5.0m

"A cylinder is released from rest at the top of a ramp and allowed to roll without slipping. What is the rotational kinetic energy?"

I already correctly solved that the total kinetic energy at the bottom of the ramp is 10.59J.
You are indeed correct, if we ignore drag then the only force acting on the sphere is it's weight, which acts through it's centre. So if the net torque is zero after the ball has left the ramp, what can you say about the angular velocity?
 
w=18.08314 and I=0.03888 and v=2.1699768.

i keep getting that the rotational kinetic energy is ~4.23, but it keeps coming up as incorrect. are any of my preliminary calculations incorrect? but i'd have know correctly, w, v, and I to solve for the distance (which i know is correct). so I'm not sure what I am doing wrong after i solve for w, I, and v.
 
smilingsteph said:
w=18.08314 and I=0.03888 and v=2.1699768.

i keep getting that the rotational kinetic energy is ~4.23, but it keeps coming up as incorrect. are any of my preliminary calculations incorrect? but i'd have know correctly, w, v, and I to solve for the distance (which i know is correct). so I'm not sure what I am doing wrong after i solve for w, I, and v.
You may want to recheck your moment of inertia calculation.
 
so i rechecked my calculations and still get the moment of inertia. how do i go from there though to solve for the number of rotations in the fall?