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Rotational Mechanics(The concept of spinning) 
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#1
Apr2309, 09:34 AM

P: 336

well this is a really good question which i just thought while playing cricket.
A sphere of mass m and radii r(a tennis ball) falls freely from a height of H and angular velocity(clockwise) about its axis omega.Coefficient of restitution is e The coefficient of friction is mu. After the sphere touches the ground it moves in the forward direction. We have to find the range of the sphere. Now we can find the velocities by energy conservation. We we can apply angular conservation about the point which touches the ground. initial translational + initial rotational = final translational + final rotational Final velocity * e = Initial velocity. Lets suppose that after collision the velocities of the sphere is vx and vy. so we have four variables final angular velocity,final velocity,Vx and Vy. But we have just two equations. we can get one more by angular conservation which i am unable to get through and we are still one short. 


#2
Apr2309, 10:47 AM

P: 2,048

Using energy to find velocities is a crock usually.
You can treat the vertical and horizontal comonents seperately. You know how high i'll bounce from COR*inital h. so you can work out the vertical component of velocity. the forward motion will depend on rotational speed, moment of inertia, radius and initial sideways velocity (which in this case is zero as its a verticl drop). I really cant be bothred deriving the equation you'll have to do that one yourself. 


#3
Apr2309, 09:02 PM

P: 336




#4
Apr2409, 04:16 AM

P: 2,048

Rotational Mechanics(The concept of spinning)
It's just that to he honest I couldnt remember the below formula and couldnt derive it when I wrote that, but didnt want to admit it! :D I've been looking through my notes on oblique impacts and found this. Vxnew = (Vx + Awr) / (1+A) where A = I / ma^2 Set the inital velocity in x to be zero due to the vertical drop and bobs your uncle. This is a model for impacts that involve rolling when contact occurs. Which is the only thing that can happen in this case due to no horizontal inital velocity. 


#5
Apr2409, 08:51 AM

P: 336

Now back to the problem. I am not able to understand how we get the formula mentioned by you. Do you know its possible derivation? 


#6
Apr2409, 11:56 AM

P: 2,048

i'll have a crack at it. It's goig to be diffucult to get across on here. it'll probably need a diagram to help explain. also can someone please check I havent messed up the derivation.
m=mass r=ballradius T=torque F=force I=moment of inertia w=angular velocity We start off with: F=ma that can be rewritten as the integral, so we can get the inital and final velocities. Fdt=mdv F=m(vi  vf) (1) Now we know that as the ball is spinning and it hits the ground, a force in x will occur. This force acting at the contact patch acts as a torque on the ball. T=Fr Like F=ma this can be rewritten as the integral to get. T=I(wiwf) Fr=I(wiwf) (2) We also know that after impact, converting angular velocity to linear velocity we have. vf=wf*r (3) What we do is then is: rearrange (2) for F and put this into the rearranged F=ma. equation 1 F=(I(wiwf))/r (I(wiwf))/r = m(vivf) This gives us a problem as we dont know either vf or wf. However we do have equation 3 so we can rwrite it as 1 unknown. As we want to know vf, we rewrite wf in terms of this. (I(wiwf))/r = m(vivf) wf=vf/r (I(wi(vf/r)) = M(vivf) (4) 4 can be expanded and rearranged to get Vi  (Iw/rm) = Vf ((I/r^2m) + 1) That is the derevation. The eqaution can be used in the form above. But to make it look tidy, we rewrite the left ahnd bracket multiplied by r/r. Vi  (Iwr/r^2m) Vi  (Iw/r^2m)r The (Iw/r^2m) was given the label A to make it easier to write. Tidying up the eqaution gives. Vf = (Vi + Awr)/(1+A) As there is no inital velocity in x. we have. Vx final = Awr/(1+A) 


#7
Apr2409, 10:36 PM

P: 336

Okey..... I got the concept. Let me derive the formula on my own from the points which you have made. And lets see if it matches yours



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