MIT OCW 8.01 PS10.6: A Massive Pulley and a Block on an Incline

[giodude]

Homework Statement

(Screen shot of question is posted below)

Consider a pulley of mass $m_{p}$ and radius $R$ that has a moment of inertia $\frac{1}{2}m_{p}R^{2}$. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass $m$ that is initially held at rest on a frictionless inclined plane that is inclined at an angle $\beta$ with respect to the horizontal. The downward acceleration of gravity is $g$. The block is released from rest.

How long does it take the block to move a distance $d$ down the inclined plane? Write your answer using some or all of the following: $R$, $m$, $g$, $d$, $m_{p}$, $\beta$.

Relevant Equations

$$\tau_{total} = I_{s} \alpha$$
$$I_{s} = \frac{1}{2}m_{p}R^{2}$$
$$a_{1} = \alpha_{1}R$$

Set up the force equations:
(1) $mgsin(\beta) - T = ma_{1}$
(2) $TR = I_{s}\alpha_{1}$

Multiply (1) by $R$ and isolate $TR$:
$R(mgsin(\beta) - T) = R(ma_{1})$
$mRgsin(\beta) - TR = mRa_{1}$
$TR = mRgsin(\beta) - mRa_{1}$

Plug $TR$ into (2):
$TR = I_{s}\alpha_{1}$
$mRgsin(\beta) - mRa_{1} = I_{s}\alpha_{1}$
$mRgsin(\beta) - mRa_{1} = \frac{1}{2}m_{p}R^{2}\alpha_{1}$

Solve for the $\alpha_{1}$:
$\alpha_{1} = \frac{mRgsin(\beta) - mRa_{1}}{\frac{1}{2}m_{p}R^{2}}$
(3) $\alpha_{1} = 2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}$

We now use (3) to solve for linear acceleration, $a_{1}$:
$2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}$
$m_{p}a_{1} + 2ma_{1} = 2mgsin(\beta)$
(4) $a_{1} = \frac{2mgsin(\beta)}{m_{p} + 2m}$

Use (4) to solve the linear kinematics equation for t:
$d = \frac{1}{2} a_{1} t^{2}$
$t = \sqrt{\frac{2d}{a_{1}}}$
$$t = \sqrt{\frac{(m_{p} + 2m)d}{mgsin(\beta)}}$$

I wonder if this solution is correct given that the time to move distance $d$ is independent of the radius $R$ of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration. However, I'm not confident in this intuition so I'd love feedback on (a) if my solution is correct and (b) if my intuition explaining the solution is correct. Thank you in advance!

 

[erobz]

Homework Statement: (Screen shot of question is posted below)

Consider a pulley of mass $m_{p}$ and radius $R$ that has a moment of inertia $\frac{1}{2}m_{p}R^{2}$. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass $m$ that is initially held at rest on a frictionless inclined plane that is inclined at an angle $\beta$ with respect to the horizontal. The downward acceleration of gravity is $g$. The block is released from rest.

How long does it take the block to move a distance $d$ down the inclined plane? Write your answer using some or all of the following: $R$, $m$, $g$, $d$, $m_{p}$, $\beta$.
Relevant Equations: $$\tau_{total} = I_{s} \alpha$$
$$I_{s} = \frac{1}{2}m_{p}R^{2}$$
$$a_{1} = \alpha_{1}R$$

Set up the force equations:
(1) $mgsin(\beta) - T = ma_{1}$
(2) $TR = I_{s}\alpha_{1}$

Multiply (1) by $R$ and isolate $TR$:
$R(mgsin(\beta) - T) = R(ma_{1})$
$mRgsin(\beta) - TR = mRa_{1}$
$TR = mRgsin(\beta) - mRa_{1}$

Plug $TR$ into (2):
$TR = I_{s}\alpha_{1}$
$mRgsin(\beta) - mRa_{1} = I_{s}\alpha_{1}$
$mRgsin(\beta) - mRa_{1} = \frac{1}{2}m_{p}R^{2}\alpha_{1}$

Solve for the $\alpha_{1}$:
$\alpha_{1} = \frac{mRgsin(\beta) - mRa_{1}}{\frac{1}{2}m_{p}R^{2}}$
(3) $\alpha_{1} = 2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}$

We now use (3) to solve for linear acceleration, $a_{1}$:
$2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}$
$m_{p}a_{1} + 2ma_{1} = 2mgsin(\beta)$
(4) $a_{1} = \frac{2mgsin(\beta)}{m_{p} + 2m}$

Use (4) to solve the linear kinematics equation for t:
$d = \frac{1}{2} a_{1} t^{2}$
$t = \sqrt{\frac{2d}{a_{1}}}$
$$t = \sqrt{\frac{(m_{p} + 2m)d}{mgsin(\beta)}}$$

I wonder if this solution is correct given that the time to move distance $d$ is independent of the radius $R$ of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration. However, I'm not confident in this intuition so I'd love feedback on (a) if my solution is correct and (b) if my intuition explaining the solution is correct. Thank you in advance!

Your solution is correct. A bit inefficient solving for $\alpha$. Just eliminate it as soon as possible by subbing $\alpha = \frac{a}{R}$ into (1) after subbing (2), And go right to $a$.

 

[giodude]

Thank you!

 

[haruspex]

I wonder if this solution is correct given that the time to move distance d is independent of the radius R of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration.

There's a bit more to it. The MoI rises as the square of R, while both the torque and the distance the mass moves per unit of rotation rise in proportion to R.
It may also seem intuitively wrong because you would expect the pulley's mass to increase too, but it is given as fixed.

 

[giodude]

Oh, I think I see. Since the mass of the pulley is fixed, as the size of the pulley increase its really just a purely proportional change because the actual mass isn't increasing (or decreasing) whereas the time dependency is "focused" on the mass of the pulley rather than the shape of it. Which we see by $m_{p}$ being in the solution for time while $R$ is not.

 

[zenterix]