Moment of inertia of disc with plasticine dropped onto it

[Skuzzy]

Homework Statement

The question involves a rotating disc, spinning about a vertical axis through its centre. We are given its angular speed, 20rpm and diameter 24cm.

A piece of plasticine is dropped onto the disc (no torques are applied by this process). We are given its location relative to the centre of the disc, 8cm and its weight 8g.

We are told that the angular speed is reduced and given its new value, 16rpm

We are asked for the moment of inertia of the disc.

Homework Equations

$$\large L=I\omega$$

$$\large I=\frac{1}{2}mR^{2}$$

$$\large I=mR^{2}$$

The Attempt at a Solution

We cannot immediatley calculate the moment of inertia as the mass of the disc is not given. I 'think' that the problem is solved by conservation of angular momentum. As no torques are applied to the system angular momentum is conserved.

L₁= I₁$$\omega$$
L₁= I₁$$(20rpm)$$

L₂= I₂$$(16rpm)$$

I₁=1/2(M)(10cm)²

I am lost... How do i calculate I₂? I can't seem to see past the unknown mass of the disc... I can see that for the plasticine we can calculate it's moment of interia if we treat it as a particle but I'm still confused.

Can someone please point me in the right direction?

 

[RoyalCat]

You are also wrong to assume that the disk has a uniform mass distribution (Which is what you did by assuming $$I_{disk}=\tfrac{1}{2}MR^2$$ )

Consider the moment of inertia of the disk, $$I_d$$ and the moment of inertia of the disk with the plasticine attached to it (Remember that the moment of inertia is an additive quantity).

As no torque acted on the system, you are right to assume that angular momentum was conserved ($$\tau = \frac{dL}{dt}=0$$)

With these two facts you can now solve for the moment of inertia of the disk.

 

[Skuzzy]

OK so if:

$$L_{1}=I_{d}\omega_{1}$$

and

$$L_{2}=I_{d+p}\omega_{2}= I_{d}\omega_{2}+I_{p}\omega_{2}$$

where $$\omega_{1} = 20rpm$$ and $$\omega_{2} = 16rpm$$

Then because angular momentum is conserved,

$$L_{1}=L_{2}$$ and $$I_{d}\omega_{2}+I_{p}\omega_{2}=I_{d}\omega_{1}$$

Is it correct to treat the plasticine as having $$I_{p}=MR^{2}$$ ?

 

[RoyalCat]

OK so if:

$$L_{1}=I_{d}\omega_{1}$$

and

$$L_{2}=I_{d+p}\omega_{2}= I_{d}\omega_{2}+I_{p}\omega_{2}$$

where $$\omega_{1} = 20rpm$$ and $$\omega_{2} = 16rpm$$

Then because angular momentum is conserved,

$$L_{1}=L_{2}$$ and $$I_{d}\omega_{2}+I_{p}\omega_{2}=I_{d}\omega_{1}$$

Is it correct to treat the plasticine as having $$I_{p}=MR^{2}$$ ?

Yep, since you weren't given any information about the piece of plasticine, I think it's safe to treat it as a point mass.

Now all you need to do is isolate for $$I_d$$, and that should prove to be simple enough. Well done!