Angular Momentum(Turntable and blob collision)

• MRMooneyham
In summary, the conversation revolves around a problem involving a small blob of putty falling onto a rotating turntable and determining the postcollision angular speed of the system. The conversation includes a discussion of the equation L = r x p = Iω and the concept of conservation of momentum. The main issue is determining the moment of inertia of the turntable and the correct equation for calculating it. The conversation concludes with the suggestion to look up a list of moments of inertia and learn them.
MRMooneyham
A small blob of putty of mass m = 7.1 falls from the ceiling and lands on the outer rim of a turntable of radius R and mass M = 7.5 kg that is rotating freely with an angular speed of 1.4 rad/s about its axis. The postcollision angular speed of the turntable-putty system is ____ rad/s.

L = r x p = Iω
this is the only equation i think i need for this problem

As far as i can tell all i would need to do is...
since R is unknown but will be a constant through out the whole problem, i give it a value of 1 for simplicity sake.
the momentum of the turntable before would be, L = (7.5 x 1^2) x 1.4
L= 10.5
since momentum as to be conserved in this problem the end momentum should also be 10.5
10.5 = [(7.5 x 1^2) + (7.1 x 1^2)] x ω
10.5 = 14.6 x ω
10.5/14.6 = ω
.719 = ω

and since this is a online assignment i can try and answer as many times until i get it correct, but i can't think of where i might be going wrong cause it says i have the wrong answer.
I don't need the problem worked out for me per say...just maybe hint at the error or explain where I am going wrong, and thank you for any help

Welcome to PF!

Hi MRMooneyham! Welcome to PF!

(try using the X2 button just above the Reply box )
MRMooneyham said:
the momentum of the turntable before would be, L = (7.5 x 1^2) x 1.4

nooo … wrong moment of inertia !

I figured that's where I was wrong at, but I don't understand how, so another hint pot favor :)?

you've used MR2 for the moment of inertia of a disc about its axis

it isn't

look it up, you need to learn the common moments of inertia

so then it would be? I am still using the ^ symbol cause i don't understand the x^2 above the reply
L = (1/4 * 7.5 * 1^2) * 1.4
L = 2.625
so after the collosion the equation would look like
2.625 = [(1/4 * 7.5 * 1^2) + (7.1 * 1^2)] * ω
so 2.625 = 8.975 * ω
and ω = .292479?

ok thanks, i got confused on that :/

1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object. It is defined as the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum conserved in a turntable and blob collision?

In a turntable and blob collision, angular momentum is conserved because the initial angular momentum of the system is equal to the final angular momentum of the system. This means that the total amount of rotational motion remains constant throughout the collision.

3. What factors affect the angular momentum in a turntable and blob collision?

The main factors that affect the angular momentum in a turntable and blob collision are the mass and velocity of the objects involved, as well as the distance of the objects from the axis of rotation.

4. How does the moment of inertia affect the angular momentum in a turntable and blob collision?

The moment of inertia, which is a measure of an object's resistance to rotational motion, directly affects the angular momentum in a turntable and blob collision. Objects with a larger moment of inertia will have a greater angular momentum, while objects with a smaller moment of inertia will have a smaller angular momentum.

5. Can angular momentum be changed in a turntable and blob collision?

In a perfectly elastic collision, the angular momentum cannot be changed. However, in a non-perfectly elastic collision, some of the kinetic energy may be converted into other forms of energy, which can affect the angular momentum of the system.

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