# Simple problem involving pulley with mass

1. Nov 14, 2013

### Saitama

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

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2. Nov 14, 2013

### Staff: Mentor

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

3. Nov 14, 2013

### Tanya Sharma

Hi Pranav...

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

4. Nov 14, 2013

### Saitama

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 $5.96 \,\, m/s^2$.

5. Nov 14, 2013

### Tanya Sharma

10.31 ms-2 is the correct answer .

6. Nov 14, 2013

### Saitama

Thanks for the check! :)

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

7. Nov 14, 2013