Question

Two discs are rotating about an axis as shown in the figure. The moment of inertia of the right disc is 2 kg m2 and it rotates with angular speed of 2000 rpm. The moment of inertia of the left disc is 5 kg m2 and it rotates in the opposite direction with angular speed of 100 rpm. The discs are suddenly clamped together.
Determine the final common angular speed of the clamped discs.

Answer 1

Answer:

Relation between war momentum and moment of inertia

U+Iω

where I

1

=?

I

2

=1×10

−3

kg−m

2

ω

1

=2rad/s

ω

2

=5rad/s

ω=4rad/s

So, J=J

1

+J

2

(I

1

+1×10

−3

)×4=I

1

×2+5×1×10

−3

4I

1

+4×10

−3

=2I

1

+5×10

−3

2I

1

=1×10

−3

I

1

=

2

1

×10

−3

=0.5×10

−3

kg−m

2

Answer 2

Answer:

We use the formula:

ω=($I_{r}$ω$_{r}$-$I_{l}$ω$_{l}$)/($I_{r}+I_{l}$)

The answer is 500rpm.

Explanation:

According to angular momentum conservation, we have:

$L_{i}=L_{f}$

$I_{r}$ω$_{r}$-$I_{l}$ω$_{l} =(I_{r}+I_{l})$ω

where I is moment of inertia and ω is angular speed.

$I_{r}=2kgm^{2}$   ω$_{r}=2000rpm$

$I_{l}=5kgm^{2}$    ω$_{l} =100rpm$

ω=$\frac{4000-500}{7}=\frac{3500}{7}=500rpm$

The final common angular speed of clamped discs is 500rpm.