Recent content by PhysicsA1

I honestly have no clue

So it will only be effecting the side of which i jump on. How in return will that actually be calculated then?

Here are the questions again. If the steel disk has mass of 200 kg and a radius of 2 meters you can make it spin by applying a force to the rim. This torque increases the angular momentum of the disk. Suppose the force is 20 Newtons. How long would you have to apply it to get the wheel...

yes. its a playground merry go round

I= 520 kg*m (Due to taking the initial mass (200) and adding 60 to it due to us jumping on the merry go round) so...I=(1/2)(260)(2)^2 40=520a SO a=.077 SO .077=.55/t; Therefore T=6.75 sec. Where did i go wrong? or misunderstand??

So with the addition of the 60kg would that then make the time be 6.75 rad/s instead of 5.2 rad/sec, which would make the rate of spin be slower!? right?

So with the addition of the 60kg would that then make the time be 6.75 rad/s instead of 5.2 rad/sec, which would make the rate of spin be slower!? right?

I figured that out exactly the same. So once we figure this out would be multiply the .083 through to find our final answer

in which equation used uses α?