Simple problem involving pulley with mass

[Saitama]

Homework Statement

The descending pulley (disc shaped) shown in the figure have a radius 20 cm and moment of inertia 0.20 kg-m². The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.

Homework Equations

The Attempt at a Solution

The moment of inertia for a disc about its CoM is: $\frac{MR^2}{2}$. Equating this with 0.2 kg-m² and substituting R=0.2 m, M=10 kg.

Let T₂ and T₁ be the tensions in the strings. T₂ on the left of disc and T₁ on right of disc.

If A is the acceleration of disc, then 2A is the acceleration of disc. Also, let $\alpha$ be the angular acceleration of disc the direction of which is anti-clockwise.

Newton's second law on disc: $10g-(T_1+T_2)=10A$ and for block: $T_2=2A$ (mass of block is 1 kg).

Torque about CM of disc: $(T_1-T_2)R=I\alpha \Rightarrow (T_1-T_2)(0.2)=(0.2)\alpha \Rightarrow T_1-T_2=\alpha$.

Next is equating the acceleration along the string. The point of contact of string with the disc on the left goes down with an acceleration of $A+\alpha R$. Hence, $A+\alpha R=2A$.

I have enough equations but solving these equations gives me a wrong answer.

Any help is appreciated. Thanks!

 

[Doc Al]

Your analysis seems reasonable. What did you get for the acceleration of the disc? (And what was the "right" answer?)

 

[Tanya Sharma]

Hi Pranav...

It looks alright to me .What is the correct answer and the answer you are getting?

 

[Saitama]

Your analysis seems reasonable. What did you get for the acceleration of the disc? (And what was the "right" answer?)

Hi Pranav...

It looks alright to me .What is the correct answer and the answer you are getting?

Hi Doc Al and Tanya! :)

I had $A+\alpha R=2A \Rightarrow \alpha=5A$. Since $T_1-T_2=\alpha$ and $T_2=2A$, we have $T_1=7A$.

From Newton's second law on disc: $10g-(T_1+T_2)=10A \Rightarrow 10g-9A=10A \Rightarrow A=10g/19$.

Hence, acceleration of block is $2A=20g/19=10.31 \,\, m/s^2$.

The given answer is $5.96 \,\, m/s^2$.

 

[Tanya Sharma]

10.31 ms^-2 is the correct answer .

 

[Saitama]

10.31 ms^-2 is the correct answer .

Thanks for the check! :)

The answer key seems to give acceleration of the disc .

It doesn't look so. The acceleration of disc comes out to be 5.15 m/s^2.

 

[Doc Al]

Hence, acceleration of block is $2A=20g/19=10.31 \,\, m/s^2$.

I agree with that answer.

The given answer is $5.96 \,\, m/s^2$.

I don't know what that's supposed to be. Did they give a solution, or just an answer?

 

[Saitama]

I don't know what that's supposed to be. Did they give a solution, or just an answer?

Only the answer. :(