How can I find the time which a block will take when descending from a support and tied to another block?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

A block which has a mass of $4\,kg$ is resting over a book stand as seen in the diagram from below. This block is tied to another which has a mass of $2\,kg$ by a thin wire which goes through a rugged pulley. This pulley is formed by a disk which has a radius of $50\,cm$ and has a moment of inertia equal to $1\,kg\cdot m^2$. Given these conditions find the time in seconds which will take the block which has $2\,kg$ to descend $4\,m$.



The alternatives given are as follows:

$\begin{array}{ll}
1.&\textrm{1 second}\\
2.&\textrm{2 seconds}\\
3.&\textrm{3 seconds}\\
4.&\textrm{4 seconds}\\
5.&\textrm{5 seconds}\\
\end{array}$

How exactly can I tackle this problem?. I'm not very sure how to use the formula for the moment of inertia to find the time. Could it be that this is related with the angular acceleration for the pulley?. The only formula which I do recall to estimate the time when a body is hanging is given by:

$v_f^2=v_o^2-2g\Delta y$

But I don't know exactly if this can be related to this problem. Can someone help me with the right approach for this problem?.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Negligible friction on the book stand?

A pulley with a diameter of 1 meter ... WOW

$\dfrac{1}{2}m_p r^2 = 1 \implies m_p = 8 \text{ kg}$

$r(T_1 - T_2) = \dfrac{1}{2}m_p r^2 \cdot \dfrac{a}{r} \implies T_1 - T_2 = \dfrac{1}{2}m_p a$


$T_1 - T_2 = \dfrac{1}{2}m_p a$
$mg - T_1 = ma$
$T_2 = Ma$
----------------------------------
$mg = a\left(\dfrac{1}{2}m_p + m + M \right)$

$\dfrac{mg}{\dfrac{1}{2}m_p + m + M} = a$

you're trying to determine time ...

$h = \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2h}{a}}$
 
Jun 2017
399
6
Lima, Peru
Negligible friction on the book stand?

A pulley with a diameter of 1 meter ... WOW

$\dfrac{1}{2}m_p r^2 = 1 \implies m_p = 8 \text{ kg}$

$r(T_1 - T_2) = \dfrac{1}{2}m_p r^2 \cdot \dfrac{a}{r} \implies T_1 - T_2 = \dfrac{1}{2}m_p a$


$T_1 - T_2 = \dfrac{1}{2}m_p a$
$mg - T_1 = ma$
$T_2 = Ma$
----------------------------------
$mg = a\left(\dfrac{1}{2}m_p + m + M \right)$

$\dfrac{mg}{\dfrac{1}{2}m_p + m + M} = a$

you're trying to determine time ...

$h = \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2h}{a}}$
Continuing the operations which had not been done:

If I'm getting the picture correctly.

Inserting the numeric values and assuming the acceleration due gravity is $10\frac{m}{s}$:

$a=\frac{2\cdot 10}{\frac{1}{2}\cdot 8+2+4}=2\frac{m}{s^2}$

To get the time it would be:

$t=\sqrt{\frac{2\times 4}{2}}=2\,s$

This formula is obtained from?

$y=y_{0}+v_{oy}t-\frac{1}{2}gt^2$

or is it from:

$x=x_{o}+x_{o}t+\frac{1}{2}at^2$

Both indicate the same concept but I'm confused about the sign next to the acceleration. In this context the acceleration acting on the block is caused by the a calculated from applying the torque and rotational inertia plus the Newton's second law.

Wouldn't it be:

$-y=0-\frac{1}{2}at^2$

Am I getting the right picture here?. :)
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
$\Delta y = v_{yo} \cdot t + \dfrac{1}{2}at^2$

if one chooses the downward direction as negative ...

$-h = 0 + \dfrac{1}{2}(-a)t^2$

multiply both sides by (-1) ...

$h = \dfrac{1}{2}at^2$

So ... one could choose the downward direction as positive and end up with the same calculation.
 
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Jun 2017
399
6
Lima, Peru
$\Delta y = v_{yo} \cdot t + \dfrac{1}{2}at^2$

if one chooses the downward direction as negative ...

$-h = 0 + \dfrac{1}{2}(-a)t^2$

multiply both sides by (-1) ...

$h = \dfrac{1}{2}at^2$

So ... one could choose the downward direction as positive and end up with the same calculation.
Despite the diameter of the pulley seemed unreal. At least I'm glad that I got right the interpretation for this equation. Btw the source of confusion is my main reference textbook "Essential college physics" by Rex Wolson does have the habit of assigning the downward acceleration with a negative sign in front of it and I came used to this standard practice.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
Despite the diameter of the pulley seemed unreal. At least I'm glad that I got right the interpretation for this equation. Btw the source of confusion is my main reference textbook "Essential college physics" by Rex Wolson does have the habit of assigning the downward acceleration with a negative sign in front of it and I came used to this standard practice.
If you're comfortable using the textbook method, and you can keep the signs in order, then do so ... I prefer dealing with magnitudes and assigning signs for direction after the fact.