Recent content by AL115

1. A spool that rotates and slides freely while pulling a mass

I just realized that I made a dumb mistake (again) in the post #14. I wrote the equation relating the tension to the linear acceleration ac as T=MRac which is obviously wrong. It is T=Mac. And we also know that T = (1/2)MRα = (1/2)Mat We get that at=2ac. Now everything makes sense. My problem...
2. A spool that rotates and slides freely while pulling a mass

I got it from the answers in the first image. Question 13- Answer C- α =(4mg)/(d(M+3m)) R=d/2 α R=(2mg)/(M+3m)=(2/3)a a = α R+ac α R is the at in the last image.
3. A spool that rotates and slides freely while pulling a mass

I chose the downward direction in the image (the direction of the acceleration of the center of mass) to be the negative direction. -α R -α R+ac -α R=(2/3)a @bold wouldn't the hanging mass accelerate downward? -T=(1/2)MRα -T=MRac are they correct? Also can you tell me why α R=2ac
4. A spool that rotates and slides freely while pulling a mass

This is what I came up with. I am not sure if this is how the problem should be solved. I also relied on the fact that At = 2Acm which I got from looking at the answers. Now I need to know why At = 2Acm.
5. A spool that rotates and slides freely while pulling a mass

I just made a dumb mistake. Acm is not equal to 3a. a is equal to 3Acm. which means At is equal to (2/3)a which means mg+ma=(1/2)M(2/3)a --->a = 3mg/(M-3m) which means I have a problem with the sign. So the equation should be T=mg-ma so that means that the mass is accelerating downward and...
6. A spool that rotates and slides freely while pulling a mass

For linear acceleration of the spool, I suppose MAcm where Acm = the acceleration of the center of mass. And the net torque is equal to Iα which should equal the force of the tension(T) times the radius (R) in other words(TR). Right? The angular acceleration is equal to the tangential...
7. A spool that rotates and slides freely while pulling a mass

I tried to solve it but no result. The spool doesn't move in an uniform circular motion when it moves linearly. But it moves in a uniform circular motion relative to the acceleration of center of mass. Right? But how can that help me when I don't know the acceleration of the center of mass? And...
8. A spool that rotates and slides freely while pulling a mass

-Oh. Pardon me then I am new here so I didn't know. -Up. Isn't the spool pulling the string? -Yes. Isn't a equal to the tangential acceleration of the disk? -Hmmm. I never considered relative motion. I will try to solve it again with this in mind. Thanks.
9. A spool that rotates and slides freely while pulling a mass

Thanks for the reply. I already thought of the bold but I still couldn't solve it. I don't know how the linear acceleration of the spool (which is it's center of mass acceleration right?) affect the tangential acceleration (which is equal to the acceleration of the bolck with mass m right?) Can...
10. A spool that rotates and slides freely while pulling a mass

My attempt to solve the first one. T-mg=ma T=m(g+a) TR = Iα I=(1/2)(M)(R^2) α=a/R TR=(1/2)(M)(R^2)(a/R) T=(1/2)Ma mg+ma=1/2Ma a=(2mg)/(M-2m) And... I got a wrong answer. As usual.
11. Conservation of momentum in a bullet-block-spring system

So if we try to describe this mathematically, we can consider that the moment where the bullet makes contact with the block (lets call it t1) and the moment where the bullet emerges from the block (lets call it t2) the same since t2 - t1 is close to zero? That makes sense. Thank you very much.
12. Conservation of momentum in a bullet-block-spring system

Yeah that is correct. But my question is that did the block move after the bullet emerged from it? mV1 + MV2 = mV3 + MV4 where m = mass of the bullet = 5g , M = mass of the block = 1kg, V1 = the initial speed of the bullet = 400 m/s, V2 = the initial speed of the block = 0 m/s, V3 = the final...
13. Conservation of momentum in a bullet-block-spring system

I don't know...
14. Conservation of momentum in a bullet-block-spring system

According to the first equation, the final potential energy is equal to the initial kinetic energy of the block. So that means that Vblok is the instantaneous speed of the block right before it moves to the right and compress the spring, right? But doesn't the second equation (The initial total...